The Standard Model reason of why the Moon is moving away from us is supposedly due to the Earth's rotation inducing the Moon with angular momentum. But if this were the case, surely you would expect the Moon to be rotating slightly, due to it's irregular orbit, if nothing else.

Cheers, AL :rolleyes:

It is spinning. It rotates once a month, more or less, as it moves around the Earth.

It does rotate slightly from our viewpoint.

It just goes back and forth.

I still don't think that the relative amount of rotation of the Earth compared to that of the Moon is enough to induce the amount of angular momentum needed to increase it's orbit by 3cm a year.

I propose that it is simply the size of the gravitational field being produced from the center of the Earth which is reducing due to the effect of CONSERVATION OF ENERGY.

A full explanation can be found in PSEUDOSCIENCE under 'Gravity Problem Solved'.

AL :soapbox:

It does rotate slightly from our viewpoint.

It just goes back and forth.Quite correct. That was why I put in the remark about 'more or less'. As a consequence, and I am sure you know this, over time we can see more than 50% of the moon's surface.

I still don't think that the relative amount of rotation of the Earth compared to that of the Moon is enough to induce the amount of angular momentum needed to increase it's orbit by 3cm a year.




You think wrongly.

The Earth slows its rotation due to interaction with the Moon by some 1.5 milliseconds per century. It is this angular momentum lost by the Earth that is given to to the Moon.

The total kinetic energy of a rotating sphere can be found by

E = \frac{\omega^2 M r^2}{5}

Where w is the angular velocity in radian/sec.
M is its mass
r is its radius

Plugging in the values for the Earth for both its present rate of rotation and the amount it would have slowed in one year, and taking the difference, we get a value of:

4.34E+18 joules.

This is the amount of energy the Earth has to transfer to the Moon.

The total energy of the orbiting Moon is found by

E= \frac{GMmMe}{2a}

Where G is the gravitational constant
Me and Mm are the masses of the Earth and the Moon
a is the average orbital radius for the Moon.

Plugging in the correct values for the Moon's present orbit and one 3.8 cm further out and taking the difference we get a value of:

3.77E+18 joules

Which is the amount of energy it would take to raise the Moon's orbit by 3.8 cm and is less than the amount of energy the Earth has to transfer in one year.

The difference is due to the fact that some of the energy given up by the Earth is lost due to tidal heating.

So yes, the Earth does give up enough rotational energy to the Moon in order to increase its orbit by 3.8 cm per year.

It is spinning. It rotates once a month, more or less, as it moves around the Earth.

I think that it fair to point out that it rotates once per sidereal month. During the synodic month (the "month" most commonly referred to.) the Moon actual rotates 1.08 times.

Janus abouth your last post:

the angular velocity of earth is: 0,0000727 rad/s
the earths mass is 5,97E27 gram
and has a radius of 6378100 meters

(0,0000727²*5,97E27*6378100)/5= 2,57E32 not 4,34 E18 joules
are you sure abouth that formula?

Janus abouth your last post:

the angular velocity of earth is: 0,0000727 rad/s
the earths mass is 5,97E27 gram
and has a radius of 6378100 meters

(0,0000727²*5,97E27*6378100)/5= 2,57E32 not 4,34 E18 joules
are you sure abouth that formula?

First, you need to use kg, not grams for the Earth's mass, which gives an answer of 2.57e29 joules for the KE of the Earth.

Second, the 4.34e18 joules is the KE the Earthloses by slowing its rotation by 15 microseconds in the course of one year. IOW, the KE difference between now and one year later when it will take 15 microseconds longer to complete a rotation.

I think that it fair to point out that it rotates once per sidereal month. During the synodic month (the "month" most commonly referred to.) the Moon actual rotates 1.08 times.I was trying to keep it simple. It also osscilates up and down as well as from side to side in a regular, but quite complex pattern. Additionally when you consider its path from the perspective of the sun things get even more complex. You are aware I am sure that the moon does not go round the Earth, but both rotate around their barycentre, which admittedly is inside the Earth, for the moment. I didn't see the need to introduce that for Common_Sense_Seeker/

Isaac Asimov once worked out that the Moon and Earth would part from each other until they rotated around each other with a period of about 50 days--i.e. about seven revolutions per year. Of course, at the same time--because the same process of the loss of angular momentum is occurring between the Sun and the planets, the Earth is withdrawing slowly fromthe Sun, and its orbit is getting longer as well.

The Standard Model reason of why the Moon is moving away from us is supposedly due to the Earth's rotation inducing the Moon with angular momentum. But if this were the case, surely you would expect the Moon to be rotating slightly, due to it's irregular orbit, if nothing else.

Cheers, AL :rolleyes:You mean like this http://en.wikipedia.org/wiki/Image:Lunar_libration_with_phase_Oct_2007.gif due to the motion of the Moon during it's orbit?

The reason the same side of the Moon always faces the Earth is due to tidal breaking, something completely explained by Newtonian gravity.

Thanks for the link, I've never seen the lunar libration in video. It looked like just the kind of motion you would expect from my theory, being due to it's irregular orbit of an uber-condensed Earth's inner core. The explanation I give of a reduced gravitational field is an elegant and simple one. What is the transfer mechanism of your angular momentum? If it is by a particle, which I propose will be proven from the 10th Sept, then it would have to have a large mass, would it not?

Do gravitons have mass?

AL :shrug:

It looked like just the kind of motion you would expect from my theoryYou don't have a working theory though. You have no quantitative predictions, no equations, no derivations, nothing but a load of BS and ignorance.

What is the transfer mechanism of your angular momentum? Newton's ego, voodoo and the Christian faith.

I thought you knew about mainstream models already? :rolleyes:

I'm not claiming to be an expert. Perhaps you can summarise the models for me?

AL

The natural decline of the Earth's gravitational field strength is a simple and elegant solution. It shouldn't be dismissed out of hand quite so easily.

AL

If the moon rotates, we should see well over 90% of its surface.

If the moon rotates, we should see well over 90% of its surface.

Try this:

Get yourself 2 coins, a large one and a small one. The large one is the Earth, the small one is the Moon. Rest them flat on a table top.

Now, with the Earth held still, move the Moon around the Earth in a circle to simulate the Moon's orbit around the Earth, with the "Heads" side of the coin facing upwards so you can see it.

When the Moon is above the Earth, would a person looking from the Earth's surface towards the Moon see the top or bottom of the "head" on the Moon? What happens when the Moon has moved to the other side of the Earth?

Now, in fact you know that we always see the same side of the real Moon. Using the coins, how can you make sure that the "head" on the Moon coin always faces the same way, as seen from the Earth?

And now the big question: do you think the Moon rotates now, or not?

And now the big question: do you think the Moon rotates now, or not?

I understand what you are saying about relative motion, but my answer would be no.

If you are sitting right on the nose of the man in the moon and looking at earth, you would see the earth rotating, but it would never be out of your sight.
I would conclude the mmon does not rotate.

Likewise, imagine you have a ball tied to a rope at the end of a string.
The ball may appear to be rotating if the observer is orbiting you at the same speed and believes that you are orbiting around the ball - but the fact remains that the ball is orbiting around you.

Am I in the Twilight Zone???

If you are sitting right on the nose of the man in the moon and looking at earth, you would see the earth rotating, but it would never be out of your sight.How would you explain the apparent progression of the sun, as you sat on the moon?
The fact that it would appear to follow the same path across the sky, behind the earth and back, once for every 29 or so apparent rotations of the earth?

I understand what you are saying about relative motion, but my answer would be no.

If you are sitting right on the nose of the man in the moon and looking at earth, you would see the earth rotating, but it would never be out of your sight.

Yes. But the only way it can do that is if the moon rotates relative to the distance stars.

I agree with one_raven. That's a great statement you make. The Moon is not spinning about it's own axis in a single direction only. I propose that this would be the case if it was gaining angular momentum from the Earth, which is spinning about it's own axis in a single direction.

Therefore the theory that the Earth is simply losing gravitational field strength seems the likeliest explanation in my mind, and so the Moon's orbit would increase slightly.

Is the velocity of the Moon increasing? Can that be determined by the laser reflector measurements?

AL

If the moon was not rotating about its own axis, it would look to us as if it were rotating.

Yeah, well my head's startin' to spin.

Do you think that the Moon's speed has increased with it's increase in orbital distance?

Common Sense Speaker. I hope you are not being disingenuous and asking naive questions to which you know the answer. People here will gobble you for breakfast.

Here's an explanation of why the moon is rotating by Rosanna Hamilton:

The Moon is 384,403 kilometers (238,857 miles) distant from the Earth. Its diameter is 3,476 kilometers (2,160 miles). Both the rotation of the Moon and its revolution around Earth takes 27 days, 7 hours, and 43 minutes. This synchronous rotation is caused by an unsymmetrical distribution of mass in the Moon, which has allowed Earth's gravity to keep one lunar hemisphere permanently turned toward Earth. Optical librations have been observed telescopically since the mid-17th century. Very small but real librations (maximum about 0°.04) are caused by the effect of the Sun's gravity and the eccentricity of Earth's orbit, perturbing the Moon's orbit and allowing cyclical preponderances of torque in both east-west and north-south directions.


I think I can explain this more simply:
The moon both rotates and always keeps the same face pointing towards the earth. To understand this, draw a large circle (earth) on a piece of paper, then take a smaller circular object (moon) with a mark to one point of the edge. Move the "moon" around the "earth" so that the mark always faces the "earth". You will see that in one full rotation around the earth, the mark has itself made one rotation. If you can imagine the moon without the earth, on its own and centred in a single space, it is spinning.

btw
Has any astrologer worked out a chart for a child born on the moon?
What are the key astrological effects of the earth?

I am being serious, alright. My science is fundamentally rusty, I agree, since it's been over 12 years since I was involved with science research.

I'm never going to convince Standard Model devotees over night, I know that. The simple question of why gravitational fields don't decay naturally still stands. It's a good solution.

I am being serious, alright. My science is fundamentally rusty, I agree, since it's been over 12 years since I was involved with science researchBull. You've never been involved in physics research.

No need to get personal. Incidentally, anybody with common sense can see that modern physics is floundering. :)

Bull. You've never been involved in physics research.
Perhaps he means psychic research.

common sense seekeer, the current explanation for the moons orbit, orbital variations and progressive movement away from the Earth are all wholly consistent with observations and standard gravitational theory. There are no paradoxes or enigmas to explain. Your explanation is an unnecessary addition, that offers nothin and likely generates contradictions for other situations.

The natural decline of the Earth's gravitational field strength is a simple and elegant solution.

AL

So was phlogiston. It was also completely erroneous.

I'm not claiming to be an expert. Perhaps you can summarise the models for me?

AL

Okay:

The Moon, due to the differential of its gravitional effect across the diameter of the Earth, raises two tidal bulges on the Earth. Ideally, these bulges would be in line with the line joining the center of the Moon adn the center of the Earth.

The Earth, however rotates, and there is a friction between the bulges and the main rotating Earth. As a result, the Earth tries to drag the bulges along wth its rotation. The Moon, on the other hand, tries to keep the Bulges aligned with itself. This results in a tug of war between the Earth and the Moon which ends in a "tie" where the tidal bulges don't rotate with the Earth, but don't stay aligned with the Moon either. Instead, they "lead" the Moon by a slight amount.

This leading the Moon by the bulges, alters the line of gravitational force between the bulges and the Moon. It no longer acts on a line going through the center of the Earth. Instead, it works on a slight angle which pulls forward on the Moon in its orbit and backwards on the bulges with respect to the Earth's rotation. These pulls act to increase the orbital energy of the Moon while decreasing the rotational energy of the Earth. greater oribital energy results in a higher orbit and lower rotational energy means a slower rotation.

One aspect of this mechanism is that if the satellite orbit retrograde or directly but with a shorter period than it take the planet to rotate, then the tidal bulges lag behind the satellite and pull backwards on the satellite in its orbit, robbing it of orbital energy and drawing it into a lower and lower orbit.

This is what is happening to Mars' moon Phobos, which orbits faster than Mars rotates.

It also explains why the majority of Jupiter's outer moons are retrograde. These bodies are captured asteroids, and since a retrograde satellite will be drawn in towards the planet and prograde ones ejected, It is easier to capture and hold bodies into a retrograde orbit.

Your explaination of decreasing gravitational strength, on the other hand, would have all satellites receding from their planet regardless of the type of orbit, and this does not match what we actually see in practice.

Your explaination of decreasing gravitational strength, on the other hand, would have all satellites receding from their planet regardless of the type of orbit, and this does not match what we actually see in practice.

Thanks for the reply in general, I'll look into it a bit more. As to the last statement, we only know that the Moon is moving away from the Earth due to precise laser reflector measurements. So how would we know whether other natural satellites are receding from their parent planet or not?

No need to get personal.There's nop need for you to lie. If you didn't lie about having done aeronautical research, that doesn't mean you've done physics research. My father has worked with Lockheed, Boeing, Airbus, the development team for the Eurofighter and even the supersonic car. He doesn't know the first thing about theoretical physics research. Doesn't even know the concepts in it!

Incidentally, anybody with common sense can see that modern physics is floundering. :)How so? You don't even know modern physics, you're making mistakes in this forum which are on material taught to people doing GCSEs and A Levels. I did circular orbital velocities when I was 16 or 17. You claim to have a degree in astronomy and you don't know about orbital velocities? I teach people doing degrees in astronomy and they know about such things!

So either you lied about your degree or you have forgotten so much of it you effectively don't have a degree in anything but in paper form only.

So how would we know whether other natural satellites are receding from their parent planet or not?We don't, not to the level of accuracy we know the Moon is. But given places like Io experience even more tidal interactions with their parent planet than our Moon does with the Earth and all models of Newtonian and Einsteinian gravity predict such behaviour and we have not found any reason to doubt such results yet, the models are not invalidated.

We have yet to even see you provide a single quantitative model for your work.

Does the moon spin on its axis?

Try reading the thread.

I'd like a simple civil yes or no answer.

Try really hard to imagine looking at the moon and not the earth, so just the moon.

If you watched it from say, a point several hundreds of thousands of Km 'above' the plane of the ecliptic, would it rotate around an axis, as well as orbiting a "non-existent" earth?
(Answer: yes, it would, just like holding a string with a ball at the end of it, and 'swinging' it around a centre means the ball rotates around it's axis as it orbits the centre).

Is it possible for an orbiting body to not rotate?

Does the moon spin on its axis?

Relative to the stars, yes, it does. Once a month.


Is it possible for an orbiting body to not rotate?

Theoretically, yes. In practice, it is quite unlikely. Some asteroids rotate very slowly, though.

Earth spins like a top. The moon doesn't. Correct?

No.

Earth spins like a top. The moon doesn't. Correct?

What do you mean?

I think it is a reference to precession.

Does the Earth rotate about its axis? Yes. As it goes about the Sun. Does the Moon rotate about its axis? Yes. As it goes around the Earth. It just happens that its 'day' (time to go around its axis) is the same length as it's 'year' (time to go around its parent).

If the Earth's day was the same length as a year, one half would always face the Sun and one half wouldn't. Doesn't mean the Earth isn't spinning on its axis.

Here's an explanation of why the moon is rotating by Rosanna Hamilton:

The Moon is 384,403 kilometers (238,857 miles) distant from the Earth. Its diameter is 3,476 kilometers (2,160 miles). Both the rotation of the Moon and its revolution around Earth takes 27 days, 7 hours, and 43 minutes. This synchronous rotation is caused by an unsymmetrical distribution of mass in the Moon, which has allowed Earth's gravity to keep one lunar hemisphere permanently turned toward Earth. Optical librations have been observed telescopically since the mid-17th century. Very small but real librations (maximum about 0°.04) are caused by the effect of the Sun's gravity and the eccentricity of Earth's orbit, perturbing the Moon's orbit and allowing cyclical preponderances of torque in both east-west and north-south directions.


It makes perfect sense to me, but it still seems that it would be relative rotation, while not "real" rotation.
Does the ball at the end of the string rotate?
Again, if you are hovering above the ball it would appear that the ball is rotating and I am orbiting around the ball, but that's just not the case.
It seems to me that since the moon is trapped in an orbit around the earth, the earth-bound observer would be the correct perspective.

Does the ball at the end of the string rotate?



The ball on the piece of string is also rotating.
If you were a tiny observer on the ball where the string is tied to the ball, the room would appear to be spinning around you.

We have yet to even see you provide a single quantitative model for your work.

I am shying away from the maths, but that's because I believe the only way to really understand whether a model is correct or not is by computer simulation. Something akin to the work being done at Durham University. Although their model seems to be an unweildy joke, which manages to produce pictures similar to those revealed by the Hubble Telescope.

I am shying away from the maths, but that's because I believe the only way to really understand whether a model is correct or not is by computer simulation. So what equations would you use in your simulation?

So what equations would you use in your simulation?

No simulations at all, I suspect, because that naturally requires equations.:rolleyes:

And speaking of suspicions, I'll go ahead and state the obvious: the poster is yet another kid trying to pose as an adult. Silly attempt, actually.

That's the first comment that's made me laugh.

AL :)

That's the first comment that's made me laugh.

AL :)

Can you show me to be incorrect? You certainly show over and over that you know practically nothing about what you claim to have a degree in, plus the fact that modeling requires equations to run and computer simulations require math - yet you say you shy away from math.

Who else but a kid could be so self-contradictory and not even realize it?

No knowledge, no math = young kid! You've exposed yourself with no help needed from others.

No simulations at all, I suspect, because that naturally requires equations.:rolleyes:Which is exactly why I asked the question and why I shall continue to ask it until one of three things happen:
1) He provides the equations.
2) He admits his level of ignorance.
3) He disappears.

Do not avoid the question CommonSenseSeeker, which equations would you use in your simulation?

The result of 25 years work isn't for free.

AL :)

The result of 25 years work isn't for free.

AL :)

Ah, well - yet another "would-be scientist" bites the dust.:rolleyes:

I wonder sometime why these forums attract so many impostors? While anonymity can allow anyone to claim to be anything they want to be, they simply aren't intelligent enough to realize that their very on words will always convict them of being a fraud.:shrug:

What a waste!

My simulation model is that of the beginning of creation followed by the big bang, incidentally.

Can we not get back onto the original thread idea?

AL :)

Not unless you post something of substance. I have seen two things from you:
a) A clear display that you do not understand basic physics.
b) Waffle about alternative physics.
The opportunity to change my perception of you is in your hands, not mine.

Not unless you post something of substance. I have seen two things from you:
a) A clear display that you do not understand basic physics.
b) Waffle about alternative physics.
The opportunity to change my perception of you is in your hands, not mine.

Ditto.

I'm not trying to impress you. My scientific mind is a good one, if a little unconventional. Most physicists at the top of their profession welcome alternative thinking. It's how most of history's great scientific discoveries were made.

AL :)

Fine. So give us something concrete on this alternative thinking you have instead of waffle.

I'm not trying to impress you. My scientific mind is a good one, if a little unconventional. Most physicists at the top of their profession welcome alternative thinking. It's how most of history's great scientific discoveries were made.

AL :)

Your blatantly obvious attempts to avoid answering direct questions brands you, without a doubt, as a mere child.

Either ask questions (people here will be happy to answer) so that you may learn something - or just go away.

Your blatantly obvious attempts to avoid answering direct questions brands you, without a doubt, as a mere child.



It's kind of sad in a way. It's like watching someone trying to bluff in poker while wearing mirrored sunglasses.

Then showing their hand before the bets are placed and saying, "are these three aces any good?".

It's kind of sad in a way. It's like watching someone trying to bluff in poker while wearing mirrored sunglasses.

Yeah - good one!;)

Then showing their hand before the bets are placed and saying, "are these three aces any good?".

HAR! Even better one!!!:D

Common_sense_seeker, give yourself and the rest of us a break. Poking fun at you is not much of a pleasure. We are doing it for two reasons, one honourable, the other dishonourable.
The dishonourable reason is that we are frustrated by your attempt to project yourself as something you are not and we are striking out. It is quite common for people on forums to pretent to be someon other than they are - and that's fine in many instances. But here, who you are, or rather what you claim to know is central to the discussion.
Rest assured we know that you know bugger all about orbital mechanics and simulation. You may have played Civilisation, but you didn't write the code. Stop pretending.

The honourable reason is that we are trying to get you to openly admit your lack of knowledge. There is no shame in that. We are all hugely ignorant of most things, but toegether we know quite a lot. We can learn from each other, but only if we are honest about the levels of our knowledge. Everyone who has been interacting with you here and on the other thread would genuinely welcome that.
At the moment we all think you are a bit of an immature prat. If you come clean I for one would be well impressed by your maturity and I am pretty sure the others would be too.

Your choice, prat or professional? Antagonist or ally? Well?

Does the Earth rotate about its axis? Yes. As it goes about the Sun. Does the Moon rotate about its axis? Yes. As it goes around the Earth. It just happens that its 'day' (time to go around its axis) is the same length as it's 'year' (time to go around its parent).

If the Earth's day was the same length as a year, one half would always face the Sun and one half wouldn't. Doesn't mean the Earth isn't spinning on its axis.

In a day, whether it be 24 hours, 1 year or 24 years, both sides face the sun. Otherwise, it is not a day.

In a day, whether it be 24 hours, 1 year or 24 years, both sides face the sun. Otherwise, it is not a day.
You are referring to the synodic day - the time from sunrise to sunrise. AlphaNumeric, on the other hand, was referring to the sidereal day - the time from star-rise to star-rise. The Earth's sidereal day is about 3 minutes and 55.9 seconds shorter than the 24 hour long synodic day (actually, 24 hours plus 2 milliseconds; the synodic day 24 hours long back in 1820).

Fine. So give us something concrete on this alternative thinking you have instead of waffle.


I imagine it sounds like a lot of waffle to a lot of people. My background is genuine. I haven't been in high-tech employment for over 12 years, and I probably sounded like a clueless idiot then too. I'm happy to continue with the thread 'Are Galaxies Expanding?' with people who are interested in toying with cosmological new ideas, which is my main area of interest.

I'm quite proud of being childish, that's just me.

AL :)

I imagine it sounds like a lot of waffle to a lot of people. My background is genuine. I don't give a damn about your background. Your ideas will stand or fall on their merit not upon your background. But so far you haven't given us any ideas, just a few lightweight juvenile speculations. Ante up.

Fine. So give us something concrete on this alternative thinking you have instead of waffle.

If the Moon is supposed to gain angular momentum via the rotating Earth's gravitational field, why can't this effect be replicated in the laboratory via experiment?

I propose that this is because the theory is complete BS. If you're a mathematician who disagrees, the onus is on you to prove it in the lab, is it not?

AL :)

If the Moon is supposed to gain angular momentum via the rotating Earth's gravitational field, why can't this effect be replicated in the laboratory via experiment?

I propose that this is because the theory is complete BS. If you're a mathematician who disagrees, the onus is on you to prove it in the lab, is it not?

AL :)

Same answer I gave you in the other thread - when someone comes along with a claim, it's up to THEM (in this case, YOU) to prove it!

It's not our job to try and disprove a negative. (Hopefully, your smart enough to realize that attempting to prove a negative is a stupid exercise anyway.):bugeye:

Same answer I gave you in the other thread - when someone comes along with a claim, it's up to THEM (in this case, YOU) to prove it!

It's not our job to try and disprove a negative. (Hopefully, your smart enough to realize that attempting to prove a negative is a stupid exercise anyway.):bugeye:

If the entire scientific consensus of humanity claims that the Moon is moving away due to induced momentum via gravity, surely it should replicate the effect in the laboratory to confirm this new 'law' of physics?

I'm not say that you personally should conduct the experiment. All I've done is highlight how preposterous the situation is, and given an alternative suggestion of why the Moon is receding from us.

In a day, whether it be 24 hours, 1 year or 24 years, both sides face the sun. Otherwise, it is not a day.

And what precisely do you think is happening at New Moon when the near side of the moon is in shadow.

Or during a solar eclipse?

If there is a side facing us, there must be a side facing away from us.
If the side facing us is in shadow, then the side facing away from us must be in sunlight.

Simple deductive logic.

Or do you think that during a new moon - when the moon lies between the sun and the earth, that both sides are in shadow?

If the entire scientific consensus of humanity claims that the Moon is moving away due to induced momentum via gravity, surely it should replicate the effect in the laboratory to confirm this new 'law' of physics?

I'm not say that you personally should conduct the experiment. All I've done is highlight how preposterous the situation is, and given an alternative suggestion of why the Moon is receding from us.

The only things preposterous here is your silly suggestion/idea.

Get over it.

The only things preposterous here is your silly suggestion/idea.

Get over it.


Do you think that the effect could be replicated on a smaller scale? Yes or No?

Do you think that the effect could be replicated on a smaller scale? Yes or No?

No.

Do you think that the effect could be replicated on a smaller scale? Yes or No?


No.

Because Gravity is too weak.

Because Gravity is too weak.


Newton's law of universal gravitation is just that, a universal law i.e. on the scale of the planets or the lab. Claiming "gravity is too weak" is the kind of response from a mathematician.

Newton's law of universal gravitation is just that, a universal law i.e. on the scale of the planets or the lab. Claiming "gravity is too weak" is the kind of response from a mathematician.

Pity i'm not a mathematician, or can you not read my member title?
Or perhaps you thought my Avatar was some kind of mthamatical equation? (Hint, it isn't).

Gravity is a weak force is a relevant statement, and conducting the sort of experiment you're talking about would require the gross domestic product of several small third world nations (and Belgium).

Or we can take the common sense approach, you know, that which you claim to be seeking, and we can make inferences based on observations, and there have been plenty of observations of tidal locking, and various other similar phenomena.

Try this:

Get yourself 2 coins, a large one and a small one. The large one is the Earth, the small one is the Moon. Rest them flat on a table top.

Now, with the Earth held still, move the Moon around the Earth in a circle to simulate the Moon's orbit around the Earth, with the "Heads" side of the coin facing upwards so you can see it.

When the Moon is above the Earth, would a person looking from the Earth's surface towards the Moon see the top or bottom of the "head" on the Moon?

The conventional wisdom is (amazingly) that the moon rotates on its polar axis one time per orbit. HOWEVER, that is simply not true, because due to tidal braking, the moon no longer rotates around its internal polar axis at all. The moon's polar rotation ground to a halt billions of years ago, so today the moon just *orbits* around the Earth-moon common-mass barycenter, which is a point located within the Earth.

The *apparent* rotation of the moon from the sidereal perspective (and your above example is from the sidereal perspective) ONLY shows a 360 degree turn as the moon orbits, which is not a true polar rotation. E.g., if you had a sidereal perspective of the Earth orbiting the sun, then you would count 366.25 rotations per year instead of 365.25, which is our Earth's 365.25 polar rotations plus one (1) 360 degree orbit.


What happens when the Moon has moved to the other side of the Earth?

The moon will then have *turned* halfway around as a result of its orbit, so as viewed from the stars (the sidereal perspective), the moon would point in the opposite direction. HOWEVER, the axis for that turn would be the Earth-moon barycenter located within the Earth and NOT a rotation around our moon's internal polar axis!

If the moon BOTH orbited and rotated around its polar axis (two axes at the same time), then you would see all sides of the moon. The Earth has two axes, its polar axial spin as well as its orbit around the Earth-sun-moon barycenter. The moon has only ONE (1) relevant spin axis as the moon orbits (revolves) around its barycenter.


Now, in fact you know that we always see the same side of the real Moon.

Yes, we do ...BECAUSE, the moon lost all of its polar axial spins billions of years ago!!


Using the coins, how can you make sure that the "head" on the Moon coin always faces the same way, as seen from the Earth?

Are you referring to a so-called "zero-rotation" moon as shown in this graphic -- NOTE, I don't have enough posts yet to post URLs, so you need to add the http part to this URL (don't add www):

community-2.webtv.net/kdine5/Lunacy/scrapbookFiles/mailedD1.gif

If that so-called "zero-rotation" moon is what you mean, then that moon actually has one clockwise polar rotation per each counter-clockwise orbit.

When viewed from a sidereal perspective, all counter-clockwise orbiting bodies that are either tidally locked, or rotating on their polar axis counter-clockwise, *appear* to have one (1) extra polar rotation, which in fact is its orbit and not a true polar rotation. Thus, the Earth has a 366.25:1 spin ratio, and the moon has a 1:1 spin ratio - BUT, the Earth still has only 365.25 polar rotations per year (366.25 - 1 = 365.25), and the moon has zero polar rotations per orbit (1 - 1 = 0).

Conversely, from the same sidereal perspective, all counter-clockwise orbiting bodies that are spinning down from the clockwise direction will *appear* to have one (-1) less polar rotation from the sidereal perspective than they actually have.

Think about it - at one time our moon did have polar axial rotation, but billions of years ago tidal braking quickly ground the moon's polar rotations to a halt ... and normally, when you lose all of something you end up with ZERO, not one!

What confuses this issue is the claim that a sidereal perspective shows some sort of absolute truth (i.e., God's eye!) But, there are no absolute reference frames, and it's easy to prove that the sidereal perspective has obvious problems.

E.g., from the sidereal perspective a counter-clockwise orbiting body spinning down from the clockwise direction will not only have a 0:1 spin rate when that theoretical moon actually still has one polar rotation left as it orbits around its larger planet, but also, a moon spinning down from that clockwise direction will have TWO (2) 1:1 spin rates!!!!

Don't believe it? It's true, and rather easy to prove!

Let's try your table top experiment again, but instead of orbiting a coin around the center-point, instead take an orange (or any other round object) and draw a happy-face on one side, and draw an X on the opposite side.

Place the orange in the 6 o'clock position with the X facing you, and then push the orange in the counter-clockwise position to 3 o'clock as you spin the orange around its polar axis one time – when you land at 3 o'clock the happy-face should again be facing the center-point and you should have counted the X one time.

Proceed in this same fashion to 12 o'clock, then 9 o'clock, then back to the starting point at 6 o'clock.

In that single orbit the orange's happy face will have faced inwards four (4) times, and you will have counted the X five (5) times! No matter how many polar rotations you make per each orbit, you will ALWAYS count the X one more time ... and, that extra X count is caused by the 360 degree orbit, NOT a polar rotation!

Try the same experiment with the same counter-clockwise orbit, but rotate the orange clockwise this time, and you'll then count the X one (-1) time LESS than than the orange's actual polar rotations - thus, the so-called zero-rotating moon is actually a counter-clockwise orbiting moon spinning down from the clockwise direction. Amazingly, counter-clockwise moon spinning down from the clockwise direction that still has 2 polar rotations left will have a 2nd 1:1 sidereal spin ratio, the same 1:1 it will have when it finally spins down.

So much for the absolute truth of the sidereal perspective!!!

Thus, despite appearing to not spin from the sidereal perspective (0:1), that zero-rotation moon is actually spinning clockwise around its polar axis one time per orbit (0 + 1 = 1):

Likewise, this tidally-locked moon with a 1:1 (sidereal) spin rate is often incorrectly cited as a one (1) rotation moon:

community-2.webtv.net/kdine5/Lunacy/scrapbookFiles/mailedD0.gif

But, the only spin you're seeing these 1-rotation models make is the moon's 360 degree orbit around an exterior axis (its barycenter.)

The moon ONLY *orbits* around its Earth-moon barycenter! The Earth both *revolves* around the sun and *rotates* around its polar axis. The Earth also wobbles around the Earth-moon barycenter.

I realize that most websites have this wrong - likely, this rotation nonsense is being taught in Astronomy 101 courses and most people don't think too deeply about it and just accept it.

The Earth was once known to be flat, too!
:shrug:

Ken





And now the big question: do you think the Moon rotates now, or not?

The Earth was once known to be flat, too!
:shrug:Can you provide a citation for that?

Can you provide a citation for that?Rand-McNally. (Has many 2D maps of it.)

Ophiolite: "Can you provide a citation for that?"

Funny!
:)

Yup, the Flat-earthers are still out there, but their numbers pale in comparison to the loony moon-rotaters still walking around loose:

en.wikipedia.org/wiki/Flat_Earth_Society

Ken

...What confuses this issue is the claim that a sidereal perspective shows some sort of absolute truth (i.e., God's eye!)
The moon ONLY *orbits* around its Earth-moon barycenter! The Earth both *revolves* around the sun and *rotates* around its polar axis. The Earth also wobbles around the Earth-moon barycenter. ...Hi Ken & WELCOME to sf.

Consider for example: A top with lots of kinetic energy so it can stand on its tip without falling over (I am avoiding the word "spinning.") and it is going around with me on a circus merry-go-round, is it not spinning? Even spinning on it Polar axis?

You understand some things well, but are a too dogmatic and provincial on others, IMHO.

There is really no good way to absolutely distinguish between your "turning", "spinning" & "rotating." Adding the word "polar" in front of "rotation" does not do it because as you say there are no absolute reference frames. In fact the only reference frame in which the moon appears not to spin is the Earth-fixed rotating frame. One fixed in the sun or Mars or any other planet or even any star does see all sides of the moon. Thus to hold your provincial POV you must ignore the POV of ALL other frames, which see the moon as spinning.

Also your provincial bias to this particular frame made you make statements such as:

"The Earth both *revolves* around the sun and *rotates* around its polar axis. The Earth also wobbles ..."

But do not state that the moon is doing the same thing. I.e. the only difference is that the Earth's "wobble" and spin rates are both larger.
A difference in degree is NOT a difference in kind.

Perhaps you, like many others, erroneously think the Earth is the main gravitational force acting on the moon and thus it orbits the Earth?

Well that is false. The sun is the dominate gravitational force on the moon. Both Earth and moon are in slightly elliptic orbits about the sun, which are always curving towards the sun. Both do "wobble" about 13 cycles as they make a complete circuit. The moon makes a 360 degree spin about 13 times and the Earth about 365 times during each trip around the sun. Neither is doing anything the other is not. The magnitudes of their wobbles and spin rates differ in value - that is all.

James was correct.* Your POV (quite a provincial one) is valid only from one special non-inertial frame whereas ALL other frames clearly see both the Earth and moon as spinning and wobbling as they orbit the sun.
----------------
*He usually is. Thus I take special delight when I catch his errors. I have been watching for more than three years and his rate is 4 errors in that period. (much lower than mine.) Take care when correcting him to make sure it is not you making the error, as you unfortunately did in your very first post.

Later by edit: Since Janus58 is still active here let me again thank him for teaching me the facts about the moon orbiting the sun, not the Earth, in an orbit that is ALWAYS curving towards the sun.

Ken, I like you to consider this image which shows a satellite orbiting a body. The satellite has a sidereal rotation equal to its orbit, but a has a 90° axial tilt. Would you consider this satellite as rotating on its axis or not? If not, why not?

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate1.gif


If yes, then consider this image, same situation but with 45° axial tilt. Does it rotate or not? If not, why, and at what axial tilt between 90° and 45° did it quit rotating?

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate2.gif

Finally, we have 0° of axial tilt. I already know that you say this situation involves no rotation of the satellite. And assuming you said yes to rotation for the last image, at what point between 45° and 0° did it stop rotating?

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate3.gif

Janus, why not go ahead and finish out the series? Show a true non-rotating moon - in which case all sides become visible during the course of one full orbit.

Janus, why not go ahead and finish out the series? Show a true non-rotating moon - in which case all sides become visible during the course of one full orbit.If we are going to make recommendations to Janus58, mine would be to show ONLY the satellite, stationary at the center of the frame, with the same rotation that it has in Janus58's first frame And ask if that satellite was spinning. Then add a mass for it to orbit and ask if that same motion of the planet was no longer a spin? (i.e. Janus58's first frame would become the second frame.)

Ken, I like you to consider this image which shows a satellite orbiting a body. The satellite has a sidereal rotation equal to its orbit, but a has a 90° axial tilt. Would you consider this satellite as rotating on its axis or not? If not, why not?

home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate1.gif

If yes, then consider this image, same situation but with 45° axial tilt. Does it rotate or not? If not, why, and at what axial tilt between 90° and 45° did it quit rotating?

home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate2.gif

Frankly, I don't see any difference between graphics #1 & #2 – even so, both moons appear to be tidally locked showing only the white face towards their primary. They both would have a 1:1 spin rate, which means they do not rotate at all around their polar axes and thus both only revolve around their common barycenter with their primaries.

Can you more clearly explain where your 90° and 45° axial tilts are??


Finally, we have 0° of axial tilt. I already know that you say this situation involves no rotation of the satellite. And assuming you said yes to rotation for the last image, at what point between 45° and 0° did it stop rotating?

home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate3.gif

While graphic #3 appears a tad different than the first two, I didn't see your 90° and 45° axial tilts, and I don't see any 0° axial tilt with this one either?

What I see are 3 tidally locked moons that aren't rotating around their polar axes, but they all do seem to have orbits with high declinations, which can contribute to libration – i.e., more than 50% of their respective surfaces may be viewable from the surface of their primaries.

E.g., due to libration, from Earth (over time) we can view about 59% of our moon's surface. Libration has various causes.

Our moon's apparent orbit can also exhibit extreme declination – e.g., google "moon's standstill" and you'll find that our moon has an 18 year cycle where it can get very high in our northern sky – as I recall, as much as 58°, and then two weeks later the moon will travel across our sky at just above the southern horizon, depending upon where the viewer is standing, of course.

Even though the moons in your graphics may show more of their surfaces (but less than all of their surfaces) to their primaries due to libration and still be considered tidally locked, a tidally locked body can even show ALL of its surface to its primary and still be considered tidally locked!

A classic example of this would be the planet Mercury, which has a 3:2 spin rate. Even though Mercury has spun down and lost all of its original polar rotation, because of Mercury's very eccentric orbit, each time Mercury orbits the sun it does a half-flip, so every two orbits mercury makes one complete rotation - thus, its 3:2 spin rate.

Even so, Mercury is still considered to be tidally locked since its spin is caused by its eccentric orbit.

As I recall, even though they are still considered to be tidally locked, harmonics may cause a few moons around the gas giants to also rotate 360°, which is caused by the gravitational effects of the other moons.

There are over 200 moons in our solar system and a lot to examine that is out of the ordinary - e.g., Venus is nearly tidally locked with the sun but still has a slight retrograde (clockwise) rotation.

Venus takes 243.01 days to spin once on its axis. With this slow rotation rate it actually takes 18.3 days longer for Venus to spin once on its axis than it takes for Venus to orbit the Sun, so from a sidereal perspective, Venus nearly has a 0:1 spin rate as it spins down to a tidally locked 1:1 spin rate.

Pluto is already tidally locked with its (relatively) large moon Charon, and both Pluto and Charon obit around an exterior barycenter, a point of their common-mass that lies in the space between them, which is very unusual for planets, but then again, Pluto has been downgraded from the planet classification.

Uranus' axis of rotation lies on its side with respect to the plane of the solar system, with an axial tilt of 97.77 degrees.

Ken

Ken, a very clearly written, accurate and informative post. I just missed the bit where you accepted that the moon rotates on its axis. :)

...I just missed the bit where you accepted that the moon rotates on its axis. :)And I missed ANY comment on post 86 about how the moon "turning" as it appears to orbit the Earth differs from the Earth turing as it goes around the sun Or even the top turing as it goes around the center of the circus merri-go-round, except these all have different rates of turning and orbit periods.

Ken you will soon learn (I hope, although some active here never have) that the best policy when you do error is to quickly convert to correct POV and thank those who have helped you do so. - That way the "beatings" stop.

I still ask (again but more specifically):

What does the Earth do as it orbits the sun that the moon does not also do as it orbits the sun?

The Earth has smaller "wobbel" and a more rapid spin, but that is not a difference in kind. They are doing the same type of thing.

Hi Ken & WELCOME to sf.

Hi, back at ya!
:)


Consider for example: A top with lots of kinetic energy so it can stand on its tip without falling over (I am avoiding the word "spinning.") and it is going around with me on a circus merry-go-round, is it not spinning? Even spinning on it Polar axis?

Why avoid the word "spinning" since it's a perfectly good word? Because there are potentially two axes involved (the barycenter & the moon's polar axis), the only problem when discussing this issue is in NOT identifying the axis of spin you are referring to!

E.g., you can say that the moon rotates, orbits spins or revolves around the common mass barycenter and I don't have a problem with any of that "spin" verbiage. For clarity, just give me the location of your spin axes! Location, location, location!

Back to your question:


Consider for example: A top with lots of kinetic energy so it can stand on its tip without falling over and it is going around with me on a circus merry-go-round, is it not spinning? Even spinning on it Polar axis?

That top will have two spin axes, the center of the merry-go-round and the top's own internal axis.

Instead of a top which is difficult to observe spinning by an observer not on the merry-go-round, instead sit your body on a swiveling bar-stool positioned on the outer rim of the merry-go-round. If you sit motionless on that counter-clockwise spinning merry-go-round, then you will have a 1:1 spin rate as viewed by the observer, but your body won't spin at all when viewed by other people on the merry-go-round.

NEXT, pick a tree off in the distance, and as the merry-go-round continues to spin, now spin your body on that rotating stool to keep facing that distant tree – from the observer's perspective you will have stopped rotating on your stool (a 0:1 sidereal spin rate), but the other people on the merry-go-round will see you spin clockwise once per each complete rotation of the merry-go-round.


You understand some things well, but are a too dogmatic and provincial on others, IMHO.

By "dogmatic" I assume you think my opinion wasn't humble enough in my certainty? Hey, I'm right (about this issue), and I've been called worse!
:)


There is really no good way to absolutely distinguish between your "turning", "spinning" & "rotating." Adding the word "polar" in front of "rotation" does not do it because as you say there are no absolute reference frames. In fact the only reference frame in which the moon appears not to spin is the Earth-fixed rotating frame. One fixed in the sun or Mars or any other planet or even any star does see all sides of the moon. Thus to hold your provincial POV you must ignore the POV of ALL other frames, which see the moon as spinning.

You're wrong since there are more than one reference frames by which it can be determined that the moon does NOT rotate on its polar axis! Two obvious reference frames would be when standing on the moon itself, or any observer standing on the Earth.

If you were inside a windowless spacecraft it would be impossible to determine if you were traveling at a steady 2 mph or 25,000 mph, but you could detect acceleration and deceleration. You could also detect a turn or spin from inside a windowless spaceship.

There are also the so-called fictitious forces, such as the Coriolis effect. E.g., a person with no other reference points could use a Foucault pendulum to determine if they were on a astronomical body that had polar axial rotation.

You can also observe deformation in the astronomical body for physical evidence of polar rotation since a polar rotating body will be oblate. WHEREAS, a tidally locked body's shape won't be oblate (like the Earth is) and often it will be in the shape of a prolate spheroid (add the httt w/o www to view this shape):

upload.wikimedia.org/wikipedia/commons/8/88/ProlateSpheroid.png

Our moon solidified into a prolate spheroid when the Earth-caused tidal bulges deformed our moon into that elongated shape.

As I stated above, I often add "polar" to help identify the particular spin axis I'm then discussing – nothing devious about that.


Also your provincial bias to this particular frame made you make statements such as:

"The Earth both *revolves* around the sun and *rotates* around its polar axis. The Earth also wobbles ..."

But do not state that the moon is doing the same thing. I.e. the only difference is that the Earth's "wobble" and spin rates are both larger.
A difference in degree is NOT a difference in kind.

You're 100% WRONG!!! The Earth-moon common-mass barycenter is located within the earth, so the earth simply CANNOT orbit, or rotate around its own internal barycenter, thus, the Earth can only wobble (i.e., deviate from a smoot circular orbit.) WHEREAS, our moon can orbit around its exterior barycenter.

The Earth rotates around its polar axis 365.25 times per complete orbit around the sun, and the moon spins zero times around its polar axis each 27.3 day orbit around the Earth. The ONLY thing there in common is, that both the Earth and the moon *appear* (from the sidereal perspective) to rotate one extra time per complete orbit, but that extra 360 degree *apparent* spin is the result of the 360 degree orbit and not caused by any extra spins around a polar axis.


Perhaps you, like many others, erroneously think the Earth is the main gravitational force acting on the moon and thus it orbits the Earth?

Well that is false. ...

I do believe that, and I'm NOT in error for believing that!!!!

Our moon has 1/80th of the mass of the Earth - YET, the effects of our puny moon's gravity upon the Earth is over twice that of the sun's!!!! I think you should study Earth tides, and in particular look at "spring tides" and "neap tides."

Because our moon is so much closer to us than our sun causing the moon's gravity to increase our ocean's tidal bulges much more strongly than the sun's gravity does, our much more massive Earth's gravity must by consequence be the main gravitational force acting upon the moon. Indeed, the Earth's gravity has distorted the moon into an prolate spheroid!


... The sun is the dominate gravitational force on the moon. Both Earth and moon are in slightly elliptic orbits about the sun, which are always curving towards the sun. Both do "wobble" about 13 cycles as they make a complete circuit. The moon makes a 360 degree spin about 13 times and the Earth about 365 times during each trip around the sun. Neither is doing anything the other is not. The magnitudes of their wobbles and spin rates differ in value - that is all.

You're wrong about most of that!


James was correct.* Your POV (quite a provincial one) is valid only from one special non-inertial frame whereas ALL other frames clearly see both the Earth and moon as spinning and wobbling as they orbit the sun.

I think James would be better off defending himself!
:)

BTW - James accurately had stated a conventional example that is often used to (incorrectly) prove that the moon rotates around its polar axis once per orbit, so if he was a good student, then that's where he got it. I don't deny that at first blush that example's logic seems valid, which is probably why most students don't challenge their teachers on it. And, the teachers learned the same thing as students and they are merely passing on the same bullshit.

I consider myself to be a person of science, and I've relied on expert scientific opinion myself many times in explaining evolution and other scientific principles to Neanderthals, but this is one time where I see supposedly reputable websites failing to grok the basic concepts affecting our own moon's motions!



*He usually is. Thus I take special delight when I catch his errors. I have been watching for more than three years and his rate is 4 errors in that period. (much lower than mine.) Take care when correcting him to make sure it is not you making the error, as you unfortunately did in your very first post.

You're wrong in your unsupported belief that I'm wrong!!!


Later by edit: Since Janus58 is still active here let me again thank him for teaching me the facts about the moon orbiting the sun, not the Earth, in an orbit that is ALWAYS curving towards the sun.

You say the moon only orbits the sun!?!?!? That's wrong – our moon both orbits the sun AND orbits the earth.

Ken

Ken, a very clearly written, accurate and informative post. I just missed the bit where you accepted that the moon rotates on its axis. :)

Thank you about the writing praise, but don't expect an admission from me any time soon that the moon rotates around its polar axis!
:)

A race-car also doesn't rotate around the axis of its mass when it completes a lap, but it does indeed TURN around its mass each lap. To actually "rotate" the car would need to spin out.

Seriously, I've discussed this rotation nonsense before, both online and in person, and I've always won. I win because I'm right!

A simple demonstration - draw a happy-face on an orange, then go into a darkened room and take the shade off of one table lamp (that will be the sun), then hold the orange on your outstretched palm with the happy-face facing you, then spin 360 around counter-clockwise on your heels.

If you do that, then you'll see all phases of the moon likewise pass across that orange.

The orange will NEVER spin in your hand, yet, its happy-face will always remain pointed towards your spinning body!

You can even see more than 50% of the orange's surface (libration) by raising the orange/moon model above and below your eye level. Viewing the orange/moon first with one eye and then the other eye will also cause some libration (caused by parallax.)

When you show a person that demonstration to prove that the moon doesn't spin on its polar axis (only around your body's axis), they will either grok it, or they'll dig their heels in and claim your arm isn't the same thing as gravity.

Of course an arm isn't gravity, but in an accurate model your arm can serve the same purpose.

Men have been making sun-Earth-moon models (called Orreries) for hundreds of years now, and no orrery ever constructed has any gearing to rotate the moon-model on its internal axis! BECAUSE, rotating the moon-model on its spindle just isn't necessary to do!!!

E.g., here's a good example of one (add the http & www to it):

.abhirjoshi.com/models/sme_model_abhir.jpg

Ken

And I missed ANY comment on post 86 about how the moon "turning" as it appears to orbit the Earth differs from the Earth turing as it goes around the sun Or even the top turing as it goes around the center of the circus merri-go-round, except these all have different rates of turning and orbit periods.

Ken you will soon learn (I hope, although some active here never have) that the best policy when you do error is to quickly convert to correct POV and thank those who have helped you do so. - That way the "beatings" stop.

I have NOT erred! You have! So, perhaps, you should follow your own advice?


I still ask (again but more specifically):

What does the Earth do as it orbits the sun that the moon does not also do as it orbits the sun?

As the Earth orbits the sun it rotates 365.25 times around its INTERNAL polar axis each orbit.

As the moon orbits the sun in that same time period, the moon instead orbits (aprox. 13 times) around the Earth-moon barycenter (which is a foci of their combined common-mass) that's locate within the Earth and EXTERIOR to the moon.

As the moon orbits around the Earth its gravity tugs the Earth slightly in the direction of the orbiting moon, and that tug causes the Earth's slight orbital wobble.

Pluto and its moon Charon both orbit around an exterior barycenter, thus allowing them to both orbit around that barycenter as well as wobble in their orbit around the sun.


The Earth has smaller "wobbel" and a more rapid spin, but that is not a difference in kind. They are doing the same type of thing.

NO – an astronomical body's orbit is NOT the same thing as a "wobble!"

E.g., the Earth and moon's combined mass orbits around another barycenter located within the sun, which causes the sun to likewise wobble in its larger orbit. Are you claiming that the sun "orbits" around this Earth-sun-moon barycenter too?

I say NO!!!

BTW - detecting the wobble of neighboring stars is one method now being used to indirectly detect other planets in orbit around these stars.

Billy, you need to do some research since you're obviously shooting from the hip!

Ken

...You're wrong since there are more than one reference frames by which it can be determined that the moon does NOT rotate on its polar axis! Two obvious reference frames would be when standing on the moon itself, ...No that is false. If standing at the North Pole of either Earth or moon you will see that the body you are standing on is rotating. In both cases you can constaantly look at the sun or any other star, but to do so you will need to constantly turn on the surface you are standing on to compensate for its rotation. The Earth rotates at about 15 degrees/ per hour. The moon at about (360degrees) / {(28days)x(24hours/day)} = 0.536 degrees per hour. Not a difference in kind only in rate of spin


... a person with no other reference points could use a Foucault pendulum to determine if they were on a astronomical body that had polar axial rotation. Excellent point and absolutely correct. That does tell when some body is rotating, but you are shooting yourself in your foot. A Foucault pendulum at either pole of the moon will appear to rotate over the surface of the moon at 0.536 degrees per hour but of course is its plane of oscillation is really fixed in space and the moon is rotating under it. - (For your hole in foot, I recommend stemming blood flow and quick visit to the physics facts hospital)


...Our moon has 1/80th of the mass of the Earth - YET, the effects of our puny moon's gravity upon the Earth is over twice that of the sun's!!!! I think you should study Earth tides, and in particular look at "spring tides" and "neap tides."No you are wrong here too. The tides are caused by the GRADIENT of gravity not by gravity itself. The gradient falls off as the CUBE of the distance, not the square. This is why the moon is much more important than the sun in producing tides on Earth.

Just do the calculations with mass of sun at 1 AU from moon and mass of Earth at one moon /Earth separation. When calculating gravity force on moon the denominator distance is squared and the sun is much more important.

When calculating the gradients the inverse CUBE makes the moon's gradient larger at the Earth than the sun's is. So the moon does dominate the tides. Do it and then you will understand why moon dominates the tides on Earth but the sun, not the Earth, controls the moon's orbit around the sun. Earth only causes an "hardly noticicable" percentage variation or "wobble" in the distance between the moon and the sun.

Another way for you to see that the moon orbit the sun, not the Earth, is make a scale drawing of the moon's 1AU orbit about the sun and the "wobble" will be almost less than the width of your pencil line! I.e. to the accuracy of your drawing, (if on 8.5 by 11 inch paper or less), the moon will have zero wobble as it goes around the sun. If this is too much for you, JUST TELL ME HOW FAR THE MOON IS FROM SUN when it is closest to the sun and ~2 weeks later when most distant from the sun. The difference is twice the wobble. Please give the wobbler as a percent of the average distance to the sun. - I dare you to do this simple calculation. If you do you will understand the moon orbits the sun with very small wobble due to the Earth.


...Because our moon is so much closer to us than our sun causing the moon's gravity to increase our ocean's tidal bulges much more strongly than the sun's gravity does, our much more massive Earth's gravity must by consequence be the main gravitational force acting upon the moon. ...You do not understand what causes the tides. I.e. that it is not gravity but the gradient of gravity.

Thus it would be interesting to hear you try to explain why there are TWO tides each day, (one on each side of the Earth), instead of just one bulge on the side of Earth towards the moon. The correct answer has is due to fact the moon's gravity is stronger on the side near to the moon than on the side most distant from the moon - I.e. depends upon the GRADIENT of the moon's gravity, not on the gravity itself.


...You're wrong about most of that! No look in the mirror to see someone making several errors. I will not comment on your "off-thread” points.

Let us hope this fascinating exchange can be continued without fatal injury to any of the parties involved. (Sure enough, I heard that in my head with the lilt of an Irish accent. You'll need to do the same to get where I'm coming from.:))

Frankly, I don't see any difference between graphics #1 & #2 – even so, both moons appear to be tidally locked showing only the white face towards their primary. They both would have a 1:1 spin rate, which means they do not rotate at all around their polar axes and thus both only revolve around their common barycenter with their primaries.

Can you more clearly explain where your 90° and 45° axial tilts are??



I have redone the images to show the polar axis of each satellite.

If you are still having trouble, here is the 90° tilt as seen from the planet surface.

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate4.gif

Does it rotate on its axis?

No matter how much or little educated or how stupid or intelligent, people say & do anything to try to justify their position. That's your story & you're sticking to it!
This is extremely simple. Truly. If you can't get this, how can you handle the extremely complicated???

Well i just would like to make mention that the earths moon is a unsual lunar satellite, it is far to large to be a considered a natural lunar satellite, although it apppears have its orgin from the earth in earths early formation.

A natural earth lunar satellite should be only about 550 miles in diameter.

The moon appears to be a apart of the earth from a time when the earth was much larger, a time when the earth was a gas giant, in which the moon seperated from the gas shell of the earth when the earth was about 20,000 miles in diameter or slightly less. Avery good example of such bodies that become abnormal lunar satellites is that of the Red Eye on Jupiter which is a body simular to the moon trapped in the gaseous body of Jupiter. eventually when the gas of jupiter has been evaporated this moon will be exsposed.
As can be seen from the Red Eye of Jupiter these bodies are tidally locked to the primary body.


DwayneD.L.Rabon

No matter how much or little educated or how stupid or intelligent, people say & do anything to try to justify their position. That's your story & you're sticking to it!
This is extremely simple. Truly. If you can't get this, how can you handle the extremely complicated???

You really need to learn to use the quote function so we can tell just who the heck you're talking to!!!

This post, as it is now, is worthless because there's NO way to tell who it's in response to. That's not too bright in anyone's book.:bugeye:

Well i just would like to make mention that the earths moon is a unsual lunar satellite, it is far to large to be a considered a natural lunar satellite, although it apppears have its orgin from the earth in earths early formation.

A natural earth lunar satellite should be only about 550 miles in diameter.

The moon appears to be a apart of the earth from a time when the earth was much larger, a time when the earth was a gas giant, in which the moon seperated from the gas shell of the earth when the earth was about 20,000 miles in diameter or slightly less. Avery good example of such bodies that become abnormal lunar satellites is that of the Red Eye on Jupiter which is a body simular to the moon trapped in the gaseous body of Jupiter. eventually when the gas of jupiter has been evaporated this moon will be exsposed.
As can be seen from the Red Eye of Jupiter these bodies are tidally locked to the primary body.


DwayneD.L.Rabon

Rabon, it's easy to see that you have no idea what you're babbling about. Every bit of your post is nothing but pure nonsense. Even a kid of 18 knows better.

Read only, your a proffessional dumb ass.


DwayneD.L.Rabon

Read only, your a proffessional dumb ass.


DwayneD.L.Rabon

Haha.

You really need to learn to control yourself.

You really need to learn to control yourself.

Who???????

Once again, by quoting nothing and addressing nothing, you are just talking to yourself alone. No one has any idea who else you might be talking to.:shrug:

Silly boy/girl/whatever.

Stranger,
Read-Only is right. Your last comment, for example. "You really need to learn to control yourself."
It follows a post from Steve100. Internet forum convention would strongly suggest your post is directed at Steve. However, it makes little sense to ask Steve to control himself. He seems perfectly in control.
It is more likely that you are talking to Dwayne. (He does need to control himself.) But how are we meant to know this? You could be talking to anyone, frankly.
Cheers
O.

I have redone the images to show the polar axis of each satellite.

If you are still having trouble, here is the 90° tilt as seen from the planet surface.

home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rotate4.gif

Does it rotate on its axis?

#1 - that graphic has nothing in common with your earlier 3 graphics which all (more or less) had kept their same side pointing towards the center.

#2 - your 4th graphic, if viewed from a planet at the center of its orbit, would be tumbling around two axes. It thus would NOT be tidally locked ...UNLESS, there were other moons close by that were causing it to tumble and keeping it from becoming tidally locked.

How do your graphics relate to the issue of whether (or not) our own moon is still *rotating* around its polar axis?

If our moon is still rotating around its internal polar axis one (1) time per complete orbit as some here claim, then please do name the force that has (for at least 3 billion years) kept our moon rotating around its polar axis against the torque of tidally locking?

Other than the Earth and moon's mutual pull of gravity on each other (and to a lessor extent of the sun & Jupiter, etc), the only remaining force I see that's affecting the moon's motion would be the moon's own orbital motion around the Earth-moon barycenter, and the force of that orbital motion is not constant since the moon's orbital motion is speeding up as energy is transferred from the Earth through tidal braking.

Pretty graphics, though.
:)

Ken

Ken your post 93has many errors, which I CORRECTED IN POST 96, but you have ignored that so I condense your post 93 to display only your errors directly related to the thread:
… reference frames by which it can be determined that the moon does NOT rotate on its polar axis! {one} would be when standing on the moon itself…a person with no other reference points could use a Foucault pendulum to determine if they were on a astronomical body that had polar axial rotation. … KenIf I sit on North pole of the moon for 28 days without moving initially facing the sun, two weeks later the sun is at my back and 28 days later the sun is again at my face again.
Did not I and the moon make one 360 degree turn about the polar axis?

I also watched the Foucault pendulum near me make one 360 degree turn during those 28 days. That you yourself CORRECTLY said was PROOF that the moon did turn on its polar axis. Are you contradicting yourself? Or just confused?

You also think the Earth’s gravity force on the moon is stronger than the sun’s gravity force on the moon is. It is so easy to show you are wrong here too, that I will:

If M & m be the masses for Sun and Earth and D & d be the distances of Sun and Earth from the moon. Thus, the ratio of Sun’s to Earth’s gravity on moon is:
{M/(D^2)} / {m/(d^2)} or (M/m) x (d/D)^2.
Now the mass ratio (M/m) = 1.3 x 10^6 and D = 150 x10^6 km. The moon to Earth separation, d, is 0.384 x10^6 km and I assume the moon is as far as it can be from the sun so D = 150.384 x 10^6.
Then (d/D)^2 = (0.00255)^2 or (2.55)^2 x10^-6 or 6.5 x 10^-6

Thus the Sun’s gravity is 1.30 x 6.50 = 8.45 times stronger force on the moon than the Earth’s gravity is. This is why the moon orbits the sun, not the Earth. All the Earth does is make the moon’s basically elliptical orbit about the sun have some small “wobble.”

How small is that “wobble”? Well that too is easy to answer: It is as 0.384 is to 150 or a ¼ of one percent variation in the moon to sun distance. So do graph it but for easy assume a circular orbit. When you draw a circle with 13 cycles of only ¼% variation in the radius you will see that AT ALL POINTS in the moon’s orbit it is ALWAYS turning towards the sun. The moon does NOT orbit the Earth it only very slightly crosses from one side of the Earth’s orbit to the other as it orbits the sun.

Summary: YOU HAVE BEEN 100% WRONG on everything stated directly related to the thread and even about the cause of the tides. You did NOT even correctly note the fact that both the Earth and moon only APPEAR* to rotate about their barycenter, which only wobbles from its elliptical orbit very little (due mainly to the perturbations of Jupiter).
--------------------------
*Appear” as neither is rotating about the barycenter, but if you were traveling with it in orbit about the sun it does APPEAR to be what they are doing; however, from ANY point fixed in space, both Earth and moon are in slightly (1/4 of a percent at most perturbed) elliptical orbits about the sun. They are NOT rotating or orbiting about each other nor about the barycenter. It only APPEARS THAT WAY. Also in case you are also confused about the sun: The sun is not rotating about the Earth either, but it too APPEARS to be rotating about the Earth from an Earth based perspective. Most educated persons now know that appearance is false. - The sun does not orbit the Earth; it only Appears to orbit the Earth. Unfortunately, most people still think the moon DOES orbit the Earth because it appears to orbit the Earth. Things are not always what they appear to be. To know the truth you need to do the calculations / analysis (as done above) not just assume the common POV and appearance are correct.

Finnally: As the moon goes around the sun in an eliptical orbit with 1/4 percent perturbation by the Earth of that orbit it is spinning about its polar axis approximately 13 times in each complete orbit of the sun. From Mars or some more distant place the telescope-aidded eye could ONLY observe the polar spin of the moon, not the tiny, slow 1/4 % wobble.

Ken,
you seem like a smart guy. Please prove it now by conceding that you have had, like many others, the wrong end of the stick.
Sincerely
O.

Ken,
you seem like a smart guy. Please prove it now by conceding that you have had, like many others, the wrong end of the stick.
Sincerely
O.

Huh? In plain English please?

Ken

Huh? In plain English please? KenI will help you by drawing a verbal picture. (I cannot post any picture for reasons not known):

Imagine an eight inch diameter perfect circle drawn on an 8.5x11 inch piece of paper with a pencil that makes a mark 0.01 inches wide. (That is about the width of: | on my computer screen.)

Now you have a "circularized" scale model of the moon's orbit going around the sun. I.e. the 1/4 of one percent "wobbles" do not go outside of the 0.01 inch wide line, but are contained entirely inside its 0.01 inch width. This is why it is correct to state the moon is in orbit about the sun, not the Earth.

It is also why it is correct to state that it spins on its polar axis as it does so. The only motion noticeable to a telescope aided eye from most everywhere in the universe is this polar spin, not these tiny wobbles. To even see them with a telescope, your position if represented on the scale drawing would need to be close to the point on the circle of the scale drawing where the Earth is.

Also I worked out for you in my last post the sun's gravity on the moon is 8.45 times larger than that of the Earth. You seem to be a very stubborn fool to ignore all the facts that I and others have explained to you and persist in beliefs that contradict these facts which are only based on appearances when viewed from the Earth.

Based on the appearances you equally well can believe the sun orbits the Earth also as it too does gives the appearance of doing so. You have only the appearance to support your POV and none of the facts. Facts you have all wrong, including the believe that gravity, not the gravity gradient, causes the tides.

If the Earth just disappeared, would the moon continue to rotate about its axis, instantly start rotating about its axis, or remain not rotating about its axis?

If you think it's the second option, what causes the moon's rotational acceleration?

If you think it's the third option, you need to get on the right stick, never mind the right end.

Nicely put questions, Steve100.

I do tend to get too detailed, give numerical facts etc.

If he still does not admit his errors he really is a "stubborn fool" not only just appearing to be one.

Huh? In plain English please?

KenYou are wrong. Completely and utterly wrong. You are more wrong than a hirsute Latvian accordionist wearing embroidered silk bloomers in a remake of The Third Man. If they gave Oscar's for wrongness you would have a clean sweep. Generations from now young children will be shown these posts in order to understand how wrong, wrong can be. Your wrongness is of such magnitude that it may even stop the LHC from operating correctly. W-R-O-N-G. Wrong!

Did that help.

I tried to be polite by using a common colloquial expression. Suggesting you had the wrong end of the stick allows the possibility that you just have the wrong perspective on things. However, as Steve100 has now touched on, it looks like you may not even have the right stick.

What I don't understand, returning to my original comments, is how someone so apparently smart could be so persistently committed to remaining wrong.

Here's some unsolicited advice. When you are in a hole that you do not wish to be in the first thing to do is to stop digging. We are all rooting for you, but only you can make the transition.

Well, heres a little in addition..... The mass for the moon according to Nasa is 1/87th the mass of the earth, which cleary would define that the moon, exerts less force on the earth than the earths magentic feild.

Also if you are going to consider the effect of the sun on the moon then you should also consider the effect of the sun on the earth. Which would result in finding out that the earth should not spin either.

Under such anaolgy the moon would not rotate it would be locked as the earth should be locked in orbit with the sun.

The earth is located just out side of a region where things stop rotating, and begin to become more like the motion of venus and mercury.
and so the moon is just small enough that if it broke away from the earth it would begin to rotate.

And so also the moon is in a normal orbit around the earth.

DwayneD.L.Rabon

Post 93 in part.
...Ken you will soon learn (I hope, although some active here never have) that the best policy when you do error is to quickly convert to correct POV and thank those who have helped you do so. - That way the "beatings" stop. ...I guess Ken is just a very slow learner (or perhaps a masochist?).

Yes. A great shame. Otherwise he seems like a real nice guy and capable of reasonable analysis, up to a point. I hope he comes around. On the plus side we got to see all those neat animations.

If the Earth just disappeared, would the moon continue to rotate about its axis, instantly start rotating about its axis, or remain not rotating about its axis?

If you think it's the second option, what causes the moon's rotational acceleration?

If you think it's the third option, you need to get on the right stick, never mind the right end.

Since the moon is revolving around an exterior axis (the barycenter within the Earth), if the Earth were magically removed, then the moon would switch over to revolving around the barycenter of the moon & sun's combined common-mass, a point near the sun's center.

Because the moon had no actual polar rotation prior to being released by the Earth's gravity, at most the moon would wobble slightly until tidal-braking by the sun stabilized it.

The proof that the moon's circular revolution around the Earth is not the same thing as polar rotation is easy. Have you never seen an athlete's hammer toss?

Prior to releasing the hammer (a round metal ball on a chain), the hammer-thrower's ball is whizzing about him (or her) at such blazing speed the ball is just a blur at a high rate of circular motion, but upon release, the ball at most does a slow tumble and does NOT spin in the air at all.

Any minor tumbling of the ball upon release is caused by minor cartwheeling prior to the release, thus any minor tumbling motion caused by cartwheeling (around its own mass) the ball would retain after its release.

Here's a video clip - there are several athletes tossing in this clip and only the last athlete's toss can be seen clearly while in the air, the blond Russian gal's toss. At the end of the clip they replay her toss in slow motion making it easy to see the ball's lack of rotation while in the air (add the http & www):

youtube.com/watch?v=Jpbgg2TRCuw

Don't believe the video? Then go to the park and fling a couple balls for yourself.

Ken

I would have to agree,
if the moon were suddenly released from earth the moon would wobble around a bit first the start in to a rotation later.
However in the release the moon could make a tumble into a orbit region where it would not gain a rotation, given that the earth is just a few million miles out side of the no rotation region of our solar system.

DwayneD.L.Rabon

Ken, again you have shot yourself in the foot (Same as with the Foucault pendulum)!

Yes I watched YouTube video (link below)* of the Russian blond's hammer throw - the slow motion frames just before the ball hit the ground. The chain and handle grip are always stretched out from the ball in straight line by the centrifugal force of the spinning ball.

This straight line of chain does revolve around the spinning ball; even despite the air drag which would eventually stop the ball's spin with the chain trailing behind if the ball did not hit the Earth. The last slow motion part of the video begins with the chain out in front of the ball, but as the ball continues its spin the chain swings, first away for the recording camera, then continues to spin around to trail behind the ball and finally as it hits the ground, the chain is pointing towards the camera.

Thus the camera only records about 3/4 of one complete rotation or spin of the ball; however the slow motion sequence is only the tiny end fraction of the ball's total flight. I would estimate that during the ball's flight it made more than dozen complete 360 degree spins, but of course it too is spinning about some point only very near the center of the ball due to the relatively slight mass of the chain.

What is it you have against the very useful term "spin" to want to abolish it or at least replace it with "rotating about a barycenter" in two body cases?

Why do you not at least admit a few of your errors? For example, start by admit the sun has much more influnce on the moon than the Earth does (8.45 times more). After you have done that I will suggest the next error to admit.
----------------------------
* http://www.youtube.com/watch?v=Jpbgg2TRCuw

PS later by edit: That the only poster to agree with you is Dwayne, should also give you reason to think again. He has his own POV which is always quite disconnected from reality.

Ken, again you have shot yourself in the foot (Same as with the Foucault pendulum)!

Billy-bob, do you have any actual proof that a Foucault pendulum would work on the moon? If so, please do post it!


Yes I watched YouTube video (link below)* of the Russian blond's hammer throw - the slow motion frames just before the ball hit the ground. The chain and handle grip are always stretched out from the ball in straight line by the centrifugal force of the spinning ball.

This straight line of chain does revolve around the spinning ball; even despite the air drag which would eventually stop the ball's spin with the chain trailing behind if the ball did not hit the Earth. The last slow motion part of the video begins with the chain out in front of the ball, but as the ball continues its spin the chain swings, first away for the recording camera, then continues to spin around to trail behind the ball and finally as it hits the ground, the chain is pointing towards the camera.

Thus the camera only records about 3/4 of one complete rotation or spin of the ball; however the slow motion sequence is only the tiny end fraction of the ball's total flight. I would estimate that during the ball's flight it made more than dozen complete 360 degree spins, but of course it too is spinning about some point only very near the center of the ball due to the relatively slight mass of the chain.

----------------------------
* youtube.com/watch?v=Jpbgg2TRCuw


Since you're such an observant chap, did you notice that the Russian gal was RAPIDLY spinning her hammer in the counter-clockwise direction prior to release, but that very slow rotation of the ball you noticed was instead in the opposite CLOCKWISE direction?????

Billy-bob, putting aside the fact that ball was spinning at high rps prior to her release, and that the ball was clearly spinning much more slowly in the OPPOSITE direction AFTER the release (I say from cartwheeling prior to the release), can you explain why the ball shifted from a counter-clockwise rotation to a clockwise rotation prior to landing?



What is it you have against the very useful term "spin" to want to abolish it or at least replace it with "rotating about a barycenter" in two body cases?

Nothing wrong with the word "spin," so please re-post any comment of mine that gave you that idea????

As for "rotating about a barycenter," while many people do improperly use "rotate" when describing that, it's better usage to say "REVOLVING about a barycenter."

As in, "the earth *rotates* on its polar axis as it *revolves* around the sun."

If you prefer saying that "the earth *spins* on its polar axis as it *spins* around the sun," then that's OK by me too.


Why do you not at least admit a few of your errors? For example, start by admit the sun has much more influnce on the moon than the Earth does (8.45 times more). After you have done that I will suggest the next error to admit.

Define "more influence?" The moon's gravity affects the Earth's tidal bulges roughly twice as much as our sun's gravity affects our oceans' tidal bulges.

The Earth's gravity likewise is the stronger force that has, over the eons, caused the moon's two tidal bulges to solidify pointing towards us and not towards the sun.

This page may help you to understand:

---snip---
Mathematical Explanation of Tides
[ ]
Since R is small compared to D (about 1/60 of it) this is nearly equal to G * Mmoon * (2 * D * R) / D4, or (a constant) / D3. Tidal acceleration is therefore approximately proportional to the CUBE of the distance of the attracting body. If the Moon were only half as high above the Earth, the tidal accelerations would be EIGHT TIMES as great as they are now!

As it happens, both the Moon and the Sun cause such tides in our oceans. The Moon is about 400 times closer than the Sun, so it causes a tidal acceleration equal to 4003 or 64 million times that of an identical mass. But the Moon's mass is only about 1/27,000,000 that of the Sun. The result is that the Moon causes a tidal acceleration that is about 64,000,000/27,000,000 or 64/27, or a little more than double of that of the Sun. That's why the tides due to the Moon are larger than those due to the Sun.

At different times of a month, the Sun and Moon can cause tides that add to each other (called Spring Tides) (at a Full Moon or a New Moon, when their effects are lined up.) However, at First Quarter or Third Quarter Moon, the tides cause by the two sort of cancel each other out (with the Moon always winning!) and so there are lower tides then, called Neap Tides.

(Add the HTTP only): mb-soft.com/public/tides.html

Is my post even here?

I said, a small change, like a moon having a moon, results in new variables so that we may not even exist.

Ken: Another way to show how silly your POV is follows:

For convenience assume the moon Earth are in “deep space” far from any other masses but all else is unchanged. I.e. the moon turns around once in exactly one orbit period, which is 28 days. Everyone but you and Dwayne describes this as “spinning” on its polar axis with period of 28 days, but you two say: “No that is not it is not spinning, it is tide-locked orbiting the barycenter without any spin.”

Now suppose a third body rapidly passed by, making a gravitational impulse (acting on the moon’s center of mass of course and applying no net torque* to the moon) which throws the moon into a new, much more elliptic, orbit about the barycenter with a period of 50 MONTHS. Now at apogee the moon is no longer keeping the same face turned towards the Earth, but turns 360 around MANY times so we see all side of the moon. (We already can see more than half of the moon as it is not now in a purely circular orbit.)

Do you at least agree it is “spinning” in its new orbit, even though it angular momentum about it polar axis is exactly what it was before the third body approached? If you do, why do you say it is NOT now spinning about its polar axis when the rotation about that axis (rate = 360degrees / 28 days) is exactly the same before as after the third body passed?

You are in the silly position of saying that the moon is spinning around it polar axis if the Moon day is 28.00001 or more Earth days long or if it is shorter than 27.999999… Earth days long; but if it happens to be rotating at exactly 28 days, then it is not spinning – it is orbiting about the barycenter!

Dwayne delights in holding silly positions, but you seem capable of more understanding. We all are trying to help you understand.
-----------------
*That is certainly possible, and the case I am assuming for simplicity, but may not be true in general.

Is my post even here?

I said, a small change, like a moon having a moon, results in new variables so that we may not even exist.

Hi - I don't think I saw any earlier post by you. However, I think your comment concerns how a moon(s) may influence a planet's ability to evolve life?

Earth's moon is very unusual in being so large relative to the Earth, and there are indeed theories that our moon's strong gravity was helpful in allowing life to form here.

E.g., the moon was likely very much closer to the Earth when life began to develop, thus creating massive tides and waves in our planet's primordial soup. While these early extreme conditions wouldn't be very friendly to later more advanced life forms – even so, these rough conditions would actively stir the soup, which may have been very beneficial in the beginning

As for a moon having its own moon, I won't say NO, but I don't recall that being observed in our solar system?

The problem I see, is that once these binary moons become tidally locked to each other (as Pluto and Charon are locked together), then their primary planet will eventually halt their revolution around their common barycenter, and eventually that would cause them to close together and eventually meld into one moon. Once that happened the evidence of a binary moon system would be lost.

The same thing is projected to happen to our Earth and our moon – as the theory goes, eventually the moon will tidally lock the Earth, and at that point they will both revolve around their barycenter, roughly once every 50 to 60 days – i.e., the Earth will have the same type of day the moon now has. Then, the moon would gradually close back in and meld into the Earth.

Of course, none of that will ever happen since the Earth and our moon will be consumed by our sun as it goes red-giant long before the moon tidally locks the Earth.

Ken
.

Billy-bob, do you have any actual proof that a Foucault pendulum would work on the moon? If so, please do post it!I do not think one has been observed on the moon for 28 days. I based my statement that it would appear to turn 360 once in 28 days to an observer sitting on the moon without moving on the fact that the Foucault pendulum keeps in the same plane (or one parallel to it) fixed in space. That is its "claim to fame." - Why Foucault invented it. Even you admit that the moon turns relative to planes fixed in space. You just want (for some silly reason - see my post 124) to claim this is not spinning on it axis IFF the spin rate is exactly the same as the orbit period.

Since you're such an observant chap, did you notice that the Russian gal was RAPIDLY spinning her hammer in the counter-clockwise direction prior to release, but that very slow rotation of the ball you noticed was instead in the opposite CLOCKWISE direction????? No, not originally, did I notice that, but on second viewing I agree that is what the camera records.

However two different cameras were used. One close to the Russian lady and viewing her swing the hammer around from BELOW THE PLANE IT WAS SWINGING IN. The second camera was high in the stands with a telephoto lens and looking DOWN ON THE PLANE the ball and its chain were spinning in. That one camera recorded clockwise and the other counter clockwise is EXACTLY what they should if the spin was always in the same direction.

You are adding yet another error with these most recent comments.
As Ophiolite advised you: When deep in an embarrassing hole stop digging!

You do not seem to know that clockwise and counter-clockwise are RELATIVE to the viewer terms, not absolute terms independent of the viewer's position. If you have a big clock, turn it face down on the floor and then you will note that your perfectly good clock now has the hands going around counter clockwise! One camera had a view from below and the other from above so of course one recorded the VERY SAME SPIN as clockwise and the other as counter clockwise. I am losing count, but think that is error six for you in this thread.


...can you explain why the ball shifted from a counter-clockwise rotation to a clockwise rotation prior to landing? I just did. - You are just adding to your list of silly statements, all based on appearances, without any understand of the facts.

You seem like a nice guy. Please stop shooting yourself in the foot and stop digging deeper in your embarrassing hole. That must hurt with both feet bleeding now. :D

You do seem to be trying to come to the correct POV, which I explained earlier (in post 96) to you that it is the gradient of gravity which falls off as the cube of the separation, which causes the moon to produce stronger tides than the sun despite the quadratic fall off of gravity, which causes the sun to control the motion of the moon thru space except for the very tiny (1/4 of one percent radial wobble).

Here is part of post 96 again:
...No you are wrong here too. The tides are caused by the GRADIENT of gravity not by gravity itself. The gradient falls off as the CUBE of the distance, not the square. This is why the moon is much more important than the sun in producing tides on Earth.
Just do the calculations with mass of sun at 1 AU from moon and mass of Earth at one moon /Earth separation. When calculating gravity force on moon the denominator distance is squared and the sun is much more important.
When calculating the gradients the inverse CUBE makes the moon's gradient larger at the Earth than the sun's is. So the moon does dominate the tides. Do it and then you will understand why moon dominates the tides on Earth but the sun, not the Earth, controls the moon's orbit around the sun. Earth only causes an "hardly noticicable" percentage variation or "wobble" in the distance between the moon and the sun.
You do not understand what causes the tides. I.e. that it is not gravity but the gradient of gravity.
Thus it would be interesting to hear you try to explain why there are TWO tides each day, (one on each side of the Earth), instead of just one bulge on the side of Earth towards the moon. The correct answer has is due to fact the moon's gravity is stronger on the side near to the moon than on the side most distant from the moon - I.e. depends upon the GRADIENT of the moon's gravity, not on the gravity itself....

Well Billy T , are you trying to say that the moon does not orbit the earth, but instead orbits the Sun in a seperate orbit from earths prime orbit around the sun.

If the moon has a spin rotation as appearantly suggested in discussion, then the moon must gain the momentum for that spin accross 638 miles of it trajectory before a cancelation of that angular trajectory which occurs every 15 minutes. So in short terms the moon acutally spins for 15 minutes before it cancels that spin motion and begins another 15 minutes of spin in the opposite direction.

DwayneD.L.Rabon

"Why is the moon not spinning then?"

The moon is not spinning (we don't see both sides) because it turns once on its axis for each orbit, so we see the same side, because its axis of rotation is aligned with ours, more or less .

This thread has spun a tale or two but it started to spin out with what apparent and actual spinning are. Or someone thinks they've found a hole in Newtonian mechanics again, but spin is a distinct kind of motion - we all know what it is and we can all see the moon doing it.

The question should be: "If the moon isn't spinning, why don't we see both sides?"

P.S.
Oh crap, back in the "the moon orbits around the sun, not the earth" ballpark.
The earth-moon system is coupled to the sun's gravitational influence, primarily. The moon is primarily coupled to the earth's. Any satellites between them can be coupled to either, primarily, or in a lagrange point, so coupled equally to both (earth and moon). We have observational craft orbiting earth, the moon, the sun, I'm sure there are one or two at these lagrange points.

Well Billy T , are you trying to say that the moon does not orbit the earth, but instead orbits the Sun in a seperate orbit from earths prime orbit around the sun.

If the moon has a spin rotation as appearantly suggested in discussion, then the moon must gain the momentum for that spin accross 638 miles of it trajectory before a cancelation of that angular trajectory which occurs every 15 minutes. So in short terms the moon acutally spins for 15 minutes before it cancels that spin motion and begins another 15 minutes of spin in the opposite direction.

DwayneD.L.RabonHard to follow you, but yes Every body orbiting the sun not in a perfectly circular orbit does have changing (linear) momentum (More near the sun than at apogee) but its angular momentum is constant.

The maxium stable orbit for moon orbit of the earth is about 331,559 miles from earths center after this point its orbit will become erratic, where it would become oblong in orbit like a comet, as if it where to strike earth or be lost to space.

The limit of this orbit is determine by the sun. the exstent of the current orbit is due to interaction with the sun, such as the rate of motion at which the moon gets farther from the, estimated to be 3 cm per year.

In mention of the moons trajectory changes relavant to spin, the moon travels to the gravitional inductance of the sun in its motion around the earth of which it meets the center point every 15 minutes (at times tis may be as short as 7 minutes) in that time the moon has traveled 638 miles (20 miles on earth surface or 0.3 degrees) when the moon meets this inductance point the inductance force is is canceled and the moon gains next motion of inductance of the sun which is traveling in opposite motion causing the moon to turn backward (simular to sine frequency).
Variations in the time of the moon to travel to the suns inductance point cause accumulation of motion in one direction or the other to be greater at different times, therorically if enough momentum is gained in one direction the moon could actually make a complete turn.

The overall motion is simular to walking. As well this same motion may the resulting cause of the number of subtle earthquakes that occur every second or minute.


DwayneD.L.Rabon

#2 - your 4th graphic, if viewed from a planet at the center of its orbit, would be tumbling around two axes. It thus would NOT be tidally locked ...UNLESS, there were other moons close by that were causing it to tumble and keeping it from becoming tidally locked.

How do your graphics relate to the issue of whether (or not) our own moon is still *rotating* around its polar axis?

Quite simply, The only difference between the graphics are the axial tilt of the satellite. The sidereal rotation and period of revolution is the same. In the first graphic (Which BTW is the fourth graphic from a different viewpoint), you would say that the satellite tumbles around two axes, but in the 3rd, which closely resembles the orbit of the Moon, you would say that it doesn't rotate at all. Yet you have still failed to answer as where exactly between 90° and 0° axial tilt does the satellite stop all rotation, while to someone outside the system the rotation of the satellite doesn't change, just the angle at which it rotates.

Also, let's go back the the situation where you say the satellite is tumbling around two axes. This situation is very like that of Uranus. It too, essentially lays on its side as it orbits the Sun with its polar axis stationary with respect to the fixed stars. If you were to watch Uranus from the Sun, its apparent motion would be like that of graphic 4, with the exception that it would rotate around its polar axis more quickly.

A rotating body, however, does not naturally rotate around two axes, unless there is precession. Precession is caused by a torque applied to the axis of a rotating body. This torque acts at a right angle to the precession.
In this case it would have to act at a right angle to the orbital plane. However, there is nothing that can cause such a torque, and without a constant torque, there can be no precession.

Ergo, no matter what it looks like from the Sun, Uranus doesn't tumble around two axes (and neither does the satellite in the 4th graphic.) Instead, its polar axis stays fixed in one direction while it rotates around it.

And what holds for an axis of a planet of satellite holds to a face of a satellite. If a satellite keeps one face stationary with respect to the stars, it does not rotate, even though it presents all sides to the planet as it orbits.

The upshot is that, despite your earlier protest, there is a preferred frame for rotation. This is easily demonstrated. Take a bicycle wheel that is not rotating with respect to the fixed stars. You can grab it by the axle and easily turn it in any direction. Now spin it on its axle, and try to twist it, it will fight you.(and will precess as discussed above). So even if the fixed stars are not the preferred frame for rotation, they are very close to it. And thus, rotation is properly judged with respect to the fixed stars in all instances.



If our moon is still rotating around its internal polar axis one (1) time per complete orbit as some here claim, then please do name the force that has (for at least 3 billion years) kept our moon rotating around its polar axis against the torque of tidally locking?

You are arguing a tautology. Essentially, you are saying that since the moon doesn't rotate as it orbits it keeps the same face towards the Earth, and since it keeps the same face towards the Earth, it doesn't rotate.
You're assuming your conclusion as your initial premise.

The reverse argument is that because the Moon rotates once per orbit it keeps one face always towards the Earth.

Tidal locking locks the Moon's period of rotation to 1 rotation to one orbit. Though it doesn't keep the rotational speed matched to the orbital speed. The radians per sec swept out by the Moon's orbits speeds up and slows down from perigee to apogee, the radian per sec due to rotation however stays constant. This is what causes the libration of longitude, at different parts of the orbit the rotation speeds ahead of or lags behind the orbit. This indicates independence of rotation from orbit.




Other than the Earth and moon's mutual pull of gravity on each other (and to a lessor extent of the sun & Jupiter, etc), the only remaining force I see that's affecting the moon's motion would be the moon's own orbital motion around the Earth-moon barycenter, and the force of that orbital motion is not constant since the moon's orbital motion is speeding up as energy is transferred from the Earth through tidal braking.

Pretty graphics, though.
:)

Ken

The Moon's orbital motion is slowing down not speeding up. Higher orbits are slower orbits and the Moon is receding from the Earth. It is gaining energy, it is just that the majority of that energy is gained as gravitational potential.

It's made by aliens and is hollow. Don't you read?!

That's why it doesn't rotate the way other moons do.

The Moon's orbital motion is slowing down not speeding up. Higher orbits are slower orbits and the Moon is receding from the Earth. It is gaining energy, it is just that the majority of that energy is gained as gravitational potential.

\omega r = v

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and \omega is decreasing.

I don't know though...but I do know what you said makes no real sense.

If the Moon had a moon, would that be spinning?

Yes.

\omega r = v

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and \omega is decreasing.

I don't know though...but I do know what you said makes no real sense.

Orbital speed does decrease and the period of the orbit does get longer, yes.

The orbital speed of any satellite can be found by:

V_o = \sqrt{\frac{GM}{r}}

where r is the radius of the orbit.
G the gravitational constant
M the mass of the Planet

As r increases, Vo decreases.

The period of the orbit can be found by

T = 2\pi sqrt{\frac{r^3}{GM}}

Again, note how the period increases as r increases.

Lastly, the energy of a satellite can be found by the sum of its kinetic energy and potential energy or:

E= \frac{mv^2}{2}- \frac{GMm }{r}

OR

E = -\frac{GMm}{2a}

Where m is the mass of the satellite and a is the mean orbital distance.

\omega r = v

Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.

Most likely the velocity is fixed and radius is getting higher and \omega is decreasing.

I don't know though...but I do know what you said makes no real sense.


I agree. Janus58 just quoting the same equation isn't a sufficient explanation.

Are you challenging the correctness of the equations? If so, why? On what grounds?
If not, what could possibly be more sufficient than equation, which explains exactly what is happening?

\omega r = v
Is rotational speed slowing down? If not and the radius is increasing the velocity is also increasing.
Most likely the velocity is fixed and radius is getting higher and \omega is decreasing.
I don't know though...but I do know what you said makes no real sense.Your “most likely” gave me a good laugh. Thanks. To give you one, I will say: “Most likely, the sun will rise again tomorrow.”
Neither is a probability thing. To start where surely you can follow:

F = ma
And for circular orbiting mass, m, which is r from mass M the F is GMm/r^2.
As m appears on both sides of the equation I will drop it and chose set of units where GM = 1 so I have:

a = 1/r^2 but the acceleration a is either v^2/r or r\omega^2 and the second is more convient now so:

a = 1/r^2 = r\omega^2 and now not bothering with the left most "a part" of the equation anymore and multiplying both sides of the rest by r^2 we have:

1 = (r^3)\omega^2 which you may not recognize is one of Keppler’s three laws. (The third , I think by memory.) Namely that the period, P, squared is proportional to the cube of the radius r.
It is actually more general than just this circular orbit case; because the period in an elliptical orbit depends only on the longest straight line segment with end points on the elliptical orbit, which I will call “L”.
(Normally this length is called the “major axis,” but as both “m” and “a” were already used, I was desperate for different letter, so I called that length “L”)

So the more general statement of that law for elliptical orbits is:
P^2 ~ L^3 or if willing to use the strange units in which GM = 1, then P^2 = L^3.
I like these units as who the hell remembers the mass of the planets, the sun etc. or G? (Except expert workers in the field, like Janus58.) But I digress. Back to:

1 = r^3\omega^2 = r (v^2) which tells you, for example, that if r doubles, then the tangential velocity goes down to the square root of what it was. But when r doubles, so does the circumference and even it the tangential velocity did not decrease, it would take longer to get all the way around the orbit. I.e. the orbital period, P, is increasing or \omega is decreasing. {Jeeezzz! that Janus 58 is a smart guy - he just knew this without all this work. hehehe.}

Switching to the general elliptical case (which includes circles) When L doubles, then L^3 goes up 8 times and the square of the period must too. Thus the period more than doubles. – it becomes 2(square root 2) times greater, (“most likely”) :D just as “most likely” the sun will rise again tomorrow, but who really knows? :shrug: :rolleyes:


I agree. Janus58 just quoting the same equation isn't a sufficient explanation.Was the F = ma start simple enough for you?

I agree. Janus58 just quoting the same equation isn't a sufficient explanation.

We all know how you feel about math, but my response was to CheskiChips, and they used math, so I responded with math.

We all know how you feel about math, but my response was to CheskiChips, and they used math, so I responded with math.


The laser reflector data analysis should be able to show whether the Moon's velocity is increasing or not. I'd rather trust that than your maths.

The laser reflector data analysis should be able to show whether the Moon's velocity is increasing or not. I'd rather trust that than your maths.

The LASER (it is an acronym after all) reflector data is useless without the maths to understand it.

The laser reflector data analysis should be able to show whether the Moon's velocity is increasing or not. I'd rather trust that than your maths.that tells the distance to the moon, to a few mm, I think, not its speed, which has an extremely low change as a percent /year. Also has the complexity of the changes due to moon not being in circular orbit. - Makes much greater speed changes than could be observed with lasers. To get the average speed from distance measurments you would need all the math I and Janus58 explained to you and more. You and some others active in this thread simply do not have the slightest knowledge of what you post about.*

The best way to get the average speed over full orbit period I bet has nothing to do with lasers and a lot to due with photographs against the back ground stars. (And correcting for Earth's motion between exposures, with more math, of course.)

--------
*Too many posting here seem to think orbital mechanic is something you guess at, or has probable answers. It was the queen of the precise sciences and certainly held that position uniquely even 2000 years ago, despite not know how it worked as we do now. Now we can send space craft to Pluto or to skim thru the solar corona! (both of which my former employer, APL/JHU has or is doing.) It also was the reason for/ motivation of / most of the advance in math were made pre 1850. (things like calculus, logrythmic computations, etc.)

Quite simply, The only difference between the graphics are the axial tilt of the satellite. The sidereal rotation and period of revolution is the same.
That's absolutely NOT true!!!

Since I don't have full posting privileges yet and can't post your images side by side for easier comparison, why don't you do that for us all!



In the first graphic (Which BTW is the fourth graphic from a different viewpoint), you would say that the satellite tumbles around two axes, but in the 3rd, which closely resembles the orbit of the Moon, you would say that it doesn't rotate at all. Yet you have still failed to answer as where exactly between 90° and 0° axial tilt does the satellite stop all rotation, while to someone outside the system the rotation of the satellite doesn't change, just the angle at which it rotates.

I dispute that your 1st and 4th graphics are the same with their only differences the observer's viewpoints (i.e., center-point versus sidereal.)


Also, let's go back the the situation where you say the satellite is tumbling around two axes. This situation is very like that of Uranus. It too, essentially lays on its side as it orbits the Sun with its polar axis stationary with respect to the fixed stars. If you were to watch Uranus from the Sun, its apparent motion would be like that of graphic 4, with the exception that it would rotate around its polar axis more quickly.

Uranus' polar axis is essentially on its side, but its polar axis is NOT "stationary with respect to the fixed stars!"

Can you provide a citation?


A rotating body, however, does not naturally rotate around two axes, unless there is precession. Precession is caused by a torque applied to the axis of a rotating body. This torque acts at a right angle to the precession.

The force causing torque can also be parallel to the spin axis, as is the case with a toy top's procession caused by the Earth's gravity.

There is both torque-free precession as well as procession caused by torque.



In this case it would have to act at a right angle to the orbital plane. However, there is nothing that can cause such a torque, and without a constant torque, there can be no precession.

Not so - since our Earth rotates on its polar axis and has constant torque being applied by the sun and moon at right angles:

---snip---
The Earth goes through one complete precession cycle in a period of approximately 25,800 years, during which the positions of stars as measured in the equatorial coordinate system will slowly change; the change is actually due to the change of the coordinates. Over this cycle the Earth's north axial pole moves from where it is now, within 1° of Polaris, in a circle around the ecliptic pole, with an angular radius of about 23.5 degrees (or approximately 23 degrees 27 arcminutes [2]). The shift is 1 degree in 180 years, where the angle is taken from the observer, not from the center of the circle.

Discovery of the precession of the equinoxes is generally attributed to the ancient Greek astronomer Hipparchus (ca. 150 B.C.), though the difference between the sidereal and tropical years was known to Aristarchus of Samos much earlier (ca. 280 B.C.). It was later explained by Newtonian physics. The Earth has a nonspherical shape, being oblate spheroid, bulging outward at the equator. The gravitational tidal forces of the Moon and Sun apply torque as they attempt to pull the equatorial bulge into the plane of the ecliptic. The portion of the precession due to the combined action of the Sun and the Moon is called lunisolar precession.

Revolution of a planet in its orbit around the Sun is also a form of rotary motion. (In this case, the combined system of Earth and Sun is rotating.) So the axis of a planet's orbital plane will also precess over time.

The major axis of each planet's elliptical orbit also precesses within its orbital plane, partly in response to perturbations in the form of the changing gravitational forces exerted by other planets. This is called perihelion precession or apsidal precession (see apsis). Discrepancies between the observed perihelion precession rate of the planet Mercury and that predicted by classical mechanics were prominent among the forms of experimental evidence leading to the acceptance of Einstein's Theory of Relativity, which predicted the anomalies accurately.[3][4]

These periodic changes of Earth's orbital parameters, combined with the precession of the equinoxes and of the inclination of the Earth's axis on its orbit, are an important part of the astronomical theory of ice ages. For precession of the lunar orbit see lunar precession.

(Add the http only): en.wikipedia.org/wiki/Precession#Of_the_Earth.27s_axis



Ergo, no matter what it looks like from the Sun, Uranus doesn't tumble around two axes (and neither does the satellite in the 4th graphic.) Instead, its polar axis stays fixed in one direction while it rotates around it.

Stays fixed in relation to what?


And what holds for an axis of a planet of satellite holds to a face of a satellite. If a satellite keeps one face stationary with respect to the stars, it does not rotate, even though it presents all sides to the planet as it orbits.

That simply is not true! Venus nearly does keep one face pointed towards the stars as it rotates clockwise (retrograde) on its polar axis nearly one time per orbit.

(NOTE: Venus completes one counter-clockwise orbit every 224.65 days and Venus rotates once clockwise every 243 days.)

In any event, I've already covered that topic well, and an astronomical body that *appears* from the sidereal perspective to not be rotating, actually does have one polar axial spin (in a clockwise direction) per each complete orbit, as can plainly be seen in the top graphic here (don't add www):

community-2.webtv.net/kdine5/Lunacy/index.html


The upshot is that, despite your earlier protest, there is a preferred frame for rotation. This is easily demonstrated. Take a bicycle wheel that is not rotating with respect to the fixed stars. You can grab it by the axle and easily turn it in any direction. Now spin it on its axle, and try to twist it, it will fight you.(and will precess as discussed above).

That's an informal explanation of precession often demonstrated in high school classes:

"In a classic beginning physics demonstration, the instructor stands on a swiveling platform and holds a spinning bicycle wheel at arm's length. The wheel is vertical and the instructor is standing still. The instructor then tilts the wheel toward horizontal. This causes the instructor to start spinning slowly on the platform. Bringing the wheel back to vertical and tilting it the other way makes the instructor spin the other way.

"Since the forces on opposite sides are in opposite directions, the result is torque. Each pair of opposite particles on the wheel contributes to the torque that causes the instructor to turn on the platform. Tilting the wheel the other direction produces torque in the opposite direction, slowing the instructor's spin and eventually reversing it."


So even if the fixed stars are not the preferred frame for rotation, they are very close to it. And thus, rotation is properly judged with respect to the fixed stars in all instances.

As for a sidereal perspective being a "preferred" perspective (it obviously has its uses in astronomy), that doesn't turn its perspective into an absolute reality.


You are arguing a tautology. Essentially, you are saying that since the moon doesn't rotate as it orbits it keeps the same face towards the Earth, and since it keeps the same face towards the Earth, it doesn't rotate.
You're assuming your conclusion as your initial premise.

Absolutely not true!!! I have pointed out that a large astronomical body rotating around its polar axis assumes the shape of an oblate sphere (as the Earth is shaped.) WHEREAS, our non-polar-rotating moon has solidified into the shape of a football as it cooled over 3 billion years ago freezing its two tidal bulges in place.

When the moon still had axial rotation its tidal bulges wouldn't have been fixed in one location on its surface. When the moon lost its polar rotation over 3 billion years ago, it wound down to ZERO polar axial rotations per orbit, NOT one (1) polar axial rotation per orbit!

Today the moon ONLY revolves around an exterior axis, the Earth-moon barycenter!

Models prove that an orbiting astronomical body CANNOT spin on two axes at the same time and still keep one of its faces constantly pointed towards the center point!

That's all quite a bit more than mere tautology!!!

A classic example of tautology would be claiming that, in order to keep one face pointing towards Earth, the moon rotates once per each orbit.

Rotates around WHAT?? If the axis for that claimed rotation is the Earth-moon barycenter, which is the same axis of the moon's orbit, then that's classic tautology!!!


The reverse argument is that because the Moon rotates once per orbit it keeps one face always towards the Earth.

That's your apparent argument, not mine!

I say that one complete moon orbit is one (1) 360 degree revolution around the Earth. The moon's ONLY axis for that orbit is the Earth-moon barycenter located within the Earth.


Tidal locking locks the Moon's period of rotation to 1 rotation to one orbit. Though it doesn't keep the rotational speed matched to the orbital speed. The radians per sec swept out by the Moon's orbits speeds up and slows down from perigee to apogee, the radian per sec due to rotation however stays constant. This is what causes the libration of longitude, at different parts of the orbit the rotation speeds ahead of or lags behind the orbit. This indicates independence of rotation from orbit.

Libration is interesting, but it doesn't account for any 360 degree rotations of the moon. What is at issue here is the location of the moon's axis as it orbits 360 degrees around the Earth.

If you claim the moon has two separate 360 degree spin axes and it still keeps one face pointed towards us, then I dispute that!


The Moon's orbital motion is slowing down not speeding up. Higher orbits are slower orbits and the Moon is receding from the Earth. It is gaining energy, it is just that the majority of that energy is gained as gravitational potential.

Thank you for correcting my sloppy verbiage – you are 100% right about the transferred energy being stored as gravitational potential as the moon slowly recedes from the Earth.

Ken

To Ken Dine:

You have a strong bias that drives you to claim the moon rotates about the Earth moon barycenter without spin. Others have one that prefers to consider the axis of rotation passes thru the center of mass of the rotating moon and when we do this we call the rotation spin, or even polar axis spin also.

In truth the moon has two types of momentum, angular and linear.

In the case of angular momentum of any body with mass M there is always an axis of rotation implied when stating the angular momentum, but where it is, is completely arbitrary in the sense that if you take each tiny differential mass element m and multiply it by its speed orthogonal to the line between m and the line A, which is the selected axis of rotation, and then sum up all such products for all m in the mass M and set it equal with some specified angular momentum, you will find (if memory serves me correctly - too lazy to do this all again, showing the proof, after so many years.) that the location of A can be anywhere. For example, I would have no trouble in using an axis of rotation for the moon, which passes thru Mars.

Now no one does an analysis of the moon's angular momentum about Mars* about an axis of rotation passing thru Mars because it does not seem that way to humans but the math is OK with that axis. People speak of what they know and what seems reasonable to them.

We all agree that the moon is turning the side never seen by Earthlings to point to all the distant stars in a circular arc. So we all agree the moon is turning. You unfortunately know about the barycenter and are fixated that must be the "true" axis of this rotation. Most people have never heard of the barycenter so they do not do that. Most people, myself included, think that any object steadily swinging around to sequentially present one side to all the distant stars along a great circle in the heavens is rotating or spinning and chose the spin axis as passing the center (of mass, if they have a little knowledge of physic) of the turning object.

Your stubborn dogmatic POV that only the barycenter is correct as the axis of rotation (another error not listed below) has caused you to make many statements which are clearly false and to ignore many simple examples which show them to be false. Instead of admitting or discussing your errors you are now engaged in discussion about what some of Janus58 drawings do or do not show.

Your more serious errors include:

(1) Treating as an absolute property of rotation the term "clockwise" and not realizing that "clockwise" switches to "counter -clockwise" as the observer changes from one side to the plane of rotation to the other. (Associated with your second "shooting of self in the foot" with a false "proof" that you were correct.)

(2) Not realizing the Foucault Pendulum operating above either pole of the moon does make a 360 circle in about 28 days - you offered the fact that it would not as if it did that would indicate moon was spinning about the polar axis. (Associated with your first "shooting of self in the foot" with a false "proof" that you were correct.)

(3) Not realizing that the sun's gravity is stronger force at the moon than the earth's gravity is. (I worked it out for you to show sun is 8.45 times stronger, but received no thanks or even an acknowledgement that you were wrong. - So I assume you still hold that false POV.)

(4) Not realizing that the tides are made by the gradient of gravity, not by the force of gravity. Or that your argument based on moon making stronger tides than the sun does not show the moon has stronger gravity at the moon.

(5) Not realizing the moon only appears to orbit the Earth for a small set of reference frames, but all others in the universe, for example for a frame based on any planet or asteroid, or the sun etc. see it is in orbit about the sun with about a 1/4 of one percent wobble. I illustrated this 1 part in 400 radial variation with 8 inch diameter circular drawn by a line only 0.01 inches wide. Something approximately this width| forming a circle filling your laptop computer screen and told you the wobble stayed inside the line's width. –I.e. was not even noticeable in that scale drawing!)

Hey, you do know what the barycenter is and most do not; but as they say: "A little knowledge is a dangerous thing", especially for one lacking the grace to admit their errors.

--------------
*Except Martians, of course.

That's absolutely NOT true!!!

Since I don't have full posting privileges yet and can't post your images side by side for easier comparison, why don't you do that for us all!

They are already posted one under the other, side by side would make no difference. I'm afraid that you are seeing what you want to see.





I dispute that your 1st and 4th graphics are the same with their only differences the observer's viewpoints (i.e., center-point versus sidereal.)

There is nothing to dispute, it is fact. These animations were done with a ray-tracer. Ray-tracing works like this: You place objects into a three dimensional area by defining their x,y and z coordinates. You also place a light source and "camera" in the area the same way. The computer then renders the image as it would appear from the "camera's" position. You can also translate and rotate objects in the scene by using a "clock" variable, causing the program to render successive frames of an animation. The only difference in scene language between the 1st and 4th graphic is the position and aiming of the Camera. The translation and rotation parameters of the satellite and its axis remain the same.

So, are you calling me a liar?




Uranus' polar axis is essentially on its side, but its polar axis is NOT "stationary with respect to the fixed stars!"

Can you provide a citation?

http://www.apl.ucl.ac.uk/iopw/uworkshop_060905.pdf

http://www.news.wisc.edu/releases/12826.html

Why would you think that Uranus would be any different from any other planet in that respect?






The force causing torque can also be parallel to the spin axis, as is the case with a toy top's procession caused by the Earth's gravity.

There is both torque-free precession as well as procession caused by torque.

if the force were applied parallel to the axis it will not precess. A toy top will only precess if its axis is tilted from vertical, in which case the force of gravity is not parallel to the axis, and there is no torque. No torque, no precession.




Not so - since our Earth rotates on its polar axis and has constant torque being applied by the sun and moon at right angles:

Wow, you can quote Wikipedia.
The Earth's precession is caused by the uneven gravitational pull of the Moon and Sun on its equatorial bulges. And as according to wiki, this precession has a period of over 25,000 years. The satellite in the graphic would need to precess with a period equal to its orbital period, quite a different matter.( and also an amazing coincidence.) Also, since the satellite in the graphic has exactly a 90° axial tilt, any uneven pull on it's equatorial bulges would cause a torque within the plane of the orbit, not at right angles, as would be needed to cause the apparent precession seen in the graphic.










Stays fixed in relation to what?

The fixed stars. And yes, Uranus probably undergoes a small precession, but since its moons all orbit in or nearly in the plane of its equator, and any torque on its equatorial bulges caused by the Sun would be extremely small (Such a torque is caused by tidal forces, and tidal forces fall of by the cube of the distance), it would have an extremely long period.




That simply is not true! Venus nearly does keep one face pointed towards the stars as it rotates clockwise (retrograde) on its polar axis nearly one time per orbit.

(NOTE: Venus completes one counter-clockwise orbit every 224.65 days and Venus rotates once clockwise every 243 days.)

which has absolutely nothing to do with the principles of the discussion at hand. You do seem to have a habit of flooding a post with irrelevant data.
Besides which, the 243 rotational period is Venus sidereal period. IOW, the time it takes for Venus to rotate once relative to the fixed stars.
Why would you think that the very astronomers that you claim are wrong about the rotation of the Moon would list a rotational period for Venus based on your perception of planetary rotation?




In any event, I've already covered that topic well, and an astronomical body that *appears* from the sidereal perspective to not be rotating, actually does have one polar axial spin (in a clockwise direction) per each complete orbit, as can plainly be seen in the top graphic here (don't add www):

community-2.webtv.net/kdine5/Lunacy/index.html

Once again, you are simply imposing your personal perception of "rotation of an astronomical body. What you "plainly see" in that graphic is not what everyone else plainly sees. We plainly see a moon that does not rotate in the top graphic, and one that rotates counter-clockwise once per orbit in the bottom graphic.


That's an informal explanation of precession often demonstrated in high school classes:

"In a classic beginning physics demonstration, the instructor stands on a swiveling platform and holds a spinning bicycle wheel at arm's length. The wheel is vertical and the instructor is standing still. The instructor then tilts the wheel toward horizontal. This causes the instructor to start spinning slowly on the platform. Bringing the wheel back to vertical and tilting it the other way makes the instructor spin the other way.

"Since the forces on opposite sides are in opposite directions, the result is torque. Each pair of opposite particles on the wheel contributes to the torque that causes the instructor to turn on the platform. Tilting the wheel the other direction produces torque in the opposite direction, slowing the instructor's spin and eventually reversing it."

Once again, some pasted material that does not add anything to the discussion. The ability to paste does not infer understanding.


As for a sidereal perspective being a "preferred" perspective (it obviously has its uses in astronomy), that doesn't turn its perspective into an absolute reality.



Absolutely not true!!! I have pointed out that a large astronomical body rotating around its polar axis assumes the shape of an oblate sphere (as the Earth is shaped.) WHEREAS, our non-polar-rotating moon has solidified into the shape of a football as it cooled over 3 billion years ago freezing its two tidal bulges in place.

When the moon still had axial rotation its tidal bulges wouldn't have been fixed in one location on its surface. When the moon lost its polar rotation over 3 billion years ago, it wound down to ZERO polar axial rotations per orbit, NOT one (1) polar axial rotation per orbit!

That is a silly argument. If enough time were to pass that the Earth becomes tidally locked to the Moon such that its tidal bulges remained fixed on its surface, would you then claim that the Earth had stopped rotating?

The fact that the moon is oblong is simple due to the fact that because of its slow rate of rotation, the tidal forces of the Earth had a greater effect on its shape. It is also related to the fact that the moon likely had a much more eccentric orbit when it cooled into its present shape.




Today the moon ONLY revolves around an exterior axis, the Earth-moon barycenter!

Models prove that an orbiting astronomical body CANNOT spin on two axes at the same time and still keep one of its faces constantly pointed towards the center point!

Again, it seems like you are using your own peculiar notion of rotation here. It can rotate around its own axis and revolve around a orbital axis while constantly facing the axis of revolution. In fact, it must rotate once per orbit to do so.


That's all quite a bit more than mere tautology!!!

No it isn't because you are assuming that your notion of "rotation on an axis" is true to "prove" that it is correct.


Rotates around WHAT?? If the axis for that claimed rotation is the Earth-moon barycenter, which is the same axis of the moon's orbit, then that's classic tautology!!!

It rotates around it own polar axis and revolves around the Earth-moon barycenter Oh, and by the way, the Moon's polar axis is tilted 6.5° to that of its orbit. Because of this, as seen from the Earth, the Moon's axis traces out a cone as with a period of once per orbit. This is what causes the libration of latitudes




That's your apparent argument, not mine!
Learn to read. I never said it was your argument, I said it was the reverse of your argument, And as just as strong as your argument on their own merits.


I say that one complete moon orbit is one (1) 360 degree revolution around the Earth. The moon's ONLY axis for that orbit is the Earth-moon barycenter located within the Earth.



Libration is interesting, but it doesn't account for any 360 degree rotations of the moon. What is at issue here is the location of the moon's axis as it orbits 360 degrees around the Earth.


What is at issue here is that that moon's path as it orbits the Earth-Moon barycenter is a separate and independent motion from its rotation about its own polar axis, just as much as if the moon traveled in a straight line rather than an ellipse. And both the librations of longitude and latitude demonstrate this independence of motion.

If you claim the moon has two separate 360 degree spin axes and it still keeps one face pointed towards us, then I dispute that!

Ken

And in doing so you are making yourself look foolish.

I'm gone for a week because some silly hurricane passed right overhead and filled my yard with trash, and this silly thread has itself become full of garbage! The normal, everyday garbage from those who don't think the Moon is "spinning" is one thing. The garbage from our more educated members is another.


Thus the Sun’s gravity is 1.30 x 6.50 = 8.45 times stronger force on the moon than the Earth’s gravity is. This is why the moon orbits the sun, not the Earth. All the Earth does is make the moon’s basically elliptical orbit about the sun have some small “wobble.”

The Moon orbits the Earth. The relative strength of the Sun's gravitational force is not the metric used to determine whether a body is a satellite of some other body. The Sun's gravitational force is also stronger than is Jupiter's for the six outermost moons of Jupiter -- yet they IAU still classifies these as satellites of Jupiter, just as it classifies the Moon as a satellite of Earth. You are using the wrong metric, Billy. The Moon's velocity relative to the Earth is well below Earth escape velocity, and the Moon's orbit about the Earth lies well within the Earth's gravitational sphere of influence with respect to the Sun.


There is both torque-free precession as well as procession caused by torque.
No torque, no precession.
Even those who dig themselves in deep embarrassing holes are right sometimes. This is one of those times. Any non-spherical body will undergo torque-free precession if its angular velocity vector is not aligned with one of the body's principal axes of rotation. Wikipedia discusses this briefly here (http://en.wikipedia.org/wiki/Precession#Torque-free). The Earth undergoes a torque-free nutation called the Chandler wobble. It is very tiny compared to the torque-induced lunisolar precession.

This should be good.

To get the average speed from distance measurments you would need all the math I and Janus58 explained to you and more. You and some others active in this thread simply do not have the slightest knowledge of what you post about.*





I'm glad I started the thread though. From what I've seen above it might be that the maths isn't quite as cut and dried as you may like to think.

I'm glad I started the thread though. From what I've seen above it might be that the maths isn't quite as cut and dried as you may like to think.
Or it might be that Billy is right and that you "simply do not have the slightest knowledge of what you post about." The math is very cut and dried, and has been since Newton's time. Read post #137, for instance, which contains the equation for the velocity of a small body in a circular orbit.

More generally, the vis-viva equation describes the magnitude of the velocity vector for the orbit of one point mass about another (or two bodies with spherical mass distribution):

v^2 = G\,(m_1+m_2)\,\left(\frac 2 r - \frac 1 a\right)

Back to the circular orbital velocity equation presented by Janus,

v=\sqrt{\frac{GM}r}

The circumference of a circle is c = 2\pi r. The time taken to complete one revolution at a constant speed v is

P = \frac c v = \frac{2\pi r}{v} = 2\pi \sqrt{\frac {r^3}{GM}}

or

\frac{P^2}{r^3} = \frac{4 \pi^2}{GM}

which is Kepler's third law. In other words, this is not just cut-and-dried mathematics, it is 400 year-old cut-and-dried mathematics.

Well, the mention of barycenter and lunar axis left me looking up a few words in the dictionary.

The overall perspective on terms could use some clearing up, relavant to the effect of various points in such interations between bodies. It also seems to be part the case in this disscussion, a little confusion about the significence of various points.

For example Barycenter in my memory meant the gravitional interaction that exist between to bodies at a point between to bodies. I have since found other defintions.

there are more than 5 points of interest in the relations of the moon and the earth.

The point of gravitional interactance that exist between the two bodies, and being in between the two physical bodies exist approximatly 17,280 miles from the moons surface. (being relative to the apogee and perigee). A point of reaction appeantly missed by a great deal of people.


DwayneD.L.Rabon

Even those who dig themselves in deep embarrassing holes are right sometimes. This is one of those times. Any non-spherical body will undergo torque-free precession if its angular velocity vector is not aligned with one of the body's principal axes of rotation. Wikipedia discusses this briefly here (http://en.wikipedia.org/wiki/Precession#Torque-free). The Earth undergoes a torque-free nutation called the Chandler wobble. It is very tiny compared to the torque-induced lunisolar precession.

Granted. I wasn't considering this precession as I was treating the satellite in my example as spherical. But even if it weren't, such a precession would not account for the preceived motion shown in the graphic as Ken is implying it could. The precession axis would have to be at a 90 degree angle to the spin axis, which give an angular velocity for the precesssion of zero.

Or it might be that Billy is right and that you "simply do not have the slightest knowledge of what you post about." The math is very cut and dried, and has been since Newton's time. Read post #137, for instance, which contains the equation for the velocity of a small body in a circular orbit.

More generally, the vis-viva equation describes the magnitude of the velocity vector for the orbit of one point mass about another (or two bodies with spherical mass distribution):

v^2 = G\,(m_1+m_2)\,\left(\frac 2 r - \frac 1 a\right)

Back to the circular orbital velocity equation presented by Janus,

v=\sqrt{\frac{GM}r}

The circumference of a circle is c = 2\pi r. The time taken to complete one revolution at a constant speed v is

P = \frac c v = \frac{2\pi r}{v} = 2\pi \sqrt{\frac {r^3}{GM}}

or

\frac{P^2}{r^3} = \frac{4 \pi^2}{GM}

which is Kepler's third law. In other words, this is not just cut-and-dried mathematics, it is 400 year-old cut-and-dried mathematics.

Lol... yes it is.

... You are using the wrong metric, Billy. The Moon's velocity relative to the Earth is well below Earth escape velocity, and the Moon's orbit about the Earth lies well within the Earth's gravitational sphere of influence with respect to the Sun. ...I think you misunderstood why I mentioned that the sun's gravity was ~8.45 times stronger at the moon than the Earth's is. There were two reasons:

(1) Ken was erroneously using the true fact that the moon makes greater tides in Earth's oceans to falsely assert that therefore the sun's gravity had to be less. (He apparently did not understand then at least, that the tides are caused by the gradient of gravity, not the strength of gravity, but I think he does now.) By calculating the ratio of sun to Earth gravity (my 8.45) I demonstrated his logic was false, as based on ignorance of what cases the tides.

(2) I want to show all, who might not already know, that the sun controls the motion of the moon much more than the earth does. I.e. the solar acceleration of the moon is ~8.45 times larger. This relates to your objection.

I did not use the 8.45 as my "metric" for statement that moon is in orbit about the sun. My metric was the fact that the moon is always curving towards the sun. The moon only curves towards the Earth when the Earth is closer to the sun. Then the constant curving towards the sun means it is also curving towards the Earth. When the Earth is near the line between sun & moon, the radius of curvature towards the sun is less than 1AU. Two weeks later the moon's radius of curvature is more than 1AU and then moon is curving AWAY from the Earth. Something orbiting the Earth should never be curving AWAY from the Earth, at least that is not what I call "orbiting." (As the Earth is also curving towards the sun, the earth/ moon separting may not be increasing despite the moon curving away from where the Earth is.)

My "metric" for stating the moon is NOT orbiting the earth is not only the fact that the moon is always a curving towards the sun but also that for more than half the time the moon's trajectory thru space is actually curving AWAY for the Earth! Here is another attempt to describe my "metric.":

If the moon's trajectory about the sun were a perfect ellipse, would you not agree it is in orbit about the sun? I will assume you agree and continue. So the question becomes to what extent can it deviate from a perfect ellipse and still be consider to be orbiting the Sun along with the Earth? Clearly a low Earth orbit satellite should be considered to orbit the Earth even though it is also circumnavigating the sun. I think even a geostationary satellites orbit the Earth. I make the division based on whether the sun or the Earth exerts the greater gravitational force.

At some altitude Earth satellite has portions of their trajectory where Earth's is the stronger (trajectory curving towards Earth) and others portions where the sun's is so it curves towards the sun. Here one could say part of orbit is about the sun and part is about the Earth. The moon however is ALWAYS TURNING TOWARDS THE SUN, thus I consider it to have such small deviations from the perfect ellipse that it must be considered to be orbiting the sun. For example, a scale drawing on an 8.5 by 11 inch paper with even a fine (0.01 inchwide) pencil lead as a PERFECT ellipse with 8 inch major axis wound not need anymodification to accurately represent the moon's trajectory - only a statement that the precise trajectory remains within the width of the pencil line and has approximately 13 complete oscillations within that pencil line's width! On my laptop screen, 0.01 inch is about the width of: | To me that is dam little diviation on an 8 inch ellipse from a perfectly elliptical orbit about the sun.

You, in contrast, seem to think that even though moon is always turning towards the sun and only accidently towards the Earth when the Earth happens to be on the sun-side of the moon, the moon is orbiting the Earth as it is bound to the Earth. This despite the fact it's trajectory deviates by only ~1 part in 400 radially from the perfect ellipse. It really comes down to the definition of "orbiting." For me an body that always curves to another is orbiting that other body, even if it is also bound to some third body. earth is also bound to this galaxy, but neither of us would say it is orbiting the galaxy, I think. Orbiting has to do with shape of the trajectory and completing the trajectory many times, for me at least.

Janus58, can if he chooses, state my metric better than me, but I think you understand it.

I of course agree the moon has less than the Earth escape velocity -i.e. is bound to the Earth. I am not sure but think it will be for long time all time. (Assuming no large third body comes near.) If we ignore the sun's gravity, then as I DERIVE from F=ma in post 140, r (v^2) is a constant. Thus, as the moon recedes from the Earth its kinetic energy, KE, goes down linearly as r goes up. Or KE ~ 1/r but this is just the way potential energy, PE, scales with r in any inverse square force field like gravity. So the ratio KE/PE is a constant independent of r for circular orbits. (That constant is -0.5, but I will not derive that now. The negative sign is, as I am sure you know, due to the fact the PE is negative - Moon is trapped in a negative energy well.) However, there is a solar gravity field and I think when the moon goes farther way from Earth there will be a time when it can escape for Earth's grasp.

When it does so it will still retain the spin about its polar axis but that spin rate will be less than the current one 360 turn in ~28 days as for quite some time moon will remain tide-locked to Earth so the spin rate will drop to the slower orbit rate and then reamain that forever (until sun goes red giant at least).

BTW do you or Janus know what approximately the spin rate of the moon will be when it is no longer bound to Earth?

It would be interesting to hear Ken Dines try to explain how it suddenly began to spin about it polar axis when it ceases to rotate about the barycenter. I will not hold my breath until he tries as I have already asked (and been ignored) essentially this in post 124, partly reproduced below:


Ken: Another way to show how silly your POV is follows:

For convenience assume the moon & Earth are in “deep space” far from any other masses but all else is unchanged. I.e. the moon turns around once in exactly one orbit period, which is 28 days. Everyone but you and Dwayne describes this as “spinning” on its polar axis with period of 28 days, but you two say: “No that is not it is not spinning, it is tide-locked orbiting the barycenter without any spin.”

Now suppose a third body rapidly passed by, making a gravitational impulse (acting on the moon’s center of mass of course and applying no net torque* to the moon) which throws the moon into a new, much more elliptic, orbit about the barycenter with a period of 50 MONTHS. Now at apogee the moon is no longer keeping the same face turned towards the Earth, but turns 360 around MANY times so we see all side of the moon. (We already can see more than half of the moon as it is not now in a purely circular orbit.)

Do you at least agree it is “spinning” in its new orbit, even though it angular momentum about it polar axis is exactly what it was before the third body approached? If you do, why do you say it is NOT now spinning about its polar axis when the rotation about that axis (rate = 360degrees / 28 days) is exactly the same before as after the third body passed? ...

I should also note that we see more than half of the moon because it is spinning about an axis that is not exactly perpendicular to the orbit plane. - The same reason the "sunmen" would see nearly all of the earth even if its polar spin (on a 23.5degrees tilted axis) were one 360 turn in 356.25 days.

To Ken Dine:

You have a strong bias that drives you to claim the moon rotates about the Earth moon barycenter without spin.

NO, I am instead saying that the moon ONLY revolves around an exterior axis once each 27.3 day orbit, and that axis the moon is revolving around is the common mass of the Earth & moon, which is called a barycenter, which is a foci located within the Earth.

AGAIN, draw a happy-face on an orange, then hold the orange in your outstretched hand and spin on your heels - to any observer watching you the orange will likewise TURN 360 since they'll see all sides of the orange once per each spin of your body, but you (the person in the center) will NOT see the orange rotate in your outstretched hand.

The orange is NOT *rotating* around its internal polar axis any more than the moon is rotating around its internal polar axis. The only necessary spin axis is the barycenter exterior to the moon!

In orbiting around that barycenter the moon merely turns around its polar axis, which is NOT the same thing as a true polar rotation. The moon lost all of its true polar rotations over 3 billion years ago!

E.g. the Earth has 365.25 true polar rotations per year, but from a sidereal perspective, the Earth *appears* to rotate 366.25 times per year. If the Earth has 366.25 sidereal rotations per year, then why do we only have 365 days per year?

Clearly, a sidereal rotation is NOT the same thing as true polar rotation, otherwise we'd have 366 days per year.


(1) Treating as an absolute property of rotation the term "clockwise" and not realizing that "clockwise" switches to "counter -clockwise" as the observer changes from one side to the plane of rotation to the other. (Associated with your second "shooting of self in the foot" with a false "proof" that you were correct.)

Please ducate yourself about the Right-hand rule:

(Add the http & www): studio4learning.tv/sub_subject.php?pl=72&id=210

(Add only the HTTP): en.wikipedia.org/wiki/Right_hand_rule#Direction_associated_with_a_rotati on


(2) Not realizing the Foucault Pendulum operating above either pole of the moon does make a 360 circle in about 28 days - you offered the fact that it would not as if it did that would indicate moon was spinning about the polar axis. (Associated with your first "shooting of self in the foot" with a false "proof" that you were correct.)

Foucault Pendulums don't spin 360 since they aren't set up on the poles (except a crude one once), so they spin less times per day at lessor latitudes, and not at all at the equator.

Set up a Foucault Pendulum at 45˚ latitude at any spot on the Earth and it will show rotation. Set up a Foucault Pendulum at 45˚ latitude at any spot on the moon and I seriously doubt that the pendulum will be affected by the moon's orbital revolution since a Foucault Pendulum wasn't designed to do that.

Unless you have proof otherwise, we can agree to disagree on that side issue.


(3) Not realizing that the sun's gravity is stronger force at the moon than the earth's gravity is. (I worked it out for you to show sun is 8.45 times stronger, but received no thanks or even an acknowledgement that you were wrong. - So I assume you still hold that false POV.)

At issue here are the effects of tidal braking, and in regards to tidal braking, the moon's gravity is OVER TWICE as strong at affecting the Earth's tides than the sun's gravity is.


(4) Not realizing that the tides are made by the gradient of gravity, not by the force of gravity. Or that your argument based on moon making stronger tides than the sun does not show the moon has stronger gravity at the moon.

Unintelligible, but see my above response.


(5) Not realizing the moon only appears to orbit the Earth for a small set of reference frames, but all others in the universe, for example for a frame based on any planet or asteroid, or the sun etc. see it is in orbit about the sun with about a 1/4 of one percent wobble. I illustrated this 1 part in 400 radial variation with 8 inch diameter circular drawn by a line only 0.01 inches wide. Something approximately this width| forming a circle filling your laptop computer screen and told you the wobble stayed inside the line's width. –I.e. was not even noticeable in that scale drawing!)

Since the Earth and moon together orbit the sun at a far greater speed than the moon orbits the Earth, you appear to be referring to this concept ("The Orbit of the Moon around the Sun is Convex!"):

(Add the http & www): math.nus.edu.sg/aslaksen/teaching/convex.html

Here's a good graph of the concept (Add only the HTTP):

wikipedia.org/wiki/Image:Moon_trajectory1.svg

That DOES NOT mean that the moon doesn't orbit the Earth too!


Hey, you do know what the barycenter is and most do not; but as they say: "A little knowledge is a dangerous thing", especially for one lacking the grace to admit their errors.
--------------
*Except Martians, of course.

I'm sure you've been told that many times, eh?

Ken

...the moon merely turns around its polar axis, which is NOT the same thing as a true polar rotation...

This is plainly wrong.

See my earlier post about shadows.

We can measure the Moons rotation in every frame of reference we choose. There is no frame in which the moon does not rotate, so, to say it doesn't is ludicrous at best.

NO, I am instead saying that the moon ONLY revolves around an exterior axis once each 27.3 day orbit, and that axis the moon is revolving around is the common mass of the Earth & moon, which is called a barycenter, which is a foci located within the Earth.
Explain lunar libration with this model. Hint: You can't. Lunar libration results precisely because the moon's rotation rate (and rotational axis) is not exactly equal to the moon's orbital rate (and orbital axis). The moon's rotational rate is nearly constant; it's orbital rate is not because the moon's orbit is slightly elliptical.

Think of it this way: An observer see would see the Sun and the stars rise and set. The Moon is rotating.


In orbiting around that barycenter the moon merely turns around its polar axis, which is NOT the same thing as a true polar rotation.
Umm, turning about a polar axis is precisely the definition of rotation.

Your problem is that you are thinking too geocentrically, as if that is the only "true" way of looking at things. The same is happening here:

E.g. the Earth has 365.25 true polar rotations per year, but from a sidereal perspective, the Earth *appears* to rotate 366.25 times per year. If the Earth has 366.25 sidereal rotations per year, then why do we only have 365 days per year?
We do have 366.24 sidereal days per year. What constitutes a "day" depends on one's frame of reference. You are assuming one and one only frame of reference is "true", and that viewpoint is simply wrong. All reference frames are equally valid, including the geocentric frame in which the Moon's motion appears to go a rather complex longitudinal and latitudinal libration and an inertial frame in which the Moon appears to undergoing nearly uniform rotation about an axis passing through its center of mass.

Ancients (and modern crackpots) viewed the Moon as wobbling side-to-side and up and down by some mystical force. Modern physicists and astronomers view the Moon as rotating about its own axis because that is by far the much simpler way to explain the Moon's motion. No mystical forces are needed.

(My graphics) are already posted one under the other, side by side would make no difference.

Your graphics #1 thru #3 are NOT posted anywhere close to your graphic #4 – it would have been easier to post one of the #1 thru #3 graphics near the #4 graphic than it is to bitch about it. What are you afraid of?

Soon I'll have full posting rights and I can do it for you.



(NOTE: Venus completes one counter-clockwise orbit every 224.65 days and Venus rotates once clockwise every 243 days.)


which has absolutely nothing to do with the principles of the discussion at hand. You do seem to have a habit of flooding a post with irrelevant data.

Besides which, the period. IOW, the time it takes for Venus to rotate once relative to the fixed stars.

YES, you are correct, Venus' 243 rotational period is as viewed from the sidereal perspective. HOWEVER, Venus' rotation has everything to do with the discussion at hand since its retrograde rotation is a prime example of how a sidereal perspective often confuses people!

E.g., google and you'll find that many astronomy websites claim Venus' day is longer than Venus' 224.65 day orbit. When IN FACT, Venus has TWO (2) complete days per each 243 day orbit – this graphic on this page clearly shows that Venus has NEARLY two full days per each orbit:

(Add the HTTP & www): geocities.com/kfuller2001/tVenus.html

If Venus' rotation were a tad 19 days faster, then Venus would have a sidereal spin rate of (REMARKABLY) 1:1!!! If Venus' rotation continues to slow down from the sun's tidal braking, then eventually Venus will reach the 0:1 (zero sidereal spin) that was on this page:

(Add the HTTP & www): geocities.com/kfuller2001/tVenus.html

CLEARLY, Venus would still have one internal polar rotation at a sidereal spin rate of 0:1! And after Venus is fully braked by the sun, then Venus will once again have another 1:1 sidereal spin - i.e., there will be TWO 1:1 sidereal spin rates as Venus winds down to ZERO actual axial rotations!

Of course, Venus may actually be speeding up and not slowing down:

---snip---
"Research suggests that {Venus'} "backward" rotation is caused by tides which are raised in the thick atmosphere by the Sun, and with friction interaction between the atmosphere and planet itself. It is hypothesized that these interactions caused Venus's rotation to slow, stop, and then reverse. This is somewhat similar to what is happening here on Earth, as our Moon's pull on our oceans causes tides whose subsequent friction is gradually slowing Earth's rotation"

(Add only the HTTP): chuckayoub.googlepages.com/venus_information__the_planets.htm
--/--

If that's the case, then Venus will speed up to a perfect 1:1 (w/ 2 solar days), then 2:1 (w/ 3 solar days.)

As that theory holds, Venus first spun down from a normal prograde rotation, stopped, and then slowly started turning in the opposite direction – and, as viewed from the sidereal perspective, that would be ("+" is prograde and "-" is retrograde):

+4:1 (w/ 3 solar days)
+3:1 (w/ 2 solar days)
+2:1 (w/ 1 solar day)
1:1 (w/ 0 solar days) – i.e., one long day on the near-side and total darkness on the far-side)
-0:1 (w/ 1 solar day)
-1:1 (w/ 2 solar days)
-2:1 (w/ 3 solar days)
-3:1 (w/ 4 solar days)

Spin it any way you like, but whether an astronomical body spins down to 1:1 from the retrograde or prograde direction, when it finally stops at 1:1, then it has LOST all of its polar axial rotations!!!

An astronomical body spinning down from the retrograde direction will also have TWO 1:1 spin rates on either side of the 0:1 sidereal spin rate - so much for the God's-eye sidereal perspective!

No one knows if our own moon spun down from a normal prograde direction or a retrograde direction, but either way, the moon stopped spinning around its polar axis over 3 billion years ago!

1:1 equals ZERO!


Why would you think that the very astronomers that you claim are wrong about the rotation of the Moon would list a rotational period for Venus based on your perception of planetary rotation?

Many websites do list Venus correctly:

---snip---
Sidereal Rotation 243 Earth days
Length of Day 116.75 Earth days
Sidereal Revolution 225 Earth days

(Add the HTTP & www):crh.noaa.gov/fsd/astro/venus.php

Many other websites list Venus' day as 243 Earth days – the sidereal perspective can be confusing if you don't understand it.



In any event, I've already covered that topic well, and an astronomical body that *appears* from the sidereal perspective to not be rotating, actually does have one polar axial spin (in a clockwise direction) per each complete orbit, as can plainly be seen in the top graphic here (don't add www):

community-2.webtv.net/kdine5/Lunacy/index.html


Once again, you are simply imposing your personal perception of "rotation of an astronomical body. What you "plainly see" in that graphic is not what everyone else plainly sees. We plainly see a moon that does not rotate in the top graphic, and one that rotates counter-clockwise once per orbit in the bottom graphic.

That's because you have poor spatial reasoning!

Whether a large astronomical body is spinning or not can be determined by its shape. No matter your reference frame, the Earth rotates and its rotation forces it into the shape of an oblate sphere.

The moon has stopped rotating and the Earth's gravity has distorted the moon's shape into a prolate spheroid (i.e., a football shape.)

Venus still slightly rotates, just enough to keep its sun-caused tidal bulges in constant motion, so of all our solar system's astronomical bodies, Venus is the closest in shape to a perfect sphere.


Once again, some pasted material that does not add anything to the discussion. The ability to paste does not infer understanding.That is a silly argument. If enough time were to pass that the Earth becomes tidally locked to the Moon such that its tidal bulges remained fixed on its surface, would you then claim that the Earth had stopped rotating?

Yes and no. If the Earth slowly ground to a halt, then when it finally stopped rotating on its polar axis, then the Earth would switch over to rotating around the Earth-moon barycenter. The Earth would then have a near-side and a far-side that faced the moon, and the Earth's new day would be identical to the moon's length of day.


The fact that the moon is oblong is simple due to the fact that because of its slow rate of rotation, the tidal forces of the Earth had a greater effect on its shape. It is also related to the fact that the moon likely had a much more eccentric orbit when it cooled into its present shape.

Billions of years ago when the Earth finally stopped the moon's rotation (at 1:1), ONLY then did the moon's tidal bulges stop and become fixed in place. HOWEVER, prior to fully stopping at 1:1, the moon was likely a perfect sphere, just as round as Venus is today (with its slow .93:1 spin rate.)


Again, it seems like you are using your own peculiar notion of rotation here. It can rotate around its own axis and revolve around a orbital axis while constantly facing the axis of revolution. In fact, it must rotate once per orbit to do so.

SIMPLE EXPERIMENT:

Spin 360 around with an orange in your out-stretched hand, and as you spin around looking at the orange (which isn't rotating around its own center of gravity), and then tell me the orange was spinning on its internal axis, or not?

Try it again, but this time, use your free hand to give the orange one 360 axial rotation as you spin your body around 360 – so, what happens if you rotate both the orange and your body at the same time? Did the orange keep one face pointed towards you?

NO!


No it isn't because you are assuming that your notion of "rotation on an axis" is true to "prove" that it is correct. It rotates around it own polar axis and revolves around the Earth-moon barycenter Oh, and by the way, the Moon's polar axis is tilted 6.5° to that of its orbit. Because of this, as seen from the Earth, the Moon's axis traces out a cone as with a period of once per orbit. This is what causes the libration of latitudesLearn to read. I never said it was your argument, I said it was the reverse of your argument, And as just as strong as your argument on their own merits.

Do the simple orange experiment and get back to me.


And in doing so you are making yourself look foolish.

That, too, depends upon one's perspective!
:)

Ken

Spin 360 around with an orange in your out-stretched hand, and as you spin around looking at the orange (which isn't rotating around its own center of gravity), and then tell me the orange was spinning on its internal axis, or not?Yes, the orange spun once on its axis for each turn of me. If it hadn't, it would have appeared to rotate as I did, but it showed no apparent rotation (this is logical since it was stationary wrt my hand), therefore it must have rotated about its own axis.

Objects rotate relative to the box we put them in.

If you paint an arrow on an orange, and sit it in the middle of a mat, the orange is not rotating WRT the mat, because the arrow doesn't change direction wrt the mat.

If I pick up that Orange, hold it at arms length, and start spinning on the spot, then the orange is spinning WRT the mat, because the orientation of the arrow painted on the orange is changing WRT the mat.

If we tie a string to the orange, and I stand there, swinging the orange around my head, after painting an arrow on it, and I happen to be standing facing the same direction as another arrow, painted on a mat, than the orange is still rotating wrt me, because sometimes the arrow on the orange is paralell to the arrow shaved into my head, and sometimes it's anti paralell.

So all that argument goes to prove is how wrong people claiming the moon doesn't spin are.

In the case of the earth-moon system, if we define a meridian, and give that meridian a direction, and paint the lines in place on the earth and the moon, than sometimes the meridians will be aligned paralell, sometimes they will be aligned anti paralell.

Therefore using ONLY THE EARTH AND MOON we can determine that the moon must be rotating as it orbits the earth.

Explain lunar libration with this model. Hint: You can't. Lunar libration results precisely because the moon's rotation rate (and rotational axis) is not exactly equal to the moon's orbital rate (and orbital axis). The moon's rotational rate is nearly constant; it's orbital rate is not because the moon's orbit is slightly elliptical. ...Excellent and irrefutable OBSERVATIONAL evidence that the moon is spinning about it polar axis, but Ken is very fixed on his POV, so I will elaborate a little on your statement.

First however, here is a link that show this libration:

http://en.wikipedia.org/wiki/Libration

The first cycle or two is slow, then the it speed up more movie like - wait for that also.

The moon is nearly a constant distance from the earth so the libration is not large, but note when the moon appears larger in the wiki illustration (because it is closer to Earth) its spin rate is not keeping up with its more rapid orbital rate. And conversely, when smallest (most distant from earth) its spin rate is greater than the orbital rate.

Just so Ken will understand your and my term "orbital rate" that is the rate of angular advance around the Earth for a person at the North Pole. (I place the observer there as that removes the parallax associated with Earth's spin carrying an equatorial observer to different position.) The AVERAGE orbital rate is ~360degrees/ (~28) days but when the moon is near apogee the current orbital rate is slower and at perigee it is faster.

The moon is spinning around its polar axis at a constant rate and in the same direction that the Earth is spinning - for an observer far out in space over the North Pole this would be a counter-clockwise spin. (Why the sun appears to rise in the East) So at apogee with moon's polar spin rate higher than the now less than average orbit rate, the leading edge of the moon in its travels around the Earth spins a little around to be more towards the Earth, not exactly the leading edge.

Likewise at perigee the moon's orbital rate is higher than the average orbital rate of advance about the Earth and the moon is not spinning fast enough to keep exactly the same face to the Earth. So what was not quite the leading edge of the moon can become the leading edge.

This is of course the same thing that I explained long ago with a more extreme example. I.e. I assumed that a third body passed by the Earth / moon system and threw the moon in to a much more eccentric orbit with a 50 month orbit period but still a 28 day spin period (no net torques applied by force acting on center of gravity of moon with a symetric trajectory right over the back-side of the moon's bulge) and noted that then at the distant apogee the moon would spin completely around (as seen from Earth) perhaps a dozen times as its spin rate would still be one 360 turn in 28 days but it would be far from the Earth for about a dozen months with little orbital advance while making these dozen full turns.

See my earlier post 124 giving more details at.

http://www.sciforums.com/showpost.php?p=2012966&postcount=124

Here is some of the text from above wiki link:

"...Libration in longitude is a consequence of the Moon's orbit around Earth being somewhat eccentric, so that the Moon's rotation sometimes leads and sometimes lags its orbital position.
Libration in latitude is a consequence of the Moon's axis of rotation being slightly inclined to the normal to the plane of its orbit around Earth. Its origin is analogous to the way in which the seasons arise from Earth's revolution about the Sun. ..."

Ken can not explain any of this* with his erroneous POV.
--------------
*Not my imagined many full turns around with little orbital advance (an extreme libration case) nor the actual case of very modest libration that the moon currently has.

...the moon merely turns around its polar axis, which is NOT the same thing as a true polar rotation...
”

This is plainly wrong.

See my earlier post about shadows.

We can measure the Moons rotation in every frame of reference we choose. There is no frame in which the moon does not rotate, so, to say it doesn't is ludicrous at best.

Trippy, what is at issue here is NOT whether or not the moon is spinning around an axis (it does), ONLY the LOCATION of that axis!!!!

Some people here claim that the moon both rotates around its polar axis one time per each 27.3 day orbit (i.e., 2 spin axes), which I admit can be construed as consistent with the conventional definition of *Synchronous rotation*:

---Wikipedia---

~Synchronous rotation

In astronomy, synchronous rotation is a planetological term describing a body orbiting another, where the orbiting body takes as long to rotate on its axis as it does to make one orbit; and therefore always keeps the same hemisphere pointed at the body it is orbiting.
--/--

I say that our moon ONLY revolves around the Earth-moon common mass barycenter, a point located within the Earth, and that Earth-moon barycenter is the moon's ONLY remaining (relevant) axis of spin!

The moon and Earth are both revolving around the sun too, but the Earth-sun-moon barycenter is not relevant to this discussion of, whether or not, the moon is also rotating 360˚ around its internal polar axis as it orbits 360˚ around the Earth.

The classical definition that it takes the moon just as long to orbit the Earth (and barycenter) as it takes the moon to rotate around that same axis, is double-speak ... UNLESS, the 2-axes lunatic spin crowd are claiming that the moon spins 360˚ on two separate axes at the same time? I.e., does the moon spin 360˚ around the Earth at the same time the moon rotates 360˚ around its own polar axis?

OR, does the moon ONLY orbit (rotate, revolve, or spin) 360˚ around one (1) axis, which is located within the Earth?

I say the latter!

WHEREAS, the Earth still does rotate around its polar axis as it orbits the sun – so, the Earth still has two spin axes, (1) its own polar axis; and, (2) the sun.


In my POST 94 I gave a simple experiment you can easily do in your own home, which not only proves my one (1) spin axis claim, this demonstration will also show the moon's phases (shadows) that you seemed concerned about.

Take a few minutes of your time and try this:

---snip from my Post 94---
A simple demonstration - draw a happy-face on an orange, then go into a darkened room and take the shade off of one table lamp (that will be the sun), then hold the orange on your outstretched palm with the happy-face facing you, then spin 360 around counter-clockwise on your heels.

If you do that, then you'll see all phases of the moon likewise pass across that orange.

The orange will NEVER spin in your hand, yet, its happy-face will always remain pointed towards your spinning body!

You can even see more than 50% of the orange's surface (libration) by raising the orange/moon model above and below your eye level. Viewing the orange/moon first with one eye and then the other eye will also cause some libration (caused by parallax.)

When you show a person that demonstration to prove that the moon doesn't spin on its polar axis (only around your body's axis), they will either grok it, or they'll dig their heels in and claim your arm isn't the same thing as gravity.

Of course an arm isn't gravity, but in an accurate model your arm can serve the same purpose.

Men have been making sun-Earth-moon models (called Orreries) for hundreds of years now, and no orrery ever constructed has any gearing to rotate the moon-model on its internal axis! BECAUSE, rotating the moon-model on its spindle just isn't necessary to do!!!

E.g., here's a good example of one (add the http & www to it):

.abhirjoshi.com/models/sme_model_abhir.jpg
--/--

Ken

Ken. I need a good laugh. Please explain libration*, either the observed one the moon has or the one I imagined with the moon thrown into a much more exccentric orbit with 50 month orbit period.

using only your uniform spins about the barycenter axis.
----------------
*see post 162 or 163

”


Trippy, what is at issue here is NOT whether or not the moon is spinning around an axis (it does), ONLY the LOCATION of that axis!!!!

Some people here claim that the moon both rotates around its polar axis one time per each 27.3 day orbit (i.e., 2 spin axes), which I admit can be construed as consistent with the conventional definition of *Synchronous rotation*:

---Wikipedia---

~Synchronous rotation

In astronomy, synchronous rotation is a planetological term describing a body orbiting another, where the orbiting body takes as long to rotate on its axis as it does to make one orbit; and therefore always keeps the same hemisphere pointed at the body it is orbiting.
--/--

I say that our moon ONLY revolves around the Earth-moon common mass barycenter, a point located within the Earth, and that Earth-moon barycenter is the moon's ONLY remaining (relevant) axis of spin!

The moon and Earth are both revolving around the sun too, but the Earth-sun-moon barycenter is not relevant to this discussion of, whether or not, the moon is also rotating 360˚ around its internal polar axis as it orbits 360˚ around the Earth.

The classical definition that it takes the moon just as long to orbit the Earth (and barycenter) as it takes the moon to rotate around that same axis, is double-speak ... UNLESS, the 2-axes lunatic spin crowd are claiming that the moon spins 360˚ on two separate axes at the same time? I.e., does the moon spin 360˚ around the Earth at the same time the moon rotates 360˚ around its own polar axis?

OR, does the moon ONLY orbit (rotate, revolve, or spin) 360˚ around one (1) axis, which is located within the Earth?

I say the latter!

WHEREAS, the Earth still does rotate around its polar axis as it orbits the sun – so, the Earth still has two spin axes, (1) its own polar axis; and, (2) the sun.


In my POST 94 I gave a simple experiment you can easily do in your own home, which not only proves my one (1) spin axis claim, this demonstration will also show the moon's phases (shadows) that you seemed concerned about.

Take a few minutes of your time and try this:

---snip from my Post 94---
A simple demonstration - draw a happy-face on an orange, then go into a darkened room and take the shade off of one table lamp (that will be the sun), then hold the orange on your outstretched palm with the happy-face facing you, then spin 360 around counter-clockwise on your heels.

If you do that, then you'll see all phases of the moon likewise pass across that orange.

The orange will NEVER spin in your hand, yet, its happy-face will always remain pointed towards your spinning body!

You can even see more than 50% of the orange's surface (libration) by raising the orange/moon model above and below your eye level. Viewing the orange/moon first with one eye and then the other eye will also cause some libration (caused by parallax.)

When you show a person that demonstration to prove that the moon doesn't spin on its polar axis (only around your body's axis), they will either grok it, or they'll dig their heels in and claim your arm isn't the same thing as gravity.

Of course an arm isn't gravity, but in an accurate model your arm can serve the same purpose.

Men have been making sun-Earth-moon models (called Orreries) for hundreds of years now, and no orrery ever constructed has any gearing to rotate the moon-model on its internal axis! BECAUSE, rotating the moon-model on its spindle just isn't necessary to do!!!

E.g., here's a good example of one (add the http & www to it):

.abhirjoshi.com/models/sme_model_abhir.jpg
--/--

Ken

You're wrong, and i've detailed why in my posts (for example, the fact that the orange doesn;t rotate WRT your hand is completely irrelevant - all that means is that it's moving and rotating with your hand).

...---Wikipedia---

~Synchronous rotation

In astronomy, synchronous rotation is a planetological term describing a body orbiting another, where the orbiting body takes as long to rotate on its axis as it does to make one orbit; and therefore always keeps the same hemisphere pointed at the body it is orbiting. ...but that keeping the SAME hemisphere turned is only possible IFF the spin about the polar axis is exactly perpendicular to the orbit plane.* This is not true of the moon. Moon's is spinning about an axis about 7 degrees tilted from the orbit plane, as I recall. So even if it were in a perfectly circular orbit (which it is not) more than one hemisphere would be visible from Earth.

If the moon's polar spin axis were 90 degrees tilted then all of themoon's surface would be visible from Earth - that was what Janu58 was trying to show you in his drawings. Surely even you would admit the moon spining at the 28 day per orbit rate was spinning about a polar axis if that axis were 90 degrees tilted. (Spin axis in the plane of the orbit.) What about if 80 degrees tilted - is it only spinning about the barycenter yet? 70 tilted, etc ...10degrees tilted? 9, 8 or down to the current 7?

As Janus tried to get thru your thick skull: At what tilt (greater than 7 degrees) did the moon stop spinning about its polar axis and begin to spin only about the barycenter?

It is a simple question with a simple numerical answer, but you will still ignore it, I bet.
------------------
*Other things also required. E.g tidal locked and exactly circular orbit to avoid longitudinal librations.

”
The orange will NEVER spin in your hand, yet, its happy-face will always remain pointed towards your spinning body!
Ken
Ken, you are being an idiot.

How do we determine if something is rotating about its axis? Try this:

As described, stand with an orange in your outstretched hand, with your head representing the earth. Now, gently drive a spike through your head into the ground and affix an arrow to the top of the spike (like a weather vane). The arrow points at a fixed mark on a distant wall. As you rotate, you keep the arrow fixed on this relatively non-co-rotating reference. Its free to translate in any direction, but not rotate.

Now, drive a similar spike through the orange and your hand. It too is free to translate but not rotate. Keep the arrow on this spike also pointed at the distant reference mark on the wall. Now, as you spin (being careful not to slip on the blood) you will indeed see that the orange rotates about it's own axis.

Yes, the orange spun once on its axis for each turn of me. If it hadn't, it would have appeared to rotate as I did, but it showed no apparent rotation (this is logical since it was stationary wrt my hand), therefore it must have rotated about its own axis.

You say, "therefore it must have rotated about its own axis."

Geeze ... how many times do I need to ask you guys to SPECIFY the EXACT location of your spin axes that you're talking about!!

You do understand the difference between *REVOLUTION* and *ROTATION*, right?

CLUE: "The Earth *REVOLVES* around the sun as it *ROTATES* around its polar axis."

Words do have meanings ... unless, you're an astronomer heavy in math and light on English!

You confusedly state that, "...the orange spun once on its axis for each turn of me. If it hadn't, it would have appeared to rotate as I did..."

So then, was your orange's *spin axis* the common-mass of the Vkothii-orange barycenter (i.e., you and the orange's common mass?)

Do you understand what a barycenter is? If not:

BARYCENTER: "Of or relating to the center of gravity."

THUS, when considering the combined mass of two (or more) objects, their combined mass will lie on a point between their respective centers.

OK then, Vkothii (YOU) and the orange had a common center of gravity, which was likely very close to your own body's center (perhaps, a hair's thickness off?)

In any event, I claim that the orange REVOLVED around the Vkothii-orange common-mass barycenter, AND FURTHER, that your orange NEVER *ROTATED* around its own center of mass in the process of that demonstration. YES, the orange revolved around your combined center of masses (which was in your body), but the orange did NOT spin or rotate around any axis internal to the orange itself.

---Wikipedia---
"A rotation is a movement of an object in a circular motion.

"A two-dimensional object rotates around a center (or point) of rotation. A three-dimensional object rotates around a line called an axis.

"If the axis of rotation is within the body, the body is said to rotate upon itself, or spin—which implies relative speed and perhaps free-movement with angular momentum.

"A circular motion about an external point, e.g. the Earth about the Sun, is called an orbit or more properly an orbital revolution."
--/--

Armed with those definitions, since you were brave enough to actually get off your tush and spin around with a dumb orange in your hand (I hope no one saw you), can you please specify EXACTLY where you believe the orange's spin axis was located at?

AGAIN, was the spin axis within you, or within the orange?

Location, Location, Location!!!!!

Ken

Ken,
when are you going to answer the libration question?

Ken, you are being an idiot.

How do we determine if something is rotating about its axis? Try this:

As described, stand with an orange in your outstretched hand, with your head representing the earth. Now, gently drive a spike through your head into the ground and affix an arrow to the top of the spike (like a weather vane). The arrow points at a fixed mark on a distant wall. As you rotate, you keep the arrow fixed on this relatively non-co-rotating reference. Its (sic)free to translate in any direction, but not rotate.

Now, drive a similar spike through the orange and your hand. It too is free to translate but not rotate. Keep the arrow on this spike also pointed at the distant reference mark on the wall. Now, as you spin (being careful not to slip on the blood) you will indeed see that the orange rotates about It's (sic) own axis.

It's interesting to be called an "idiot" by someone that doesn't know the correct forms of "IT!" Don't they teach that in jr. High?

*it's* is the contraction of the two words, *it is*.

For example, when you wrote, "you will indeed see that the orange rotates about It's own axis."

You actually wrote:

"...you will indeed see that the orange rotates about it is own axis.

I'm not a grammar Nazi and I don't normally care about typos and grammar as long as ideas are clearly expressed (I make errors too), but calling someone an "idiot" while expressing an incomprehensible and a bloody example, means you're fair game.

Not to mention, this grammar post gets me closer to 20 and I'll soon lose my newbie wings and be able to post URLs and images! So, thanks for that!
:-)

At least, now you can say an *idiot* once taught you something!

Seriously, if you can politely express a clear non-bloody statement or question (even with typos or other errors), then I'll honestly try and respond to your posting.

Ken
(aka the idiot)

If "The Earth *REVOLVES* around the sun as it *ROTATES* around its polar axis."
It also spins around its own axis (a polar axis) as it rotates (orbits) the sun. Both are called angular momentum.

Words do have meanings ... unless, you're an astronomer heavy in math and light on English!

You confusedly state that, "...the orange spun once on its axis for each turn of me. If it hadn't, it would have appeared to rotate as I did..."

So then, was your orange's *spin axis* the common-mass of the Vkothii-orange barycenter (i.e., you and the orange's common mass?)No, it was the orange's axis, the one parallel to my vertical axis of rotation.

Ken, you are being an idiot.


I FORGOT - since *it's* is the contraction of the two words *it is*, *ITS* (without the apostrophe) is the possessive form of *IT*

Likewise, there are other possessive words that don't use apostrophes, such as *HIS* *HERS* and *YOURS* since the words themselves are possessive.

*her's* or *your's* may not look too funny at first blush, but *Hi's* is definitely funny looking!

Another post closer to 20!

Ken
(aka the idiot)

If I had a perfect bearing tied to a rope with something inside it, and started to spin the rope around, the object would not spin just because it is "orbiting" me.

Ken Dine --- You do understand the difference between *REVOLUTION* and *ROTATION*, right?

====Evidently not.

Ken,
when are you going to answer the libration question?

If you do, then you'll (short for you will) be only one post away from twenty.

Here's one more puzzle for you, Ken, my man.

A solid cylindrical bar of metal 0.5m in length, weighing 400gm, and with a diameter of 2cm is tack-welded to a small sleeve bearing, and the assembly is press-fitted at the bearing end to a stub-axle, which is attached to a plate bolted to an upright frame, so the bar swings freely from the bearing through a vertical arc:

i) estimate the period of the swinging bar, for small displacements.

ii) write a formula for the relationship between the angular momentum of a cross-section of the bar near the free-swinging end, and a cross-section near the bearing end.

iii) if the plate with the attached cylindrical bar on its bearing, is bolted to a horizontal surface, what happens to the relationship in the previous question?

Well ken, if you're capable of just a bit more than correcting silly late-night grammar mistakes, can you address the essence of my post? I.e. that the moon or any orbiting body rotates about its(!) own axis whether you like it or not?

Remember, idiocy can be a temporary condition. You've rejected several mathematical and intuitive explanations that show clearly how the moon is rotating about it's(!) axis, yet you still blindly insist it is not. Why? Do you enjoy coming across as an idiot?

If I had a perfect bearing tied to a rope with something inside it, and started to spin the rope around, the object would not spin just because it is "orbiting" me.

Spin with respect to what? Are you going to argue that the earth is not rotating about it's own axis? Clearly it's not, right? It's the cosmos that's rotating about us, yes? Unfortunately for you though, theres a distinct centripetal acceleration at the equator that demonstrates axial rotation (same for the moon).

Your object-in-a bearing is not spinning with respect to the bearing because they share the exact same rotation! Reference the axis of the bearing and that of your body to a common fixed point and see that they are both rotating about their axes!

It's called choosing a common reference. You can't just decide to choose one reference for one part of a system and a diffrent one for another part and then argue based on this. Well, you can, if you want to look like an idiot.

The pot calls the toilet black.

The pot calls the toilet black.
Huh? :bugeye:

I think he means you're full of shit.

Not my opinion. I don't have one on this issue. Just trying to help out with translation difficulties and inject a minor amount of humour. Have you upset him previously?

I think he means you're full of shit.

Not my opinion. I don't have one on this issue. Just trying to help out with translation difficulties and inject a minor amount of humour. Have you upset him previously?
Nope. Don't even know the dude/chick (dude + chick = dick???).

Anyway...

To an outside observer the object in the bearing does not rotate.

That question is the sort of thing a high-school student should know how to answer.
If a first-year student had no idea, he'd probably fail the rest of the exam.

What happens if instead of a solid metal bar, it's made of something bendy like a soft plastic say (with a significant spring constant)?
How would the relationship described in part ii change (starting with the upright pendulum in SHM case)?

Ken:

Enough of this with orange in your hand nonsense, which all distant observers see as spinning about its polar axis. Just explain the OBSERVED librations of the moon (or the multiple 360 rotations about its axis it would have at apogee if torque-free scattered into the 50 month orbit by a passing third body.)

Or answer Janus58’s question asked long ago but now twice repeated without drawings for you to attack in your posts.

I.e. You agree that a planet’s moon could be spinning about its polar axis if that spin axis is in the plane of its orbit. (There is no way that can be attributed to spin about the barycenter.) If the polar spin axis is tilted up 1 degree from the orbit plane, or 2 degrees up etc. it is still spinning about the Polar axis, yet when 83 degrees tilted up, as the Earth’s moon is, according to you, it is not spinning about its polar axis. “It is only rotating about the barycenter.” At what number of degrees up tilt did it stop to spin about it polar axis?

The reason you continue to ignore these two different questions, both repeatedly asked, is that they are facts of nature INCOMPATABLE with your false POV.

If I had a perfect bearing tied to a rope with something inside it, and started to spin the rope around, the object would not spin just because it is "orbiting" me.

Moving at a steady rate it will not rotate out there. However, if you accelerate it, then it will spin until slowed by friction. Once friction has stopped its rotation, then some sort of force would need to be applied to re-spin it.

E.g., take a bicycle wheel and grasp its axle with both hands, and hold it axle up (wheel parallel to the ground), and start spinning your body counter-clockwise, and the acceleration will cause the wheel to spin clockwise (as viewed using the right-hand rule.)

If you keep spinning your body, then friction will soon stop the wheel from spinning, and when you decelerate your body's spinning, the wheel will then reverse direction and spin the other way, until again slowed by friction.

While you're still spinning your body 360˚, and after the wheel stops spinning 360˚ on its axial (from friction), you of course will then be viewing a non-spinning wheel since some force would be required to start it spinning again.

HOWEVER, an observer watching you (a *sidereal observer *) will still see the wheel *appear* to spin 360˚ once each time you spin your body around 360˚, even though for you in the center position, the wheel has stopped spinning.

Of course, the wheel (like the moon) stopped spinning on its axis due to friction, so the wheel's only remaining 360˚ movement the *sidereal observer * is actually viewing would be the wheel now revolving around your body's center of mass and NOT any circular 360˚ rotation around the wheel's own axial.

Billions of years ago when it formed our moon didn't start spinning on its polar axis due to orbital acceleration as the bicycle wheel would spin when accelerated in a circle, but a similar force may have started our moon spinning, much as an ice-skater accelerates her spin around her body's mass when she pulls her arms closer into her body:

"The Earth spins on its axis because of conservation of angular momentum. The classic example of this is a figure skater. When a figure skater pulls in her arms, she spins faster. The Earth formed when gas left over from making the Sun condensed into the planets. As this gas cooled and condensed, it started to spin faster. Now that it is spinning (and not condensing any more), it will keep spinning at a steady rate unless something stops it."

(add HTTP): imagine.gsfc.nasa.gov/docs/ask_astro/answers/961107a4.html

If the moon didn't form when spinning dust aggregated under the force of gravity, then an Earth collision with a large astronomical body may have started the moon spinning – no one knows for sure how it started rotating, just that the moon likely did at one time rotate around its polar axis.

Regardless of how our moon started rotating around its polar axis, billions of years ago tidal braking (friction) quickly stopped the moon's polar rotation, just as friction stopped the bicycle wheel, so today our moon ONLY orbits (or revolves) around the Earth.

Just like the bike wheel in that demonstration, our moon succumbed to friction and stopped rotating on its polar axis. Today, the moon only orbits. Case closed!

Of course, we have other viewpoints here from the spatially challenged:


I am Superluminal of the Alien clan of Usa, Nordamerica, a Terran, of Sol, and I say:

It's called choosing a common reference. You can't just decide to choose one reference for one part of a system and a diffrent one for another part and then argue based on this. Well, you can, if you want to look like an idiot.


INDEED!

Up until now Ken, you have appeared stubborn. This is not a bad quality. (In some reference frames it is called confident and strong willed.) However, you have now blatantly failed to answer this:
Just explain the OBSERVED librations of the moon (or the multiple 360 rotations about its axis it would have at apogee if torque-free scattered into the 50 month orbit by a passing third body.)

Or answer Janus58’s question asked long ago but now twice repeated without drawings for you to attack in your posts.

Now it is clear that you are evasive. (In some reference frames that is called dishonest.) Now prove me wrong by answering the questions.

Someone forgot to add a couple of things:

Just like the bike wheel in that demonstration, our moon succumbed to friction and stopped rotating on its polar axis [with respect to the earth's surface]. Today, the moon only [rotates once for each one of its] orbits. [But we see more of the moon's surface than exactly one half of a sphere, because of other influences that haven't been mentioned]

---snip from Ken's Post 94---
A simple demonstration - draw a happy-face on an orange, then go into a darkened room and take the shade off of one table lamp (that will be the sun), then hold the orange on your outstretched palm with the happy-face facing you, then spin 360 around counter-clockwise on your heels.

If you do that, then you'll see all phases of the moon likewise pass across that orange.

The orange will NEVER spin in your hand, yet, its happy-face will always remain pointed towards your spinning body!

You can even see more than 50% of the orange's surface (libration) by raising the orange/moon model above and below your eye level. Viewing the orange/moon first with one eye and then the other eye will also cause some libration (caused by parallax.)

When you show a person that demonstration to prove that the moon doesn't spin on its polar axis (only around your body's axis), they will either grok it, or they'll dig their heels in and claim your arm isn't the same thing as gravity.

Of course an arm isn't gravity, but in an accurate model your arm can serve the same purpose.
---/---


You're wrong, and i've detailed why in my posts (for example, the fact that the orange doesn;t rotate WRT your hand is completely irrelevant - all that means is that it's moving and rotating with your hand).


You say that, "all that means is that {the orange} is moving and rotating with your hand?"

OK, but if your hand and orange were both rotating around the orange's center-axis (instead of ONLY around your body's axis as I claim), then wouldn't that hurt when your hand snapped off at the wrist and started rotating along with the orange around the orange's axis?

OR, are you claiming a different spin axis than the orange's center axis?

Trippy, there are only two possible locations for a spin axis in this demonstration, either the orange spins around its own center-mass, or around your body's mass.

If you want to spin the orange around two axes at the same time, then try it again, but this time use your free hand to keep the orange pointed at the same wall as you spin your body 360˚ counter-clockwise. What happened?

From your center-point perspective, did the orange rotate one time clockwise, even though the orange didn't rotate from the sidereal perspective since it remained facing the wall? That's the 0:1 spin ratio. Spin the orange 1.93 times clockwise as your body rotates counter-clockwise, and that's Venus' spin ratio, a .93:1.

Spin the orange clockwise a tad faster to rotate it a full two times per each 360˚ counter-clockwise spin of your body, and you'll have another 1:1 spin rate, the same 1:1 rate you'd have if you stopped rotating the orange and just spun your body 360˚ counter-clockwise.

Isn't that interesting, that you can have two 1:1 spin rates on both sides of a 0:1 spin rate when astronomical bodies spin down from the clockwise direction? That's basically what Venus is now doing, so it's more than a hypothetical spin-down!

These demonstrations work better if you're honest about the results, and if you also let the results make you think about these relative motions, which are difficult to fully grok until you actually try doing these simple moon models.

Ken

either the orange spins around its own center-mass, or around your body's mass.
Depends if it's free to move around another axis. If I'm standing with an outstretched hand with nothing on it, it's stationary wrt me. If I start revolving (spinning, rotating) because I'm on a platform that turns, my hand turns too.
If I move my wrist from side to side, or up and down my hand turns some more - there's a name for that I think you should know (hand waving). Although it's a conscious movement it does model what an elastic bar would do in SHM, and it models what the earth-moon system does: librate

You're getting a sore head by now, surely?

Moving at a steady...

A perfect bearing does not have friction.

You say that, "all that means is that {the orange} is moving and rotating with your hand?"

OK, but if your hand and orange were both rotating around the orange's center-axis (instead of ONLY around your body's axis as I claim), then wouldn't that hurt when your hand snapped off at the wrist and started rotating along with the orange around the orange's axis?

OR, are you claiming a different spin axis than the orange's center axis?

Trippy, there are only two possible locations for a spin axis in this demonstration, either the orange spins around its own center-mass, or around your body's mass.

If you want to spin the orange around two axes at the same time, then try it again, but this time use your free hand to keep the orange pointed at the same wall as you spin your body 360˚ counter-clockwise. What happened?

From your center-point perspective, did the orange rotate one time clockwise, even though the orange didn't rotate from the sidereal perspective since it remained facing the wall? That's the 0:1 spin ratio. Spin the orange 1.93 times clockwise as your body rotates counter-clockwise, and that's Venus' spin ratio, a .93:1.

Spin the orange clockwise a tad faster to rotate it a full two times per each 360˚ counter-clockwise spin of your body, and you'll have another 1:1 spin rate, the same 1:1 rate you'd have if you stopped rotating the orange and just spun your body 360˚ counter-clockwise.

Isn't that interesting, that you can have two 1:1 spin rates on both sides of a 0:1 spin rate when astronomical bodies spin down from the clockwise direction? That's basically what Venus is now doing, so it's more than a hypothetical spin-down!

These demonstrations work better if you're honest about the results, and if you also let the results make you think about these relative motions, which are difficult to fully grok until you actually try doing these simple moon models.

Ken

You do understand what the following sentence means don't you?

Moving and rotating at the same time.

A perfect bearing does not have friction.

A perfect frictionless bearing would still need some force to either start or stop it.
E.g., the Earth & moon have the braking friction caused by gravity.

Even though astronomical bodies have potential frictionless axes, all are subjected to tidal locking (friction) to some degree. Jupiter, is so far from the sun, and so massive with relatively tiny moons, that not much tidal-braking force has been applied to Jupiter over the eons, so Jupiter still retains most of its original rotational speed.

If the bicycle wheel had a truly frictionless bearing, then air resistance would eventually stop the wheel from spinning.

If you had a non-spinning wheel with a frictionless bearing in a vacuum chamber spinning at the end of a boom, then it would still likely need some exterior force applied to the wheel to start it spinning.

Ken

My example was going to move on and ask more questions that would hopefully make you understand, but I'll not bother.

My example was going to move on and ask more questions that would hopefully make you understand, but I'll not bother.Good. Everyone should stop providing new analogies for Ken to discuss and thus allowing him to post but avoid two OBSERVATIONS, which are irrefutable proofs that his POV is silly. Namely KEN CAN NOT:

(1) Explain the OBSERVED libration of the moon.

Or

(2) Tell the tilt up angle at which a moon initially spinning about its polar axis in its orbit plane switches to spin ONLY about the barycenter as the polar spin axis is tilted up.*
-------------------------
*The moon's polar spin axis is tilted up about 83 degrees. For Ken's POV not to be nonsense, conflicting with observations, requires the polar axis of the moon be exactly perpendicular (90 degrees tilted up from the orbit plane.) There is also the latitudinal libration observed and it is caused by the not 90 tilt up of the moon, but is also small. Little wonder Ken has avoided answering either question despite many of us asking several times.

I also asked (1) differently in several ways, including imagining a third body rapidly passing near moon scatters the moon, without applying any torque to it, into a much more elliptical orbit with 50 month period. Then near apogee, when it orbital advance is much slower, it unchanged 360 degree in 28 days turning makes it show all sides to the Earth at least a dozen times (I.e. clearly is spinning about its polar axis in an “extreme libration”) – Moon is in an almost circular orbit so the current libration as it spins around the polar axis only shows about 59% of the surface to earth. To see more by larger libration, the eccentricity of the orbit must be larger. The size of the longitudinal libration increases with eccentricity, but LIBRATION IS CAUSED BY SPINNING ABOUT THE POLAR AXIS with a constant spin rate, which cannot be same as the changing orbital angle advance rate as moon goes around the elipse of its orbit.)

Ken is just a stubborn fool, and a dishonest one too, as long as he ignores these REAL OBSERVATIONS and speaks only of oranges in his hand, etc.

Each planet & each moon (that we know of) revolves around another body. Each 1 burmolnes except Earth's moon. Why doesn't Earth's moon burmolne???

burmolnes??

Well, the earth and the sun create the same gravitional cycle which results in the completeion of a cycle every 638 miles every 15 minutes.
These two motions result in a hap hazard motion of the moon, referred to as liberation, where at times as much as 59% of the moon surface can be seen of a assumed 30 years.
the build up or decrease of the cycyles causes for various motions of the moon.


DwayneD.L.Rabon

Each planet & each moon (that we know of) revolves around another body. Each 1 burmolnes except Earth's moon. Why doesn't Earth's moon burmolne???This is either very subtle and clever, or very dumb and stupid. Waiting to find out which is quite exciting.:)

Well, the earth and the sun create the same gravitional cycle which results in the completeion of a cycle every 638 miles every 15 minutes.
These two motions result in a hap hazard motion of the moon, referred to as liberation, where at times as much as 59% of the moon surface can be seen of a assumed 30 years.
the build up or decrease of the cycyles causes for various motions of the moon. DwayneD.L.RabonDwayne, I do not think your hearing well any more. The lunch bell at your institution has already rung. Better get there before they stop serving.

Run along now, and take Ken with you.

Run along now, and take Ken with you.
Billy, you forgot to tell Dwayne to take "common sense seeker" in tow as well.

Billy, you forgot to tell Dwayne to take "common sense seeker" in tow as well.No, just cautious. He lives and hangs out in another ward, called: http://www.sciforums.com/showpost.php?p=2019716&postcount=35 . I was afraid that Dwayne and Ken would get lost in space if trying to orbit by there first. It is the old 3-body problem. I did not know what to expect except chaos.

Ken. My dear Ken.

Try this little thought experiment.

Imagine you are sitting on the side of the moon facing away from the earth. The stars are bright and steady, yet over a 28 day period they move across your sky. You decide to take a long exposure photograph of the stars and you find that they appear to rotate about a common lunar axis. This is of course, self evident.

From this, you conclude which of the following:

A) The cosmos is rotating about an axial of the moon
B) The moon is rotating about it's own axis.

Or something else?

Your orange-in-hand analogy is completely flawed because your hand is a co-rotating reference with the earth and the orange, err, moon.

The moon clearly does not rotate about an axial reference that is itself locked to the earths axis (your arm and palm in your analogy). However it does exhibit libration, which can only occurr if the rotation of the moon about it's own axis occasionally lags and leads this rotation as it orbits the earth. Otherwise, how do you explain the periodic axial oscillations of something as massive as the moon? By what mechanism does the enormous mass of the moon stop and reverse and then stop and reverse again???

So what are we missing ken? You need to do a better job of explainig why we are all wrong.

Eagerly awaiting some better explanatory discourse from you. Thx.

Well Billy T.
I can not possibly understand what you have said about the lunch bell insitution, i assume that you think that such comments get you some kind of brownie, or other acknowledgement in argument.

Appearantly from the jest of things at conversation, you are determined to uphold your position. Thats fine by me.

The subject of topic was the spin or otherwise no spin of the moon. The moon its motion and other features have been a debate across the world, and new information sparks even newer debates. at one time the moon was thought to be only 116,000 miles away and so on.

For us to consider in argument the spin or not spin of the moon in any account various forces would have to be looked at. So can you define the primary sources to the minium forces in their order of effect, a example would be the sun first, the moon second, the earth third, and any or none of the secondardy forces that effect the motion of the moon.
Because without those assignments really defining the spin motion,liberation, nodulation, apogee become difficult.

Maybe i should look back at what you have written, but i thought you said that the moon rotates around the sun as well as revolves around the sun.

If the moon has a spin axis, it is a spin axis that is in constant motion and in moves in a circle, located in the south east of corner of the moon on the back side of the moon, haveing a near center at the crater Lemaitre
or the crater Minkowski, Give or take some range in direct area.
Such a spin axis is in constant motion around this center point and has covered a wide area in the histroy of the moon, it appears that this axis would have to at this time in lunar history have a wide arc of motion simular to the circle trace out around the celestial pole every 25,800 years.
I think that makes it clear that i will look at the evidence that supports or is relative to your claim of the spin of the moon.


DwayneD.L.Rabon

...i will look at the evidence that supports or is relative to your claim of the spin of the moon.

DwayneD.L.Rabon
Ok then.

Ken. My dear Ken.


Try this little thought experiment.

Imagine you are sitting on the side of the moon facing away from the earth. The stars are bright and steady, yet over a 28 day period they move across your sky. You decide to take a long exposure photograph of the stars and you find that they appear to rotate about a common lunar axis. This is of course, self evident.


From this, you conclude which of the following:

A) The cosmos is rotating about an axial of the moon
B) The moon is rotating about it's own axis.
also
Or something else?

C) – The moon is ONLY *revolving* around the combined mass of the Earth and the moon, which is called a *barycenter*. The Earth-moon barycenter is a foci (or axis) which is within the Earth – in this graphic the red + marks the foci of common mass of the Earth-moon barycenter:

http://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Orbit3.gif/160px-Orbit3.gif

The larger white body (representing the Earth) is also rotating around its polar axis as the moon orbits around the Earth. The smaller white body (the moon) is NOT rotating around its polar axis as it orbits the Earth – if the moon *still* both rotated as it orbited, then we would see more than one face of the moon.

{NOTE – until it was tidally-braked, billions of years ago, our moon did both orbit (revolve) as it rotated around its polar axis.}

If you still don't understand the differences between *REVOLVE* and *ROTATE* (two different types of circular motions), this next graphic shows a sphere *ROTATING* around its internal polar axis:

http://upload.wikimedia.org/wikipedia/commons/0/02/Rotating_Sphere.gif

---Rotaion & Axis---
"A rotation is a movement of an object in a circular motion. A two-dimensional object rotates around a center (or point) of rotation. A three-dimensional object rotates around a line called an axis. If the axis of rotation is within the body, the body is said to rotate upon itself, or spin—which implies relative speed and perhaps free-movement with angular momentum. A circular motion about an external point, e.g. the Earth about the Sun, is called an orbit or more properly an orbital revolution." (Emphasis added.)

http://en.wikipedia.org/wiki/Axis_of_rotation
--/--


Your orange-in-hand analogy is completely flawed because your hand is a co-rotating reference with the earth and the orange, err, moon.

NO, the orange (and your hand) are instead *REVOLVING* 360˚ around the combined mass of your body and orange, and that axis of common-mass would be a foci within your spinning body. The moon's circular motion is exactly the same as our moon's circular motion around the Earth.


The moon clearly does not rotate about an axial reference that is itself locked to the earths axis (your arm and palm in your analogy).

The moon does not *ROTATE* around any axis, but the moon does *REVOLVE* around a point exterior to its mass, and the center-point of that orbit is the barycenter within the Earth.

Sloppy verbiage is a sign of sloppy thinking – "The Earth *ROTATES* on its polar axis as it *REVOLVES* around the sun."

If you insist on using "rotate" (which is a circular movement around an INTERNAL AXIS) when the word "revolve" (around an EXTERNAL AXIS) would be a more suitable choice of words, then please do me the courtesy of identifying the EXACT location of the rotation's axis that you're then talking about.

For example, your "B" choice above could mean anything:

superluminal: "B) The moon is rotating about it's (sic) own axis."

If you mean the moon's internal axis, then NO!

If you instead mean the axis of the moon's circular motion is the barycenter within the Earth, then YES!


However it does exhibit libration, which can only occurr if the rotation of the moon about it's own axis occasionally lags and leads this rotation as it orbits the earth. Otherwise, how do you explain the periodic axial oscillations of something as massive as the moon?

As I noted early on in this discussion, Libration is interesting, but it has nothing to do with whether, or not, the moon is turning a full 360˚ around its INTERNAL polar axis as the moon also *REVOLVES* (orbits) around the Earth each month.


By what mechanism does the enormous mass of the moon stop and reverse and then stop and reverse again???

So what are we missing ken? You need to do a better job of explainig why we are all wrong.

Eagerly awaiting some better explanatory discourse from you. Thx.

You could start by polishing up on your English skills in order to correctly use the words "rotate" or "revolve" where suitable.

If you can master English, then we can compare the relevant circular motions involved, from both the sidereal and center-point perspectives, to see how they differ, and then perhaps we can agree whether this is a zero-rotating body, as it's labeled to be (a 0:1 spin rate):

http://cluelusshusbund.250free.com/MoonParts/RedGreen0RotationLabel.gif

The moon in that so-called "Zero rotation" graphic is actually *ROTATING* around its internal polar axis clockwise, one time per each counter-clockwise orbit.

Interestingly, the planet Venus' retrograde rotation is only a tad faster than that zero-rotating moon's rotation:

http://www.geocities.com/kfuller2001/tVenus.html

Thus, if Venus continues to slow its rotation due to tidal locking with the sun, then Venus will spin down from its current .93:1 rate (just short of two 360 ˚rotations per orbit) and eventually Venus will slow to a 0:1 spin rate. HOWEVER, since a sidereal spin rate of 0:1 means Venus will still be rotating one time per orbit (as viewed from the sun), Venus will continue to spin down until Venus finally stops at a 1:1 sidereal spin rate.

When Venus finally stops at a 1:1 spin rate (as viewed from the sidereal perspective), the tidal bulges caused by the sun's gravity will finally stop traveling around Venus' circumference and become locked into Venus' surface facing towards the sun. Until that happens, Venus will remain the most perfectly rounded sphere in our solar system. AFTER Venus finally stops rotating at a 1:1 spin rate, at that point Venus will loose its perfectly round shape and form a prolate spheroid:

http://en.wikipedia.org/wiki/Prolate

Ken

C) – The moon is ONLY *revolving* around the combined mass of the Earth and the moon, which is called a *barycenter*.
Pretty graphics, but they are wrong. You need to answer the questions about libration and the angle between the Moon's rotational and orbital planes.

C) – The moon is ONLY *revolving* around the combined mass of the Earth and the moon, which is called a *barycenter*.
...

The larger white body (representing the Earth) is also rotating around its polar axis as the moon orbits around the Earth. The smaller white body (the moon) is NOT rotating around its polar axis as it orbits the Earth – if the moon *still* both rotated as it orbited, then we would see more than one face of the moon.

{NOTE – until it was tidally-braked, billions of years ago, our moon did both

orbit (revolve) as it rotated around its polar axis.}

Wow. You good at engrish. Still, you wrong. The moon and earth orbit or *revolve* about their BC, and they also bothe *rotate* about their own axes of *rotation*. You just don't have the intellect to grasp such a simple concept.



If you still don't understand the differences between *REVOLVE* and *ROTATE* (two different types of circular motions), this next graphic shows a sphere *ROTATING* around its internal polar axis:
...

---Rotaion & Axis---
"A rotation is a movement of an object in a circular motion. A two-dimensional object rotates around a center (or point) of rotation. A three-dimensional object rotates around a line called an axis. If the axis of rotation is within the body, the body is said to rotate upon itself, or spin—which implies relative speed and perhaps free-movement with angular momentum. A circular motion about an external point, e.g. the Earth about the Sun, is called an orbit or more properly an orbital revolution." (Emphasis added.)

Wow. That pretty good.



NO, the orange (and your hand) are instead *REVOLVING* 360˚ around the combined mass of your body and orange, and that axis of common-mass would be a foci within your spinning body. The moon's circular motion is exactly the same as our moon's circular motion around the Earth.

YES, the orange is *REVOLVING* about your head, but is also *ROTATING* about itss'ss own axis of rotation. If the earth suddenly disappeared and the moon went on about it's's's's tangential path, it would still *ROTATE* once in approximately twenty eight days.



The moon does not *ROTATE* around any axis, but the moon does *REVOLVE* around a point exterior to its mass, and the center-point of that orbit is the barycenter within the Earth.

This is moronic.



Sloppy verbiage is a sign of sloppy thinking – "The Earth *ROTATES* on its polar axis as it *REVOLVES* around the sun."

Sloppy thinking is one sign of a moron.

"The Moon *ROTATES* on its polar axis as it *REVOLVES* around the earth." only at a rate equal to its's's's orbital period.



If you insist on using "rotate" (which is a circular movement around an INTERNAL AXIS) when the word "revolve" (around an EXTERNAL AXIS) would be a more suitable choice of words, then please do me the courtesy of identifying the EXACT location of the rotation's axis that you're then talking about.

For example, your "B" choice above could mean anything:

superluminal: "B) The moon is rotating about it's (sic) own axis."

If you mean the moon's internal axis, then NO!

YES! Then how do the stars appear to *ROTATE* about the moons's's's polar axis as seen from the moon itself?



You could start by polishing up on your English skills in order to correctly use the words "rotate" or "revolve" where suitable.

Every "ken" I've ever known has been a dick. The trend continues...



If you can master English, then we can compare the relevant circular motions involved, from both the sidereal and center-point perspectives, to see how they differ, and then perhaps we can agree whether this is a zero-rotating body, as it's labeled to be (a 0:1 spin rate):

If you can suck your own dick, then you can compare the relevant sucking styles of you and all your boyfriends.

http://www.jimloy.com/astro/moon0.htm

http://www.newton.dep.anl.gov/askasci/env99/env093.htm

http://www.physlink.com/education/AskExperts/ae390.cfm

http://wiki.answers.com/Q/Does_the_moon_rotate_on_its_axis_like_a_planet_or_ does_it_just_revolve_around_Earth


The Moon makes one complete rotation during one complete orbital revolution around the Earth (an effect called tidal locking) so that the same side of the Moon always faces the Earth (the other side is called the far side of the Moon).

Wiki.


http://www.badastronomy.com/bad/misc/tides.html

So, despite all of this, you will, of course, continue to "believe" that the moon does not *ROTATE* about itssss own axis (even though it's must) as it'ss's *REVOLVES* around the earth.

Now, as a dick, you probably don't realize just how many people, including grade school students, are laughing at you. I however, realize that there is a mental disorder that limits the ability of certain people in visualizing certain kinds of motions. You clearly have this disorder, and I feel for you. But only a true dick such as yourself ignores the counsel of so many others on such a simple topic. Just as a color-blind person constantly insisting that apples are blue, when corrected by fully sighted persons, is a dick.

So, kendick, I will leave you to your auto-oral pursuits and masturbatory fantasies about *ROTATING* objects.

its, it's, it has, itsh, been fun.

…The moon does not *ROTATE* around any axis, but the moon does *REVOLVE* around a point exterior to its mass, and the center-point of that orbit is the barycenter within the Earth.

Sloppy verbiage is a sign of sloppy thinking – "The Earth *ROTATES* on its polar axis as it *REVOLVES* around the sun."…Ken I agree on need to use terms correctly but your physics about moon’s rotation or spin being zero is wrong, still.

I will use “spins” to more compactly state: “*ROTATES* on its polar axis” for rigid bodies, like the moon. (Spin is more general than that as an ice skater can “spin” without any “polar axis.” In general “spin” just means to present different aspects of the spinning object to many observers in different locations that are relatively fixed wrt each other, even if separated by great distances as the "fixed stars" are. These observers can be in ANY non-rotating frame but locations near the polar axis extended see only very small changes in the surface of the spinning body. For them it is almost a "pure spin" view.)

Spin is most easily (and only way for rigid body)* detected by an observer fixed on the body that may be spinning by observing these “fixed objects” APPEAR to rotate about extension of the polar axis. For example someone in the USA can photograph all the stars rotating about point near Polaris star (and here in Brazil they APPEAR to rotate about point in the “Southern Cross”) and confidently conclude the Earth is spinning about a polar axis which is extend approximately between Polaris and the Southern Cross. Likewise if observing from Venus, one could confidently conclude that Venus is spinning about its polar axis which if extended passes near the stars Vn and Vs (I do not know their names, so assigned these symbolic ones.) In fact EVERY planet, MOON, and asteroid, with name “X” is spinning about its polar axis which is approximately line between stars Xn and Xs.

How can I be so sure that EVERY one is spinning about some polar axis? Well the answer is very simple: The fraction of possible “spin space” with EXACTLY zero spin rate has, to use mathematical terms, “zero measure.” I suspect that in our solar system there may be some bodies that have very small, but non-zero spin rates. Perhaps not even one complete turn wrt the fixed stars even in 1,000,000 years, but they are still spinning. What you are asserting is that the moon is an exception to this, even though a “moonman” could photograph the APPEARANT circular movement of the stars and see that the center of these circles passes very near star Mn (when he takes picture from the northern hemisphere) and about Ms (when photographing from the southern hemisphere of the moon)

Please note that Mn is about 7 degrees displaced from Polaris and likewise, Ms is displaced about 7 degrees for the point in the Southern Cross the Earth’s polar axis passes thru. That is to say that the polar spin axis of Earth and Moon are two non- parallel lines in space , call these lines E & M. Any line passing thru a point on E and parallel to M is inclined about 7 degrees to line E.

If you find any error in this post please tell me where it is. Also please explain why the moon is the only body in the entire solar system NOT spinning on it polar axis, as has been EXPERIMENTALLY CONFIRMED for Earth, Moon and Mars, by photographing the APPEARANT rotation of the fixed stars with time exposures.

You were a fool not to follow the advice I gave you following your first erroneous post, suggesting that you quickly stop your nonsense and thank those correcting you.
----------------------
*If not aided by some instrument, but as you pointed out, the Foucault pendulum can also be used to detect the spin about the polar axis. Then in embarrassment about shooting yourself in the foot, you quickly asserted that the Foucault pendulum was not known to turn relative to the moon’s surface. (I had to admit that it never had been observed.) Your stupidity and stubbornness will no doubt also allows you to also state that a time exposure photographing fixed stars made on the moon will NOT have the circular star tracks as that also proves the moon is spinning about its polar axis. The rate of turning of the Foucault pendulum is identical with the apparent rate of fixed star movement if the Foucault pendulum is set up on either pole of the body being tested for spin. I.e. two independent measurements both can confirm the SAME spin rate for the moon.

Again, of all the fools active here you are the clear leader, both in stupidity and stubbornness! But you can insert drawings, which for some unknown reason I cannot.

For non rigid body, with significant spin rates or very small masses, the polar axis spin can in principle be detected by the near parabolic surface shape of liquid surfaces. If spin rate is low or body is too massive then on surface of the body, the curved liquid surface will be nearly spherical and it is not practical to move to the center of mass to eliminate this gravitational effect.

Each planet & each moon in this solar system except our moon does something which our moon does not. In the universe I lived in until recently, it was called rotating. Evidently, you don't want to call it that. WHAT do you want to call it???

WHAT do you want to call it???I like to call it 'turning on its axis', but 'revolving as it orbits' does the trick too.

It (our moon) also spins, once for each orbital rotation, how about that?

More of the same doubletalk BS.

List of Known Tidally Locked Worlds

From Wiki;
Solar System
Locked to the Sun Mercury (in a 3:2 rotation orbit resonance)

Locked to the Earth Moon

Locked to Mars Phobos Deimos

Locked to Jupiter Metis Adrastea Amalthea Thebe Io Europa Ganymede Callisto

Locked to Saturn Pan Atlas Prometheus Pandora Epimetheus Janus Mimas Enceladus Telesto Tethys Calypso Dione Rhea Titan Iapetus

Locked to Uranus Miranda Ariel Umbriel Titania

Locked to Neptune Proteus Triton

Locked to Pluto Charon (Pluto being itself locked to Charon)

Extra-solar Tau Boötis is known to be locked to the close-orbiting giant planet Tau Boötis Ab.

...Locked to Pluto Charon (Pluto being itself locked to Charon) ...Interesting. As both are small (low gravity) we could connect them with cable I bet, if anyone can think of a reason to.

Here is one for the 22nd or 23d century:
After we have them connected, we just slightly winch them closer together. As both are spinning about their polar axis and we have slightly changed the orbit period about the barycenter, the cable will wind up on them and their mutual orbit period will decrease more – a self accelerating process. The cable was designed not to break before Charon has Pluto's and solar escape velocity. If we cut the carbon fiber cable at just the right time, Charon with its installed nuclear reactors etc. for energy can be humanity’s interstellar spaceship to colonize some other star’s planet. There should be enough mass for the rail gun launchers to guide it to the desired planet or one of its moons. Farfetched, yes, but about the only way humans will get there.

Have any better ideas?

As both are small (low gravity) we could connect them with cable I bet, if anyone can think of a reason to.



No. Can't think of a good reason, but the idea appeals.
Must be useful for something or other.

It is spinning. It rotates once a month, more or less, as it moves around the Earth.

It rotates once a DAY! Gee, guys, nobody corrected this yet??? Thread is too long to read it...

It rotates once a DAY! Gee, guys, nobody corrected this yet??? Thread is too long to read it...

http://www.otherwhirled.com/wp-content/uploads/2007/10/troll3.jpg

It rotates once a DAY! Gee, guys, nobody corrected this yet??? Thread is too long to read it...

Ophiolite is going to fry you alive when he reads this.

Should we post under the language forum the non-difference between rotation and spinning?

Anyway, the real reason is for that, because there is an alien observatory on the moon and this way they only had to build one...

It rotates once a DAY! Gee, guys, nobody corrected this yet??? Thread is too long to read it...

technically, it takes the moon about 27 days to rotate around the earth once??? not exactly sure, correct me if im wrong... which means it would take 27 days for it to make one full spin along its axis, (much like pluto and one of its close satellites charon)

this is known as a 1:1 spin-orbit coupling ratio... look it up if you still dont think the moon both rotates and spins

More & more absurdity.

It rotates once a DAY! Gee, guys, nobody corrected this yet??? Thread is too long to read it...
No my dear. The Earth rotates once a day. What makes you think the moon rotates once a day? Senility? Premenstrual tension? Menopause? Poor reading comprehension? Trolling?

if the moon is rotating around us, and we see the same side of the moon always... THEN ITS SPINNING... think about it, if you hold the moon facing one direction while it rotates around the earth, (lets say side A of the moon is facing toward the sun), then you will see side A of the moon when the earth is between the sun and the moon, but when the moon comes to the other side of earth, then you would see side B, the other side of the moon if it did not spin... and thats deffinently not the case

No my dear. The Earth rotates once a day. What makes you think the moon rotates once a day?

Clear thinking???? :eek: Now I was wrong about the once a day, it is 29 days, but it does rotate around its axis...(hey, I am only an Alaskan governor, honest mistake)

Look at the moon from above and you will see that it ROTATES around it axis...
For idiots:

http://en.wikipedia.org/wiki/Lunar_day


No go down in flames, honeybun.... :)

http://www.jimloy.com/astro/moon4.htm

if the moon is rotating around us, and we see the same side of the moon always... THEN ITS SPINNING... think about it, if you hold the moon facing one direction while it rotates around the earth, (lets say side A of the moon is facing toward the sun), then you will see side A of the moon when the earth is between the sun and the moon, but when the moon comes to the other side of earth, then you would see side B, the other side of the moon if it did not spin... and thats deffinently not the case

So when you rotate a rock around you on a string, is the rock rotating independently of the rotation of the string? It can, and can't; moderately independently of the rotation.

So when you rotate a rock around you on a string, is the rock rotating independently of the rotation of the string? It can, and can't; moderately independently of the rotation.

If you take a point of view that isnt on the rock or anywhere along the string, (that is to say some distance away from it), then the rock would be rotating, yes... its a 1:1 spin-orbit coupling ratio. Is it really that hard to see what im saying?

Im bad at explaining things, but at least i can see it clearly
Ask an astrophysicist, he can explain it better than me. Here are a few links by the way that might convince you:

http://curious.astro.cornell.edu/question.php?number=142
http://www.solarviews.com/eng/moon.htm

and the best link, because it has a diagram, is:

http://physics.fortlewis.edu/astronomy/astronomy%20today/chaisson/AT308/HTML/AT30803.HTM

This thread reminds me of the Plane on a conveyor belt thread.
A misconception based on a false assumption that has to be explained again and again, ad nauseam.

Perhaps, when the US gets its moon mission together we could send the Mythbusters to the moon to do some big experiment.

...Perhaps, when the US gets its moon mission together we could send the Mythbusters to the moon to do some big experiment.May not be needed as Ken Dine seems to have stopped posting his nonsense after my post 209.

But you continue yours.

But you continue yours.Tell where there is an error by citing and correcting it, if you can.

hello every body
thank you for everybody to put in evidece this fact
the fact that the only aster that have a proper spin period wich is equal to the rotation period*( Ps=Pr)

i ask a supplemeent question

IMAGIN for one reason the moon is divided in two halves ....

the two halves are seen in earth to be too fixed position halves
in all other eventualities (when Ps s different to Pr the too halves are seen to be in mouvement with apeed = to Ps-Pr )


a proposal answer for the general question !


IMAGIN that you have too forces you can fixe the position of the moon spin toward the earth mouvement
the second force wich can be involved is magnetic one
so imagin that the seen side of the moon is rich in iron!
thank you very much

I rest my case.
Plane on a conveyor belt all over again.
No matter how many times you explain things to people, some of them will
never understand.

Including most who are certain they do understand.

POACB. Told you.

So no 1 knows why Luna doesn't spin.
1111

I rest my case.
Plane on a conveyor belt all over again.
No matter how many times you explain things to people, some of them will
never understand.

Exactly. That's the thing about idiots - they aren't intelligent enough to realize that they ARE idiots!!:D

So no 1 knows why Luna doesn't spin.
1111
I believe this was your first contribution to this thread:

"If the moon rotates, we should see well over 90% of its surface."

It is apparent you have learned nothing in the last sixteen (?) pages. May I be so bold as to ask what educational level you reached?

I believe this was your first contribution to this thread:

"If the moon rotates, we should see well over 90% of its surface."

It is apparent you have learned nothing in the last sixteen (?) pages. May I be so bold as to ask what educational level you reached?

It is apparent you are missing the point.
1111

Very well, what do you think the point is?

Syzyqys. Hey, genius, look at the links you posted. Wiki says a lunar day is the Moon rotating on it's axis with respect to the Sun. NOT the Earth. The Moon is gravitationally locked one side, as shown in your other link. The images show the side facing the sun will change as it moves around the Earth, with one face always facing the Earth, with different parts of it in sunlight and in darkness as it moves around the Earth, giving the ILLUSION that the Moon rotates on it's axis. That is why as the second article says that the Earth will always appear overhead.

But the moon isn't rotating on it's own axis? It's rotating on a 'different' one?
Of course it's relative to the sun, so is the earth's rotation - however both bodies have "rotations" wrt each other. It's how gravity works - you need 2 bodies with mass, right?

You have to explain how come we don't see the entire surface of the satellite.
If it didn't rotate in two different ways, we would. It rotates around the earth (that's one of the motions it can have).

If the moon had no other rotation, like all bodies moving through space have, we would see the entire surface. We don't see the entire surface, though.

Why not?

But the moon isn't rotating on it's own axis? It's rotating on a 'different' one?
It is rotating on its own axis. The motion of any solid body can be described as a rotation about an axis passing through the body's center of mass and and a translation of the center of mass. This is simple geometry: Euler's fixed point theorem.


Of course it's relative to the sun, so is the earth's rotation - however both bodies have "rotations" wrt each other. It's how gravity works - you need 2 bodies with mass, right?
You are confusing rotation and orbital motion (or orbital revolution). Rotation is the motion of a body about its center of mass, while orbital revolution is the translational motion of the body's center of mass.


You have to explain how come we don't see the entire surface of the satellite.
If it didn't rotate in two different ways, we would. It rotates around the earth (that's one of the motions it can have).
You once again are confusing rotation and revolution.


If the moon had no other rotation, like all bodies moving through space have, we would see the entire surface. We don't see the entire surface, though.
Believe it or not, the Earth is not the center of the universe.

Look at it this way. Suppose you are speeding along some highway. You drive past another vehicle with a relative speed of 15 miles per hour, at which point the other vehicle turns on its pretty red and blue lights. The judge will not accept your argument that you were only going 15 miles per hour.

I have absolutely no problem with people who don't understand basic facts about some aspect of science. And we are all there for most of science. I do have a problem when someone listens to the explanation then declares, for no good reason other than incredulity, I don't believe it. And then maintains that position in the face of everything.

In this thread the only way to describe it is lunacy.

Time for a lockdown..........pretty please.

D H, you are confusing a pedantic obsessive condition with correcting an invalid argument.

You cannot possibly present a logical argument that supports a single thing that you attempt, and fail utterly, to refute. The earth is the centre of the universe for the moon's gravitational potential.

Notice how there's 1 moon and 1 earth, which makes 2 bodies, as I say above? Did you work out that they both have two kinds of motion, as I also say, and you then appear to repeat, apparently confusing the idea of "one kind of motion" with some other thing?

Since you're such an observant chap, did you notice that the Russian gal was RAPIDLY spinning her hammer in the counter-clockwise direction prior to release, but that very slow rotation of the ball you noticed was instead in the opposite CLOCKWISE direction?????

Billy-bob, putting aside the fact that ball was spinning at high rps prior to her release, and that the ball was clearly spinning much more slowly in the OPPOSITE direction AFTER the release (I say from cartwheeling prior to the release), can you explain why the ball shifted from a counter-clockwise rotation to a clockwise rotation prior to landing?


Well Ken Dine... sinse you finaly seem to except that the hammer does rotate... can you relate the hammers angular momentum to the moon... an understan that the moon also has angular momentum cause it also rotates on an internal axis.???

cluelusshusbund's quote of Ken Dine's post includes Ken's observation (from two different cameras, one at ground level near the toss-site; the other from high up in the stands looking down on the landing ball with its attached chain), which were directed at me.

Yes, it is true that film from one camera shows clockwise and the other counterclockwise rotation, as if angular momentum were not conserved.

I again note that this is a false conclusion is caused by camera perspective change. To easily understand, imagine you had a completely transparent but otherwise normal clock laying face down on the floor. From you POV, standing above it, its hands are moving counter clockwise.

Whenever you switch the side from which you are observing the rotation from, clockwise and counter clockwise reverse with no change in the actual rotation.

The nature of the moon's rotation is not the only thing Ken does not understand. Rather than thank the many who tried to help him, Ken just went away after post 109. (Another mistake, IMHO.) Perhaps he will come back when more mature. He did know some things.

The nature of the moon's rotation is not the only thing Ken does not understand. Rather than thank the many who tried to help him, Ken just went away after post 109. (Another mistake, IMHO.) Perhaps he will come back when more mature. He did know some things.

Over a period of mor than a year... me an several others have thoroughly discussed the moon rotaton issue wit Ken Dine (in 4 diferent groops that i know of)... an as you have observed he seems dead serous about his moon ideas... hes college educated an knows lots of thangs... but when it com to the moons rotaton... even a 9th grade dropout (as he refers to me... lol) such as myself can understan synchronous rotaton... an that the latitude an longitudinal libration of the moon which we observe from earf coud not occur unless the moon rotated on an internal axis.!!!

I even preformed an esperiment in which i atached a strang to an object an swung the object in a circle overhead an then released the strang... an no mater what object i used (tennis ball... steel pipe... ect)... when i released the strang the object flew off in a strate line but it also rotated on its axis... an he rejected my results as a faulty esperiment... an his evidence that my esperiment was faulty was that video of the Russian hammer thrower... in which he said the hammer clearly does not rotate when released.!!!

But i tell you what... not only has learnin about the moon an doin the esperiments been interestin... its also been interestin watchin an educated man reject all reason about the moons rotaton.!!!

You think wrongly.

The Earth slows its rotation due to interaction with the Moon by some 1.5 milliseconds per century. It is this angular momentum lost by the Earth that is given to to the Moon.
Is the angular momentum loss by earth measured or calculated from theory? If you have a reference I's like to see it.


The total kinetic energy of a rotating sphere can be found by

E = \frac{\omega^2 M r^2}{5}

Where w is the angular velocity in radian/sec.
M is its mass
r is its radius

Plugging in the values for the Earth for both its present rate of rotation and the amount it would have slowed in one year, and taking the difference, we get a value of:

4.34E+18 joules.

This is the amount of energy the Earth has to transfer to the Moon.

The total energy of the orbiting Moon is found by

E= \frac{GMmMe}{2a}

Where G is the gravitational constant
Me and Mm are the masses of the Earth and the Moon
a is the average orbital radius for the Moon.

Plugging in the correct values for the Moon's present orbit and one 3.8 cm further out and taking the difference we get a value of:

3.77E+18 joules

Which is the amount of energy it would take to raise the Moon's orbit by 3.8 cm and is less than the amount of energy the Earth has to transfer in one year.

The difference is due to the fact that some of the energy given up by the Earth is lost due to tidal heating.

So yes, the Earth does give up enough rotational energy to the Moon in order to increase its orbit by 3.8 cm per year.
Question I: I don't question your math or physics but I do wonder what the mechanical link might be such that the transfer of energy from earth to moon directly affects the moon's rotational velocity or results in a new orbit trajectory. Energy being a scalar it isn't obvious to me how a transfer of energy can alter the orbit or rotation unless the energy transferred is "directional" in some way. I was on another mission when I bumped into this thread. I had just been locked out of a thread regarding the Shell theorem and the gravity topic caught my eye.
:shrug:

...Question I: I don't question your math or physics but I do wonder what the mechanical link might be such that the transfer of energy from earth to moon directly affects the moon's rotational velocity or results in a new orbit trajectory. Energy being a scalar it isn't obvious to me how a transfer of energy can alter the orbit or rotation unless the energy transferred is "directional" in some way. ...I think I know the answer to your question. The Earth is not a rigid body. The moon's gravity gradient cause tides, which are bulges of mass that tend to keep their same relationship to the moon. Think of them as fixed relative to the moon and Earth rotating under these bulges. There are different land masses spinning underneath these bulges so they cannot keep exactly the same relationship, but this is not important if we think of their long term average.

Now to make it more clear, lets conceptually divide the Earth into two parts: One is a perfect rigid sphere and the other, to be extreme in our model, is just a dumbbell, (rotating with the Earth's 24hour day but fixed wrt the moon, representing the tides).

I will type draw the dumbbell as: #----------#

And the moon as: O

This crude arrow:
^
|
|
indicates the moon's velocity.

Now because I think you will agree that the rigid sphere part of the Earth cannot accelerate the moon, to change it orbit, even if it is spins, let’s ignore it. I.e. all of the moon's orbit change must come from the dumbbell, but it is not spinning when viewed from the moon. Wobbling a little bit, due to the effect of the land masses the tides encounter but I will make discussion simpler by having a model of Earth where there is no land mass - just a uniformly deep single ocean covering the Earth. Most of this ocean water is also spherical, so it too can be ignored as not able to change the moon's orbit.

SUMMARY (thus far): Only the tidal dumbbell is available to change the orbit of the moon and your question is how does it do it? What is the mechanism?

If the spatial fixed relationship of the moon and dumbbell were like this:


^
|
|

O................................................. .................................................. ........................................a#----------#b

where I have separately labled the two masses of the dumbell for later reference. (Ignore the ........... as they are needed to keep Sciforum's computer from compressing many spaces down to only one.)

Then the dumbbell also cannot change the moon's orbit as 100% of the forces from both of the # masses acts only along the line from Earth to Moon; Helping make it orbit the Earth. (Moon's velocity turning clockwise in my typed drawing but not changing in speed.) However, that is not their spatial relationship to the moon because it takes some time to lift up a tidal bulge. I.e. the max tide is NOT on the Earth / moon line but delayed. I.e. the near moon bulge peak is a little west of the Moon Earth line as seen from the moon. (The moon is traveling Eastward. I had to give the view as if seen from far below the south pole as I needed a point for my arrow and only the ^ is available.* Thus, in this "southern view," drawing the Earths spinning surface and the moon are going "clockewise.") Thus the typed drawing should be more like:


.................................................. .................................................. .................................................. .....................#b
^
|
|

O



.................................................. .................................................. .................................................. ...........a#

Recall the crude arrow pointing up above the moon indicates the moon's velocity. Also note the delay required for the bulge to reach its peak and Earth's rotation have rotated the dumbbell counter closkwise wrt the first drawing's in line position ( a#----------------#b )

Now I need not show the dumbbell’s bar -------------- to keep them apart and it would be hard to get correct even if I used: / to make a sloped line between the two #s because what I type on my computer displays very differently when posted.

Also imagine that the two #s were about 100 yards more to the right and thus both making a very small angle triangle with the moon at the apex.
The scale here is very wrong as the distance between the two #s is the Earth's diameter and the ....... needs to be very much larger as it is the moon Earth separation. I.e. both #s are only a small angle off the moon Earth, center to center, line with the near one's peak displaced a little west of the Earth moon line.

Now both of the #s do exert a force on the moon in the direction of its travel. The more distant # is trying to speed moon's velocity and the closer # is trying to slow it down. Because the closer one has the stronger force, it wins and the moon's velocity slows down.

With this slower velocity the "centrafugal force" is less so to still be in a circular orbit that is stable the moon must be in weaker Earth's gravity (or further away). BTW please note that there is no such thing as "centrafugal force." - It just feels that way on a merry-go-round, etc.

Initially it seems this is wrong as Janus said the Earth was transferring energy to the moon (and that is true) yet I am claiming the moon is slowing down! I.e. that the moon's kinetic energy has Decreased!

Resolution of this paradox requires a little more understanding of orbits. I.e. the moon is in a gravitational well made (by the Earth). It takes energy to climb partially out of that well and be further away. I.e. the Moon gained Potential Energy, PE, and lost Kinetic Energy, KE is the answer to the paradox. Here it is in detail:

Well it turns out that the total Energy, E, is negative for any bound orbit (that is what "bound" means.) and for circular orbits the magnitude of the PE is twice that of the KE. So to make this very clear I will switch to a numerical example with the moon assumed to be in a circular orbit (to keep it simple):

A long time ago when the moon was closer and more tightly bound to the Earth, here is the "energy story":

E = -51 = PE + KE = -102 + 51

Now the Energy story is:

E =-50 = PE + KE = -100 + 50

Note that the moon, now farther away, is only 50 energy units bound. I.e. it is escaping from the Earth's "gravity grasp" and has gained one unit of energy (from the Earth's spin - The days are getting longer.). The moon has move up out of the potential well by 2 units of energy and lost one unit of kinetic energy.

I hope and trust you now understand both the mechanism and that energy are being transferred to the moon, despite it slowing down.

You may not know that Pluto is going slower than the Earth. But this follows from Keppler's law that the square of the period is proportional to the cube of the orbit radius (actual the semi major axis of the ellipse, but I am sticking with circular orbits to keep it simple.) I will write that as:

T^2 ~ R^3 although instead of R one traditionally uses "a" for the semi major radius)

Thus if R were doubled, then T is 2(square root of 2) or more than doubled.

Yet the circumference around the twice larger orbit is only doubled. I.e. it takes more than twice as long to make the full trip around but the path around has only doubled. Thus, the planet (or moon) must be going slower in the larger orbit.

I.e. the whole "story" is consistent and here is a quick summary:

Because the two tides are not on the Moon / Earth line and the near to moon one is making a stronger force on the moon than the further one and a component of that force is slowing the moon down the moon is moving to a more distant orbit and the day is getting longer. There is energy being transferred to the moon, two units of which are decreasing the negative potential energy and one unit is decreasing the positive kinetic energy for a net transfer to the moon of one positive unit of energy, which previously was in the rotational energy of the Earth's spin.
---------------
*I could have given a view from far above the north pole if I had put the moon on the right edge of your screen, and then that same "up arrow" would give counter clockwise rotations one often sees in such drawings, but that is very hard to get correct when typing and using ............ to defeat the sciforum's computer compressing spaces down to only one. The V could have been used as the arrow point down instead of up and keep the moon O at the left edge still but I could not get it to center on the | line. Besides I live in the southern hemisphere now so it seems OK to me. :D

PS I am sure Janus 58 could explain it perfectly with 1/10 the words but some people may follow a long winded and detailed version better.

-----------------
On your first question: What is measured is the slowing of Earth's spin and the increasing in the Earth moon separation. (Two independent measures that agree but either could be used to calculate the other - so this is a very good test.) I imagine by assuming perfect consevation of total angular momentum this data allows a good measure of the Earth's moment of inertia to fall out.

The laser retro-reflectors left on the moon and very brief (or modulated for precise time ref) laser reflections lets one measure the earth moon separation.
The cumulative change in the spin rate times the total time lapsed (an angle) is best measured, I think from old total moon eclipses. For example, assume there is a constant spin, rate and calculate in which Chinese city an ecllipse of ~3,000 years ago was visible. Well it was not seen there, but in a citry X miles to the west according to the temple astronomers records and local stories of dragon that ate the moon etc.

From X and the radius of the Earth, you get the "correction angle" the Earth failed to rotate thru in these 3,500 years because it has not been spinning at a constant rate, but slowing.

Energy being a scalar it isn't obvious to me how a transfer of energy can alter the orbit or rotation unless the energy transferred is "directional" in some way. Gravity is a force, it is vectorial. A transfer of momentum, linear or angular, results in a change in energy. It is not that the energy of the Moon somehow increases but nothing else changes, there's a corresponding change in angular and linear momentum, both in terms of the Moons orbital angular momentum and its rotational angular momentum.