Jupiter's Orbital Velocity & Equatorial Velocity cancel?

You seriously do not see that you are just picking numbers that have no relationship to each other and manipulating them to give another number that has no relationship to the others. What does the distance of earth from the sun have to do with the speed of the suns orbit around the galaxy and the suns rotational velocity? None that I can see. Why pick the earth? Why not Mars or Venus? Looks like numerology to me.

So it is your contention that the speed of a stars orbit around the galaxy divided by the rotational velocity of that star multiplied by the diameter of the star will equal the distance to the 3rd planet of that star system?????
Seriously don't you think that sound at least a little bit foolish?

 

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Thank you for your concerns, The answer is no. Both rotation and revolution of the planets are stable. As would be the speed of a car on a great circle road, and the rotation of its tires. but where the rubber hits the road, the velocity of the thread is zero, and it is double the speed of the car inside the fenders.*** A rotating sphere too, has zero velocity at the point of contact, and double on the very top. In a roller bearing this fact is exploited, with many zero velocity contact points.
The zero velocity zones in orbital mechanics are not always coinciding with the surface, and that led to the interesting results of the (Vo:Vr)xR formula for the ZVZ having the Moon and Earth coming up in a "zero match" , Venus and Mars ditto,
Sirius , like a railway wheel rolling on its axis. so,
yes you can maintain your rotation, but have zero and double contact speeds.
***which leads to the conclusion that there are points in systems like the ecliptic, or the galactic plane, where matter actually is at a standstill with respect to the whole. respect that.
In such areas, gravity would have free reign, no "centrifugal force" to oppose, cancel it

 

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That's good. So that means that you no longer think this statement is true?
The band around the equator of Jupiter, and the Ring region of Saturn spend more time in the sunlight, than any other, and it is not because of the angle of incidence only, but because it is at least twice longer in the warming sunshine than the colder night side.

 

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[QUOTE="origin, post: 3548153, member: 143078"
no longer tru? The band around the equator of Jupiter, and the Ring region of Saturn spend more time in the sunlight, than any other, and it is not because of the angle of incidence only, but because it is at least twice longer in the warming sunshine than the colder night side.[/QUOTE]

It is also a question of quality time. Working with an analogy: The tire thread on the road and/or on top, under the fender, are in the same time frame of the moving car. , but with respect to stationary road, any section of the pavement, --, the rubber hitting the road is at a standstill , lingers on., whereas for the same road distance ** above, in the fender, it is a blur. Very little time spend during that distance at double the forward speed.
With your computer savviness you could find the 'moving tire' example that exist, and post it for us, (giving proper credit of course, or NotEinstein will have your head) .
**Any distance of the road of the analogy being the equivalent of the solar energy energy available in any area of the ecliptic. It pays to linger at cancelled velocities. imho.

 

Let's go one step at a time.

Does that mean that you no longer agree with this statement:
The band around the equator of Jupiter, and the Ring region of Saturn spend more time in the sunlight, than any other, and it is not because of the angle of incidence only, but because it is at least twice longer in the warming sunshine than the colder night side.

 

I probably stand corrected referring to just "Time" in that statement. I should have qualified that and linked it to intensity of exposure time with the road. thank you for showing that error. Going even further, it can be stated that the nightside of any planet spends more time in the shadow during the year, because the sidereal year, or the radius, circumference of the night side is longer than the inner, sunny side.
In the car analogy, the wheels clearly spend the same time above the axles than below them. Yet with respect to the road a point on the thread is starting slow down when pointing. pointing forward, come to a standstill at the pavement and slowly speed up again to full speed when positioned at a position at axel heights, straight back. Here is a link,
http://www.animations.physics.unsw.edu.au/jw/rolling.htm
with due credit, with animations of the rotation , translation movements.
Particular attention should be given to the last clip of a flanged railway wheel. It illustrates the Saturn situation, where the ring radius point loops actually backward to give even more exposure time.

 

Let us modify that thought experiment and bring these two satellites from your L1 and L2 spots into the other 2 stable, the Trojans position, but closer to Jupiter, and have precise radar/ laser beams pointed at a prominent feature near the equator of like the Red Spot.
According to the present Zero Velocity Zone hypothesis, the Velocity of that spot would read zero at the noon position, the horizon, facing the Sun , and would, 5 hours later, read ~ 20 km/sec, twice the orbital speed at the outer, midnight spot.
That does not mean that the rotational speed is variable, but that, as far the ecliptic field is concerned, a feature near the equator of Jupiter or Saturn can at noon, be stationary, have zero velocity. or:
Driving, as you speed, and see a radar trap ahead, ask the police lady to point that new radar gun at the lowest point of the tire, near the road, and you will get a really lower, even zero reading, different from your speedometer.

 

That is a good thought problem illustrating another feature of this alternate theory.
If you put such a laser pointing straight up corrected for spin axis inclination on Venus, operating at midnight, such pointing beam would have zero velocity with the Mars orbit, possibly ,meeting it one fine day with a laser from Mars shining down at noon when Mars, Venus and the Sun are lined up.
Venus rotating retrograde and slowly, cancels Vo and Vr at night time, way up. near Mars. as pointed out in post#130,page #7

 

False.

 

because?

 

The velocity is not zero.

 

so, what is it then, give or take a km or two per second?
what is that between friends, once they are in the general No Velocity Area , with Vo and Vr cancelled?

 

This shows a complete misunderstanding of the situation. A speed gun pointed the lowest point of the tire, near the road will NOT get a really lower, even zero reading

 

Of course, that radar, laser - like pinpointing has not been developed yet. but any device capable to isolate the foot print moment of the tire though, will show it to be at rest with the road. The flange of a railway wheel will even go backward from the speeding train's motion.

 

There really seems to be no getting through to you. It is true that there is no velocity difference between the road and the tire. You take this fact and go off on your flights of fancy that are absurd.
You said that a speed gun would show that tire is not moving at the point it touches the road on a moving car. That is wrong. The only way that would be true would be if the speed gun was moving at the same speed as the tire. I guess if the speed gun was rotating at the speed of the tire and it was moving at the same speed as the car, it would read zero otherwise it would not.
You also try and apply this concept to planets which is wrong and leads to absurd conclusions. But you simply accept the absurd conclusions and double down on your misconceptions.
I know I am just wasting my breath....

 

If I am correct, a radar device can be used at any speed, even inside a speeding cruiser . The speed of light is equal for all observers, regardless of their own velocity. It is not like shooting bullets.
For imaging the zero velocity at the rolling contact point., --point a camera, and get a non blurred image of the slightly flattened part, you might even capture the grooves catching a pebble. at a standstill.both.
If these car analogy situations are real , have consequences, why should it not be so, --at the planetary level?
If it is not fun, torture, for you, please leave me, seriously, to it.

 

Radar guns measure the frequency shift to determine the speed of an object. So the relative speed between the gun and the target is what is measured.

Wrong. There is no zero velocity between someone holding a camera and the wheel. You are just plain wrong on your thinking.

They are not.

Your understanding tire/road situation is severely flawed. Even in the correct interpretation of the tire/road situation is not applicable to planets.

The unfun part is that you refuse to understand this stuff.

 

Hang on a minute. I may be being thick but surely at the instant of contact with the road or rail, the point of the wheel in contact is at zero velocity relative to the road or rail, is it not? Each point on the periphery is under constant acceleration of course, so it is only an instantaneous zero value, but it is zero at that instant, surely? http://www.animations.physics.unsw.edu.au/jw/rolling.htm

 

Yes, that is correct. The velocity difference is zero between the road and the tire for the instant they are in contact, unless your tires are in a skid or you are spinning your tires in a burn out.
The issue is that nebel seems to think that the tire is not rotating at that point or some such nonsense. He seems to be totally confused about reference frames.
He then goes on to say that since a planet rotates 'like a wheel' and the sun rotates or the planet orbits the sun, then the planet is facing the sun for up to twice as long as the planet is facing away from the sun.

 

The 6 inches that touch the road at standstill, clearly are exposed to the road [view] more intensely, longer, than the similar 6 inches whizzing forward in the fender above. .
The tire or planet Jupiter's equator has twice the orbital speed at nigh time, "fender time" , than in the daylight, while the rotating speeds stays constant