Assume 1000Kg satellite "A" is placed into geostationary orbit, (Eastward lauch from equator and this slightly slows Earth's rotation as more of initial angular momentum is in the orbiting satellite.) Then its twin, geostationary satellite "B," is launched from adjoining pad with same eastward roll over etc. but a little longer rocket burn, because after launch of "B" the Earth is spinning still more slowly and B's altitude must be SLIGHTLY higher than A's to be exactly sychronous with the new longer day.(Satellite A is no longer exactly sychronous once B is in orbit. Sorry SL if I did not make that completely clear in first version of this post. MAKE LARGEST YOU THINK REASONABLE GUESS AS YOUR VOTE.)

Wait a sec. Maybe I'm missing the point, but if the definition of geostationary is that the satellite orbits in perfect synchrony with the earth's rotation, then (assuming the earth is perfectly round) that altitude will be the same for any satellite, no matter what the distribution of angular momentum is in the system. Right?

Ok. So I see my error. You are assuming that A may no longer be in geostationary orbit due to B's launch and you want to know what the new altitude difference is for geostationary orbit. I'd guess... Option 3.

I'll go with 1.0E-4 m , as that is what I said in the other thread that this pole was prompted by. Might as well be consistant, hey Billy. ;)

BillyT, I am shocked and astounded by the options you have provided for your questions, especially as you claim to be a scientist. All of the answers are correct except the last one. For example, 0.000000001 metres is less than 1,000m and thus answer 4 would be correct.

This is an excellent illustration of
a) the problem with polls of any kind.
b) the strong influence of the character of any 'question' upon the resultant 'answer'. It is exactly analogous to poorly structured experiments that produce misleading results.

Shame on you. :)

The length of the day has some "jitter" anyway. That is going to be a much greater number than the number due to previous satellite launch. But given the premise of an otherwise perfectly stable day my guess is that the difference is on the order of angstroms.

-Dale

BillyT, I am shocked and astounded by the options you have provided for your questions, especially as you claim to be a scientist. All of the answers are correct except the last one. Most people understood, correctly that you were to:
MAKE LARGEST YOU THINK A REASONABLE GUESS AS YOUR VOTE.*
but now I have explicitly added this to the first post which explains "SLGHTLY"
--------------------
*I had this explicitly in the options I wrote in first post,when I did not know next step in setting up a poll would provide boxes to type options in. (First time I have set up a poll.) I did not find place to but above caps in at that later stage. I have edited first post, for you and anyone else, not making the obvious assumption. I stuck in "circular" also for same reason. Some things should be obvious, but it does not hurt to make them explicit.)

PS I just now voted 1M, mainly because now that some have voted, I will not be the worst vote, when someone "cheats" and calculates the answer.

I understood exactly what you meant, but this is a science forum and you are a scientist. I think it is appropriate to expect precision in such an environment, from such an individual. If if was from the likes of Gustave I wouldn't have blinked an eye.

I understood exactly what you meant, ...I thought anyone could.

The posting instructions tell you to keep the option statement short. This means one can and should asssume a reasonable level of competence on the part of the voter. Yes I was not complete and precise. E.g. I did not define "satellite", "geostationary"* etc. tell what is to be assume about the launch rocket , many things. If you think what you state, start a poll and see how many agree with you that I was not as precise as a scientist should be.
------------------------
* I did briefly think this might lead to some confusion, especially for Russian readers, as with the northern location of their major cities, "geostationary" may also refer to the highly eliptic orbits they use with polar apogees for communication relays. (Those "geostationary" satellites dwell most of their orbit more distant from Earth, but can be seen from all high Russian latitudes etc.) I decided it was not necessary to precisely define everything. I.e. I decided to follow the poll's instructions. Sorry if you find this "unscientific."

I have no interest in how many people agree or disagree with me. It is an absolute that the manner in which you structured your poll was flawed and ambiguous. I really did not expect you to make a big deal of it and start defending the indefensible.

I have found all of your posts till now, many of which I have read and enjoyed, to be very sensible and well structured. I have differed from you on a couple of minor points, but nothing significant.

I regret that you have chosen to react to this particular issue in such an emotional, defensive manner. I guess I should have used an even larger smiley after my original post.

...I regret that you have chosen to react to this particular issue in such an emotional, defensive manner....I regret you insistence on complete precision, full details, etc. These polls are not science. (Voting never decides that.) I will not reply more, but let others decide who is unreasonable on this issue.

let others decide who is unreasonable on this issue.Sounds like a perfect opportunity for another poll! :D

-Dale

I am not insisting on precision or full details. I am insisting upon a meaningful poll.
Less than 1000 metres includes
Less than 100 metres, which includes
Less than 1 metre, and so on.

That is not a lack of precision BillyT. The way you have structured the poll is simply wrong. It is meaningless. I am sorry you are too stuborn to see this. Goodnight.

Sounds like a perfect opportunity for another poll! :D - DaleBe my guest, now that we both know how to set one up. :bugeye:

I for one saw the flaw and immediately understood the intent. Given Billy's history, why would we take him to task for this? I am wondering who here would not understand the intent? I can't wait for the answer! I have been tempted to go and calculate the answer... Maybe I'll do that tonight.

I for one saw the flaw and immediately understood the intent. Given Billy's history, why would we take him to task for this? I am wondering who here would not understand the intent? I can't wait for the answer! I have been tempted to go and calculate the answer... Maybe I'll do that tonight.Please keep answer to yourself until few days before poll closes.

I wish I could add: "Choose largest answer you think possible." where instead above the poll you see only: "(In first post of Billy T, if not here)" This was my first poll, I did not know what to expect or what could be changed later as I proceeded thru the set up proceedure, so I followed the instruction to keep options statements short instead of writting something like:

Greater than 1.0E-4Meters and less than, or equal to, 1.0E-2Meters

Which is unambigious, but excessly long by my understanding of the instrucion to be short. As this was first time I set a poll up, I feared (and still do) the above option statement might end up appearing as:

"Greater than 1.0E-4Meters and less than, or equ"

and that would be confusing. I. e. I gave credit to the voters to use a little common sense as you and all but one have.

It is cheating to CALCULATE the answer? That does explain some things.

But, I burned up my weejee board decades ago.

In a trash fire, not from excess use.

I like the idea of guessing at it. My first reaction was that it must be extremely tiny. The earth is HUGE after all, and a 1000kg satellite is NOTHING. But this mass is moved very far away. It gives an insight into how fallible our intuition is when it comes to things far outside of our experience. I suspect the answer will be larger than we think... Or not! :D

Maybe so!

And maybe not, I guess.

Actually, the answer I really wanted was on the order of the Plack length. But maybe I should be disqualified, as I once did a vaguely related calculation that I meantioned in the other thread. :)

Planck length is closer than the provided choices, in my guessy opinion.

The answer really does not need any calculation. We do need to ignore obfuscative red herring(s). And to make sure that our angulars are momenting the right way.

The less than 1.0E-4 people are 2/3 as I post. I do not know the answer, but guessed less than 1 meter (more than 1cm) which is mighty small range (99cm) to hit. I based my guess on fact that the angular momentum difference between two "geostationary" orbits only 10s of cm different in circular altitudes, must be "dam small" also. So for me it boiled down to which "dam small" is roughly the same as the "dam small" change in Earth's angular momentum associated with putting much, much, larger (than these "dam smalls") angular momentum into orbit. I can hardly wait to learn what answer is, but as usual I am much too lazy to calculate it when I thought of this way to get someone else to do it. :cool:

The 1cm to 0.1mm range (a 1cm band, essentially*) was too small a range for me to chose it. I predict few will chose it. (Unless it is the correct answer and we have a few cheaters among us.) :p

The 1.0E-4 range is "open bottomed" (if I may coin a phrase, without being accused of some new form of preversion. :eek: ) but it is after all only a band of o.1mm and I did not like that narrow choice.

I must admit I do admire the courage of the guy/gal who went for "greater than 1000M." (I can no longer see voter IDs that I saw before voting, but I guess this is necessary as if written under the color bars there might not be room. - Is there a way to see who is guessing what still?)
---------------------------------------
*"essentially" added by edit as I do not want to be jumped on again for poor science methods etc.

Ok. I have calculated the answer. That is all I will say for now.

Ok. I have calculated the answer. That is all I will say for now.As only 10 days before poll closes, feel free to give the angular momentum of Earth. (In Kg meter^2 / second.) That is the only part requiring any real effort, at least as I would do it. If you give that to me, I will calculate the answer too. - I am lazy, but not that lazy and also very courious to know the answer.

Billy,

I can do that when I get home. I have the numbers there.

Here you go.

The moment of inertia of the earth - a solid sphere - is:

I = 2/5 M_E R_E²

= 9.7212563816824800E+37 Kg m²

(I kept the satellite a point mass: M_S d_S²)

The angular momentum is of course I x w but if you do it as a ratio of w_final to w_initial to find the change (as I did) you don't really need w.

You will need a precision calculator. I found some on the web. Excel only keeps 15 digit precision...

It'll be interesting to compare results and methods!

...It'll be interesting to compare results and methods!Thanks. Yes it will be. I have not done calculation yet, but method I have in mind does not requre much accuracy (No small differences between two big, almost equal, numbers. - I think, my old slide rule will get answer to two places OK.)

I will just call I = 9.72E37 Kg M^2 and suggest others do the same (unless some one thinks SL's number is significantly wrong.) Surely it is not known to all those significant figures - I.e. was yours the summer or winter value? :rolleyes:

Perhaps someone will be kind enough to give us the nominal geostationary distance from center of Earth (Not altitude) so all can use that as a standard value as well.

Every one should of course follow SL and assume satellite is point mass, without rotation, I might add.

I suspect sl's number is high. Maybe as much as 20 or 30 percent. Remember the earth's core is a lot denser than the mantle and crust. I believe the formula he's using is for a sphere of uniform density.

kevinalm, you are probably right, but I was unable to find a reference for an empirically determined moment of inertia or spin angular momentum. In such cases an approximation like this is the best you can do (like the famous spherical cows approximation). Since the various answers are generally separated by 2 orders of magnitude even a 30% error should be ok.

However, if you are able to find an empirical reference please post it and we will gladly use that.

-Dale

I also could only find the form of I for a uniform solid sphere. I also don't think 20% - 30% matters (and I personally don't think the earth is that non-uniform).

Agreed. I'm an engineer so these kinds of approximations/assumptions are just standard operating procedure. I was planning on using the solid uniform sphere approximation myself whenever I did the calculation. I don't think you could make a better approximation, only a measured value of the spin would be preferable.

-Dale

...only a measured value of the spin would be preferable.-DaleAny ideas as to how the Earth's angular momentum (or actually the moment of inertia, I, as rotation rate is very well known) can be measured? Is any method accurate enough to actually mesure the seasonal difference as ice above sea level melts and runs into ocean, etc?

Nope. Maybe MacM's NASA buddy would know :)

-Dale

Well, the density of most rock is somewhere around 2 to 3 g/cc and heavy metals like Fe and Ni around 5 or 6 g/cc. Actually I'm not suggesting that we should try to correct the constant density model, just that we be aware of the limitations. SL's figure should be taken as an upper bound for the Earth's rotational inertia.

I think that the precession of the equinoxes should give a measure of moment of inertia. Hang on, I'll ask Google :)

Surely it is not known to all those significant figures - I.e. was yours the summer or winter value? :rolleyes:


Yeah, well... :m:

Here are all the numbers I used:

M_E = 5.9742 x 10²⁴kg

M_S = 1000kg (given)

R_E = 6378 x 10³m

D_s = 35786 x 10³m (above the surface, not earth centered. You can figure it that way if you want...)

All courtesy Google, the oracle of our age.

OK, this site (Precession (http://www.geol.binghamton.edu/faculty/barker/demos/demo10.html), by Dr Jeffrey Barker, Associate Professor of Geophysics at Penn. State U.) says that because we know the Earth's precession and the torque applied by the Moon reasonably accurately, we can pin down the Earth's Moment of Inertia to 8.07 E37 kg-m^2, or about 17% less than the uniform sphere estimate (kudos to kevinalm!)

But like others said, for the purpose of the exercise, I think that a nice round figure of 10^38 kg.m^2 will be fine for our purposes today.

I note that we're assuming that all fuel expended from the launching rocket returns to Earth, otherwise more torque is applied to the satellite and less (or none!) to Earth.

Is this reasonable?

I note that we're assuming that all fuel expended from the launching rocket returns to Earth, otherwise more torque is applied to the satellite and less (or none!) to Earth.

Is this reasonable?

Sure, but it won't matter. Trust me.

...we can pin down the Earth's Moment of Inertia to 8.07 E37 kg-m^2, or about 17% less than the uniform sphere estimate (kudos to kevinalm!)


Yes indeed! Good call K. But it won't matter...

To badly misquote the delightful old song: You can use any oment of minertia you want, in Alice's Restaurant. Even including the right one.

By the way, the thread starter did not specify the faze of the moon. Another deliberate and disgustingly obvious obfuscatory maneuver?

To badly misquote the delightful old song: You can use any oment of minertia you want, in Alice's Restaurant. Even including the right one.
You just made A.G. proud. ;)

I love Arlo.

Singingwise.

SINGINGWISE!

I love Arlo.

Singingwise.

SINGINGWISE!
Granted - not much to look at. :D

Do you happen to recall the line where he had been "inspected, infected, injected, rejected..." and about two or three more "-ected's" ?

When he was joining the Marines. So he could KILL, KILL, KILL?

....But like others said, for the purpose of the exercise, I think that a nice round figure of 10^38 kg.m^2 will be fine for our purposes today.As starter of the poll (if that confers any authority) until I reverse or modify my view, lets all follow Pete's suggestion, but you may use 5E37 kg.m^2, if only that smaller value gives you the personnal satisfaction of being "correct" as it may be true.
I also suggest that we use the radial distance SL has found in Google, etc. if your method requires any of these more detailed values. All this is open to critical comment, and potentially changeable, if you are unhappy with these "standard numbers."

I notice the "bottomless band" has recently lost relative support. I don't expect any of the more recent voters to admit to cheating. - I still do not know what the correct answer is, so perhaps none have. I.e. the "bottomless band" may be correct answer, as the majority still thinks.(An alternative possiblity, if the bottomless band is correct, yet losing support, is that we now have a few "inept cheaters", who miss-calculated, but I doubt this.)

In any case, it would be interesting to hear why most newer voters are avoiding the "bottomless band." - I gave my reasons for guessing the answer is in the 99cm of the 1M band some time back. - I am courious if any of the newer voters were infulenced by that post as well as what other thoughts had influence on them.

Yes, Pete and probably all others, are correct to assume everything except the 1000Kg of the satellite has fallen back to earth. (Any "global wind" the exhaust gases made has also disipated in friction with the ground. Likewise, the Earth's core is co-spinning with the crust. - I am confident, based on SL's comments after he found the answer, that none of this matters, but just trying to completely define the question.

Interesting that no one yet likes the 900Meters target available in the 1000M band.

There is a philosophical sublety here. If I ask BillyT," What is 1 plus 1?", he may get his calculator and perform calculation(s), check, and then give his answer. If I ask Stephen Hawking the same question he might immediately mentally calculate the answer so quickly and reflexively that even if he wanted to, he could not avoid "calculating" the answer.

PS Nobody has told me yet what the faze of the moon is during the science experiment.

I find this thread to be highly entertaining. It's fascinating to test your intuition against reality. Sometimes we are astounded to find that our intuition can be SO wrong. I voted after about two minutes of just visualizing the masses and distances involved.

I think the faze(sic) of the moon during the experiment should be assumed to be waxing gibbous, 2 days past half moon. :m:

There is a philosophical sublety here. If I ask BillyT," What is 1 plus 1?", he may get his calculator and perform calculation(s), check, and then give his answer. NO, that is too easy. I do not think, even my quick correct answer (2) is a "calculation." - That would be more "recall," but you provoke an interesting question: What is "calculation?"

For example, to be thread related:

I still do not know the answer, and do not think I have done any calculations, but I did go thur the thought process below, partially to just pose the question and set up some answer choices, but also to cast my vote. Tell if you think I did any “calculations” in the following 13 steps:

(1) Earth spins slower because of launch of Satellite "B." - Only physics, I think, no calculation.

(2) Therefore: Satellite B is higher than A, which was geostationary with the pre-B launch spin rate of Earth. (I happen to already know that a very low Earth orbit’s period is about 90 minutes and moon's is much longer, so B is in between A and the moon.) - Any calculation yet? I think not.

(3) Decrease in Earth's angular momentum is exactly the angular momentum of B. - Pure physics, no calculation here, surely.

(4) Angular momentum of B is mass, velocity, radius product. - more definition than calculation, I think, but does concern “product.”

(5) B's radius will be bigger than A {only (2) again}. Thus it will travel farther each orbit. - Perhaps a little bit "calculation"? What do you think?

(6) For B the day is longer, so it has more time to travel this greater distance. - Calculation?, I think not, just logic.

(7) In view of (6), speed will not be greatly different from A's speed. At least not a linear change with radius, anyway. - How can I be thinking about "linear" etc. without at least some "calculation"? Yet nothing numeric here. - Not a calculation, I think.

(8) If speed of B were same as A, then B's angular momentum is greater than A's linearly in the radius. - some "calculation."? Yet again, nothing numeric.

(9) Angular momentum of Earth is HUGE, compared to B's and only changed by this tiny "B amount" - physics, general knowledge only, no calculation, I think.

(10) As B is higher {5 & 2 again} and angular momentum is stronger function of radius than of speed, plus vague and uncertain recall that speed also drops with altitude as angular rate certainly does, perhaps, I may need to boost radius not only to make more angular momentum in B than in A, but also too compensate for slightly lower speed of B.

(11) Despite (9), (7) with (8) tend to tell me most of the change in angular moment is going to come from increasing the radius and in view of (10), perhaps by more than my gut feeling tells me. Although my gut feeling, and fact I have never heard anyone mention that the geostationary altitude is a function of total mass already in geostationary orbit, both support the "bottomless" range choice, that may be wrong. It seems true that I must make most of B's greater angular momentum by its greater radius. Perhaps the difference in radius is not extremely small after all. Lets look at the other choices. - Mainly physics, logic and "betting strategy", I think, not calculation?

(12) The 9mm wide band is a "dam small target" to hit. If I am going to let (11) over rule my gut feelings, the 99cm wide target of the 0.01 to 1 Meter band looks like a better bet. (I am usually not far off in my gut feelings, so better not ignore them entirely and select an even greater target.)

(13) I'm sure many will go with the "bottomless band" and if they are correct and I choose a much greater difference, I will look silly. Ok that is it. - It is the 1M band for me, lets vote. - No calculation here, surely.


If I define "calculation" as mathematical processing of numbers, but not including recall of previously calculated results, then there is no calculation in steps (1) thru (12) above, nor in 1+1 = 2 or 13 is a prime etc. For me, at least deciding if 345* is a prime, definitely is calculation, but I am sure that for some it is only "recall."

Any other comments on what is "calculation" or my mental "vote selection" process?
------------------------------------------
*now changed to 567, thanks to Dale - see my later reply to him.

For me, at least deciding if 345 is a prime, definitely is calculation, but I am sure that for some it is only "recall."It's not prime, it is divisible by 5. :)

-Dale

It's not prime, it is divisible by 5. :)

-Daleouch, that truth hurt. :( (I had first typed 234, but noticed that it was even, so made it 345, without any more thought as that solved the "its even problem.") let me replace with 567. Ok?

I don't think you "calculated" but just "already knew", that if it it ends in 2, 4, 5, 6, 8 or 0, it is not prime.

PS this is a good example of how one can know something, but fail to apply it when needed.

ouch, that truth hurt. :( (I had first typed 234, but noticed that it was even, so made it 345, without any more thought as that solved the "its even problem.") let me replace with 567. Ok?


I can't resist sorry, 567 is divisible by 3. I didn't actively think about any calculation, it just sort of happened, the divisibility by 3 test was just obvous here. Maybe something like 23415167 would be a better choice :)

I can't resist sorry, 567 is divisible by 3. I didn't actively think about any calculation, it just sort of happened, the divisibility by 3 test was just obvous here. Maybe something like 23415167 would be a better choice :)Thanks. You are much better mathematician than me, but I was not making any claim about 567's status as a prime, only noting that for me to find out, I would require what I call a "calculation" - as experienced as you are, perhaps it does not require any calculation for your - it just "pops out at you" that it is not any prime you recognize, so probably is divisible by 3 (only other choice is 7 and if not by 3 then can't be by 9 and now list of possibilites is exhausted, at least usually I think for such small numbers, but may be something bigger, but less than square root, perhaps?) and you can not help but immediately realize that it is divisible by three, skilled mathematician that your are.

In some ways, you are helping to prove my main point: Namely that what is a "calculation" is both hard to define and may be a calculation for one person and only "recall" for another. So I give you thanks again.

PS - Did you vote? I am courous what a mathematician thinks about this so tell you selection, if you do not mind. The ID of the various voters and their choicers were available prior to voting. Perhaps someone who has not voted can still see who voted for what and will post that information before voting.

PS2 - Perhaps someone thinks a new thread "what is calculation?" is desirable? If so, then it will be at least the "grandchild" of the thread that started this one, but I do not remember which thread would be the "grandfather," (father of this pole thread.)

... - it just "pops out at you" that it is not any prime you recognize, so probably is divisible by 3 (only other choice is 7 and if not by 3 then can't be by 9 and now list of possibilites is exhausted, at least usually I think, but may be something bigger, but less than square root, perhaps?)

If using brute force, you have to check for divisors up to the square root of a number to see if it's prime, the worst case for this kind of procedure would be the square of a prime as you won't find any divisor before that. Of course finding a divisor earlier, you'd stop and declare it composite.


In some ways, you are helping to prove my main point: Namely that what is a "calculation" is both hard to define and may be a calculation for one person and only "recall" another. So I give you thanks again.

Absolutely. I think the spirit of your 'no calculation' rule was clear, I was just being difficult :).


PS - Did you vote? I am courous what a mathematician thinks about this so tell you selection, if you do not mind. The ID of the various voters and their choicers were available prior to voting. Perhaps someone who has not voted can still see who voted for what and will post that information before voting.

When I first saw the poll I was going to vote for the "less than 1km" option, taken literally it looked to have no chance of being incorrect and I was going to be difficult :) . Then I saw the discussion with Ophiolite get a little out of hand and thought the better of it. Again, the spirit of your poll's options were clear and I am perfectly willing to forgive the form of your options given the medium as well as it being your first go at formatting a poll (lots of other things to worry about). I didn't really have any intention of voting, besides being difficult, so I haven't bothered.

I don't think you "calculated" but just "already knew", that if it it ends in 2, 4, 5, 6, 8 or 0, it is not prime.Correct, it was recognition not calculation.



I was not making any claim about 567's status as a prime, only noting that for me to find out, I would require what I call a "calculation" For me also. I would have to fire up Mathematica, which undoubtedly would qualify the rest as a calculation.

-Dale

I have avoided this thread because I work with spacecraft. Why isn't "less than 0" one of the answers? It should be.

Suppose two objects are rigidly coupled and that the combined objects have an angular velocity of w about the combined center of mass. The velocities of the centers of mass of the individual objects in the combined center of mass frame is be w x r_i, where r_i are the coordinates of the i^th object's center of mass with respect to the combined center of mass.

Now suppose the objects detach from each other. At that instant, each object will still have an angular velocity of w about their individual centers of mass and will still have a velocity of w x r_i with respect to the combined center of mass. Spacecraft launch does not change the Earth's rotation rate one iota.

With an eastward launch, the accelerating spacecraft also accelerates the atmosphere. The momentum added to the atmosphere will eventually be transfered to the Earth, speeding up the Earth's rotation rate by a tiny, tiny amount.

Thus I vote for a tiny, tiny decrease in the geosynchronous altitude.

Spacecraft launch does not change the Earth's rotation rate one iota.

Conservation of angular momentum says it must. Any system with angular momentum that changes it's moment of inertia must have a resultant change in angular velocity(w) since the mass of the system is constant. By distributing a portion of the mass of the system away from the initial center of rotation, the system must, as a whole, decrease it's angular speed.

(In reply to my post, in which I stated that spacecraft launch does not change the Earth's angular velocity by one iota) Conservation of angular momentum says it must. Any system with angular momentum that changes it's moment of inertia must have a resultant change in angular velocity(w) since the mass of the system is constant. By distributing a portion of the mass of the system away from the initial center of rotation, the system must, as a whole, decrease it's angular speed.

(Post prior to edit:
Bear with me. This will be tough without LaTeX support. I will edit this reply once I have the math worked out with the primitive tools available here.)


We have two objects to consider. I will denote the properties of the combined objects with a suffix c and the individual properties with suffixes s and e for spacecraft and earth
w_c is the angular velocity of the combined Earth+spacecraft system prior to launch
I_c is the inertia tensor of the combined Earth+spacecraft prior to launch
L_c is the angular momentum of the combined Earth+spacecraft prior to launch: L_c = I_c w_c
m_s and m_e are the masses of the spacecraft and Earth
I_s and I_e are the inertia tensors of the spacecraft and Earth about their respective centers of mass
L_s and L_e are the angular momenta of the spacecraft and Earth about their respective centers of mass
x_s and x_e are the position vectors of the spacecraft and Earth centers of mass in the combined center of mass frame
r_s and r_e are the magnitudes of the position vectors x_s and x_e
v_s and v_e are the linear velocities of the spacecraft and Earth centers of mass in the combined center of mass frame
L_s,v and L_e,v are the angular momenta of the spacecraft and Earth centers of mass moving in the combined center of mass frame
w_s and w_e are the angular velocites of the spacecraft and Earth about their respective centers of mass after launch.


I will do my work in the nonrotating combined Earth+spacecraft center of mass frame, which I am assuming is an inertial frame.

State prior to launch
The position vectors and masses are related by
m_cx_c + m_ex_e = 0

The spacecraft and Earth center of mass velocities are
v_s = cross(w_c,x_s)
v_e = cross(w_c,x_e)

The combined inertia tensor is, by Steiner's parallel axis theorem,
I_c = I_s + m_s(Ir_s - x_s^Tx_s) + I_e + m_e(Ir_e - x_e^Tx_e)
where I is the 3x3 identity matrix.

The combined angular momentum is
L_c = I_c w_c
= I_sw_c + m_s(Ir_sw_c - x_s^Tx_sw_c) + I_ew_c + m_e(Ir_ew_c - x_e^Tx_ew_c)

The vector triple product helps simplify this ugly mess:
A x (B x C) = (A . C) B - (A . B) C
Denoting A=C=x_s, B=w_c, then
Ir_sw_c - x_s^Tx_sw_c = dot(x_s,x_s)w_c - x_sdot(x_s,w_c) = cross(x_s,cross(w_c,x_s))

Thus
L_c = I_sw_c + I_ew_c + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

The kinetic energy prior to launch is purely rotational:
E_c = 1/2*dot(L_c,w_c)

Conservation of linear momentum
Prior to launch, the combined Earth+spacecraft system has zero linear momentum in the combined CoM frame. Linear momentum will be conserved across the very brief (instantaneous) interval between pre-launch and post-launch, and thus the velocities remain unchanged:
v_s = cross(w_c,x_s)
v_e = cross(w_c,x_e)

Conservation of angular momentum
The angular momentum of the spacecraft and Earth after launch is the sum of the individual angular momenta about their centers of mass plus the angular momenta of the spacecraft and Earth masses about the combined center of mass:
L_post-launch = I_sw_s + I_ew_e + m_scross(x_s,v_s) + m_ecross(x_e,v_e)
= I_sw_s + I_ew_e + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

Equating the pre-launch and post-launch total angular momentum,
I_sw_s + I_ew_e + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e)) = I_sw_c + I_ew_c + m_scross(x_s,cross(w_c,x_s)) + m_ecross(x_e,cross(w_c,x_e))

or

I_sw_s + I_ew_e = I_sw_c + I_ew_c

Conservation of energy
The above equation indicates that any change in the rotation rate of the Earth must be compensated with a corresponding change in the rotation rate of the spacecraft. The final step, which is just too ugly to do without LaTeX, is to realize that the kinetic energy of the combined system does not change between the instant before and instant after launch. With this,
w_s = w_e = w_c

i.e., there is no change in the Earth's rotation rate due to launch.

With an eastward launch, the accelerating spacecraft also accelerates the atmosphere. The momentum added to the atmosphere will eventually be transfered to the Earth, speeding up the Earth's rotation rate by a tiny, tiny amount.
What about the rocket exhaust?

Bear with me. This will be tough without LaTeX support. I will edit this reply once I have the math worked out with the primitive tools available here.

Okey dokey.

What about the rocket exhaust?

There are two effects. When the rocket first takes off, the plume from the exhaust impacts the Earth. The spacecraft launches vertically and the plume is thus directed through the Earth's center of masy. The pluming therefore changes the Earth's translational velocity by some amount, but does not change the rotation rate by much at all (torque = cross(r,F), and r is nearly normal to F).

After the spacecraft has taken off and turned to the East, the spacecraft exhaust does have some component normal to its position vector. This exhaust is moving faster than the atmosphere (in inertial space). The exhaust soon transfers this energy to the atmosphere, increasing the Earth's angular velocity (by a tiny, tiny amount).

Edited to add:
The exhaust is moving slower than the atmosphere until the rocket crosses the sound barrier (the ideal exhaust speed is the speed of sound). Once the rocket crosses the sound barrier, the exhaust is moving faster than the atmosphere inertially.

DH,

Thanks for the analysis. My understanding is this. Please correct me if I am wrong.

The fact that mass is moving away from the earth is exactly compensated for by the effect of the energy of the rocket exhaust being transferred to the atmosphere (indirectly to the earth). Yes? So, if we were to launch from the moon, with no atmosphere, simple conservation of angular momentum would dictate that the angular speed of the system must slow. Yes?

DH,

Thanks for the analysis. My understanding is this. Please correct me if I am wrong.

The fact that mass is moving away from the earth is exactly compensated for by the effect of the energy of the rocket exhaust being transferred to the atmosphere (indirectly to the earth). Yes? So, if we were to launch from the moon, with no atmosphere, simple conservation of angular momentum would dictate that the angular speed of the system must slow. Yes?

No. Once the masses are separated they do not interact rotationally (not quite true, but close enough). There is no change in angular velocity because of launch. The atmosphere is a secondary effect. Gravity gradient torque (something I did not discuss) is a secondary effect.

Remember that to conserve of angular momentum you must also consider that just after launch, the spacecraft and planet centers of mass have linear velocities that are normal to the spacecraft and planet position vectors relative to the combined center of mass. This adds to the angular momentum of the combined planet+spacecraft system. The math is a bit hairy, but the result is very simple. There is no change in angular velocity.

Think of the planet as a very large spacecraft. NASA knows quite well how docking and undocking change vehicle states; they have been doing this since the 1960s. While docking changes the angular velocity, undocking does not.

Mainly to D.H.:

I have only skimmed very briefly first part of your long post, and did not even read most of it, because I think you are concerned with the angular momentum of the Earth + satellite system about the sun. If this is the case, I agree with you that is will not change "one iota" by placing a satellite in orbit about the Earth.

We are concerned with the spin rate change of Earth alone, about its axis, when some angular momentum (originally part of the Earth's total) is removed from Earth by being placed into orbit.

I too have worked a little with satellites. I assume you know well the de-spin system often used, which has two small symetric weights tied at ends of cords, which are initially wrapped around the satellite (and held so until one wants to "de spin" satellite).

For benefit of those not so familiar with this cheap, fuel-less highly reliable de-spin system: When the weights are released, they move away from the body of the spinning satellite, but much more slowly because they are partially restrained by the force in the cords tying them to the circumference of the satellite. This force, times the radius of the satellite, provides a torque to de-spin the satellite (unless your technician wrapped them the wrong way around. :mad: Then as my space group would say: "you are in deep yogurt" with a more rapidly spinning satellite)

Now imagine that the iron core of the Earth is all that will remain of "Earth" and we throw all the rest of current Earth off into space in two symetric halves, (like the de-spin weights just discussed, but without any cords. For example, as if some "dark energy" condensed on the iron core and cancelled the gravity of the iron core on the higher layers so they just went flying off into space.) Most of the original angular momentum of current Earth would be found in the throw off crust parts and the iron core would be "de spun" just as a satellite can be.

Our case is not this extreme de-spining of "remaining Earth" case - just the opposite. Only a very small part of the Earth's angular momentum is removed from what is a reduced Earth Mass. Answer is made harder to guess as it is expresed in altitude change of the geostationaly orbit with a slower spinning "remainder Earth"

Answering you, has made me realize, consciously for the first time, that the moment of intertia of "Remainder Earth" has decreased as the 1000Kg of satellite was originally at full Earth radius - so inertia "I" of "remainder earth" will be very slightly less. This reduction of I tends to speed up Earth if it were the only thing happening.

This stimulates, in me, an possible new pole for guessing:

How does the current tidal slowing of Earth by moon compare to the slowing man has made by digging mines?

(All mined matterial assume to be now on the surface of a perfectly spherical Earth.) I will let someone else set that one up, if the want to do so.

D. H. - If you were concerned about our problem (and not the angular momentum about the Sun problem, where "not one iota" is correct) please say so, and I will fully read your long post.

DH and Billy,

Thanks very much. I am an engineer, but I deal largely with things like volts, ohms, bytes and code loops.

Billy T,

First, I did not mention the sun at all. In fact, I ignored that the sun even exists. I am concerned with the rotational and translation states of the Earth and satellite as observed in the non-rotating reference frame with origin at the Earth+satellite center of mass. To answer your final question, I am concerned with the problem posed in this thread.

Your de-spin example works because the weights are tethered to the satellite and because no external forces and torques are involved. The weights are still coupled to the spacecraft. Strings are attached!

This example does not apply to a spacecraft launch. The spacecraft, once it launches, is no longer connected to the Earth. The spacecraft exerts no torque on the Earth (well, almost no torque; I ignored gravity gradient torque exerted on the Earth by the spacecraft).

No strings are attached -- how could the spacecraft exert any torque? Because the spacecraft exerts negligible torque on the Earth, it doesn't matter where the spacecraft goes -- LEO, GEO, or Mars. The question then becomes how does spacecraft launch change the angular momentum of the Earth?

As I have shown, the Earth's rotation rate at the instant after launch is exactly the same as it was just before launch. Spacecraft launch does not change the Earth's rotation rate by one iota.

For your final example, suppose the Earth is quartered like an apple, with the core remaining intact, and the four quarters go flying off with the linear velocity w x r. The inner core would not be "de-spun" unless strings are attached.

I voted for the first option, because the rocket exhaust accelerates a section of air, which doesn't affect earth's rotation too much.

I guess my answer is for a satellite with a long string attached then.

No strings are attached -- how could the spacecraft exert any torque?
Through the exhaust. The rocket pushes on the exhaust, the exhaust pushes on the atmosphere, the atmosphere pushes on the Earth.


The exhaust is moving slower than the atmosphere until the rocket crosses the sound barrier (the ideal exhaust speed is the speed of sound). Once the rocket crosses the sound barrier, the exhaust is moving faster than the atmosphere inertially.
Before the rocket reaches the exhaust speed, the exhaust is acting to slow the Earth through the atmosphere.

After reaching the exhaust speed, subsequent exhaust will act to speed up the Earth, but the total speed-up must be smaller than the total slow-down. Do you see why?

Think about the momentum in the Earth-atmosphere frame of: 1) slow exhaust 2) fast exhaust 3) the rocket. They must sum to zero, right?

Think of the planet as a very large spacecraft. NASA knows quite well how docking and undocking change vehicle states; they have been doing this since the 1960s. While docking changes the angular velocity, undocking does not.

I suspect that undocking would change vehicle states if the undocking ship fired it's boosters in the direction of the larger ship. However in the case of the rocket and the planet - it seems like gravity would counteract much of the force of the blast. Gravity is pulling the two objects together (but it should be said that the energy of attraction was greater at launch because of the closer proximity), chemestry is seperating the two. I think the net effect would be close to neutral depending how the smaller vehicle accelerated to excape velocity and if it accelerated to excape velocity. The planet may not excape a net effect if the rocket left very quickly in a straight line away from the planet. In other words, the effect of a rocket varries depending on how it leaves the planet - (it could rotate the planet picking up speed for example but that would change the rotational velocity of the planet).

But the point I wanted to make is that the pull of gravity is a substantial element and is pulling the planet toward the rocket ever so slightly as the rocket tries to excape.

Wait. Hold on. Is this true or not:

Conservation of angular momentum is defined by:

I_initW_init = I_finalW_final

Right?

So, the final angular velocity of anything is:

W_final = I_initW_init / I_final

Yes? So, since I is dependent on mass, if we pitch some mass away from the earth, normal to it's surface, it must spin faster. Right?

I think DH started out with this. I also think DH said that once the rocket leaves the surface it no longer has any effect whatsoever on the L of the earth.

Now, if a rocket starts out normal to the surface and stayed that way, the earth would spin up by the equation above. If the rocket starts normal, then slowly tips over high up in the atmosphere, the reaction of the exhaust (opposite to the earths rotation of course) will tend to slow the rotation of the earth, but the transfer of linear momentum must be incredibly small on the thin atmosphere.

Wait. Hold on. Is this true or not:

Conservation of angular momentum is defined by:

I_initW_init = I_finalW_final

Right?

So, the final angular velocity of anything is:

W_final = I_initW_init / I_final



No. Three things are wrong here. First, the pre-launch inertia tensor is a combination of the Earth inertia tensor, the satellite inertia tensor, and the separation of Earth and satellite centers of mass from the combined center of mass (Steiner parallel axis theorem).

Secondly, the system after launch comprises two objects. You have too account for the angular momentum from each. Both the Earth and spacecraft are rotating after launch.

Finally, you also have to account for the post-launch angular momentum that results from the Earth and satellite velocity components that are normal to their position vector with respect to the combined Earth+satellite center of mass.

Through the exhaust. The rocket pushes on the exhaust, the exhaust pushes on the atmosphere, the atmosphere pushes on the Earth.


Before the rocket reaches the exhaust speed, the exhaust is acting to slow the Earth through the atmosphere.

Agreed. I edited my post on the effect of the exhaust to add that qualification.


After reaching the exhaust speed, subsequent exhaust will act to speed up the Earth, but the total speed-up must be smaller than the total slow-down. Do you see why?

Think about the momentum in the Earth-atmosphere frame of: 1) slow exhaust 2) fast exhaust 3) the rocket. They must sum to zero, right?

I don't see that. So many factors come into play:
Atmospheric pressure degrades rocket performance. The exhaust speed (relative to the rocket) is lower at high pressure that at vacuum. The exhaust velocity increases as the rocket gains altitude and gains speed. This argues for increasing the Earth's rotation rate.
Escape velocity versus exhaust velocity. The exhaust velocity is ideally the speed of sound at the temperature and pressure at the bell of the rocket. This is on the order of 1000 mph, much lower than escape velocity. The rocket spends more time getting from the exhaust velocity to the escape velocity than it does in getting to the exhaust velocity in the first place. This argues for increasing the Earth's rotation rate.
Atmospheric friction. The spacecraft is drilling a hole through the atmosphere. This argues for increasing the Earth's rotation rate.
The rocket equation. It takes a lot of energy just to get the rocket off the launch pad. The acceleration increases as the rocket depletes mass. This argues for decreasing the Earth's rotation rate.

In any case, the total effect is tiny.

In any case, the total effect is tiny.
Well, we can certainly agree on that much.

Just one question:
Atmospheric friction. The spacecraft is drilling a hole through the atmosphere. This argues for increasing the Earth's rotation rate.

Isn't this force canceling out some of the force of the launch? I mean if you fired a projectile from the earth the resistence from the atmosphere would keep that energy in a closed system?

But I guess that the atmosphere is a barrior and if the rocket was pushing against it, that could affect rotation velocities. But the gas that was accelerated out of the rocket has an equal and oppisite effect doesn't it?

Just one question:

Isn't this force canceling out some of the force of the launch? I mean if you fired a projectile from the earth the more resistence from the atmosphere would keep that energy in a closed system?

This isn't a closed system. The rocket engines are firing.

Can we agree that if a rocket was up in the atmosphere at a 70 degree angle or so, and it encountered a barrior, and was pushing and pushing against it - the net effect would be 0. The particles from the engine effect the earth when they or particles they have influenced hit the earth. Can we agree on that?

Wow. I'm learning more about angular momentum than I expected. It's interesting that I can find no examples of how mass loss effects (or dosen't effect) the angular momentum of a rotating body. So, if I am spinning in a chair holding a weight, and I drop the weight (no torques applied) my rotation rate will remain unchanged. So, if we do our experiment on the moon (no atmospher), and the rocket launches normal to the surface, the answer to the poll is exactly zero change. Yes?

Several things DH states I can agree with; for example, that without "strings," and gravity weakening so that current spin of Earth was able to throw off parts of outer shell of Earth, then the iron core remaining would maintain its current spin rate (assuming no "parting torques" during the separation of the outer layers)

But on the main point I think Pete (and I) are correct. I.e. There is a very tiny decrease in the spin rate of Earth, cause by placing the satellite into orbit.

Below I prove DH is wrong (I think) by simply considering the "before launch" and "after in orbit" states, without any need to go into details of the exhaust gases speeds etc. So let me offer a simple proof that there is a change in spin rate:

Let the total angular momentum of the pre-lauch system, Ao, be decomposed into two parts, Ae, is everything but the rocket and satellite angular momentum, Ars. I.e. Ao = Ae + Ars, but for the initial conditions, subscript "o," the satellite is in the rocket on the launch pad, waiting for launch.

Let the initial tangential speed of the of the launch pad, (which happens to be flush with surface of the Earth, for people who like to quible), in a non rotating reference frame fixed in the center of a spherical Earth of radius "r" be called: "Vo" and this speed is Vf after the satellite is in geostationary orbit at radius "R" Clearly R > r, as the notation implies. The question being discussed with DH is which of the following is true:
(1) Vf = Vo
or
(2) Vf < Vo.

The thread has assumed that (2), not (1), is true and DH has questioned, to say the least, this assumption. DH claims (1) is true.

All agree angular momentum must be conserved, I.e. that Af = Ao so lets calculate (actually just express) both:

Ao = Ae + Ars = f.r.Me.Vo + (r+h).Mr.Vo{1+ h/r}

Where to avoid use of "I" and "pi" I have imagined that the entire mass of the Earth, Me, is traveling at speed Vo perpendicular to a radial line from center of the Earth, but is only f.r distant from the center of the Earth. I.e. f < 1, and its value makes the first term on the right hand side of equation above exactly Ae. Periods in equations represent multiplication. (Note Me does not include, Mr, the mass of the rocket plus satellite on launch pad.) For computational convenience, the satellite is assumed to be at the center of mass of the rocket and both are h above the surface of the spherical Earth, whose radius is r.

The strange term inside { } is to account for fact that the center of mass of the rocket is not only farther than r from center of Earth, - i.e. at (r+h), but also has slightly greater tangential speed than Vo.

The final angular moment of the system is:

Af = f.r.Me.Vf + r.(Mr - 1000Kg).Vf + R.(1000Kg).Vg.....*

Where Vg is the speed of the 1000Kg geostationary satellite in orbit at radius R from center of the Earth and as before, since the satellite is geostationary:

Vg = Vf {R/r} so with this subsitution in last equation above and subtracting two equations:

Af - Ao = 0 = f.r.Me.(Vf -Vo) + r.Mr.Vf - r.(1000Kg).Vf + R.(1000Kg).Vf.{R/r}
-r.Mr.Vo.{1+ h/r} - h.Mr.Vo.{1+ h/r}.
or
0 = (f.r.Me + r.Mr)(Vf -Vo) - Mr.Vo.h - h.Mr.Vo.{1+ h/r}
- (1000Kg).Vf.{r - (R^2)/r}

but the two terms of the first line of equation immediately above prefixed with negative signs are:

- Mr.Vo.h - h.Mr.Vo.{1+ h/r} = - Mr.Vo.{2h + (h^2)/r)}

and separating out the negative part of the final term and changing signs of all terms (one sign changed by moving first term to "zero side" of equation)
we have:

(f.r.Me + r.Mr)(Vf -Vo) = Mr.Vo.{2h + (h^2)/r)} + (1000Kg).Vf.{r - (R^2)/r}

or "Eq. A" is:

(Vf - Vo) = [Mr.Vo.{2h + (h^2)/r)} + (1000Kg).Vf.{r - (R^2)/r}]/(f.r.Me + r.Mr)

For DH to be correct, both sides above of "Eq A" must be exactly zero.
For Pete & me (and several others I think) to be correct, both sides must be negative. (Assuming I have not made mistake in algebra, which is surely possible.)

The numerator in right side of "Eq.A" above has two terms. The first is clearly positive and because {r - (R^2)/r} = {r^2 - R^2}/r and R is much greater than r, the second term is clearly negative.

Thus it is clear that DH can not be correct as if (Vf - Vo) were zero, then these two opposite sign terms in "Eq A" must ALWAYS satisfy:

Mr.Vo.{2h + (h^2)/r)} = - (1000Kg).Vf.{r - (R^2)/r}
or if Vo = Vf as DH contends, then

Vo/Vf = 1 = - (1000Kg).{r - (R^2)/r} / Mr.{2h + (h^2)/r)}
or, multiplying both sides by r:
r = - (1000Kg).{r^2 - (R^2)} / Mr.{2h + (h^2)/r}
or
r = (1000Kg/Mr)[(r+R)(R-r) /{2h + (h^2)/r)}] = (1000Kg/Mr)[K], where K is some constant.

The radius of the Earth certainly is NOT a linearly function of the mass of the satellite to mass of rocket ratio!

Thus, either I have an algebraic mistake (quite possible, but not likely to change the conclusion) or DH is wrong.

Showing that Pete and I are correct, by this approach may more difficult, but I will work on it IF I find time.
---------------------------------------------
*In this equation, I am being kind to DH by assuming that all of the exhaust has condensed back onto the surface of the Earth, as if the rocket fuel were oxygen and aluminium dust and the resulting Al2O3 was resting on the ground in the final state. In reality the raising the fuel mass, converted to exhaust gases in the air, to some average altitude will by itself slow the Earth spin rate, just as a fast spinning ice dancer, with arms at her side, slows her rotation rate when she extends her arms out again.

Below I prove DH is wrong (I think) by simply considering the "before launch" and "after in orbit" states ... All agree angular momentum must be conserved, I.e. that Af = Ao

Angular momentum is most certainly not conserved between the "before launch" and "after in orbit" states. The spacecraft throws itself into orbit by firing its rockets. Those rockets change the energy of the system; the conservation laws do not apply.

Angular momentum is only conserved between the "instant before launch" and "instant after launch" states because the rockets haven't had any time to do any work.

Angular momentum is most certainly not conserved between the "before launch" and "after in orbit" states. If that were true, then you should be able to tell what "outside of system" torque has been acting. Recall that all of the exhaust gases and solid rocket parts are back on surface of Earth (by my "kind" aluminum dust+ oxygen fuel assumption discussed in footnote).


If you need to, imagine that pre-launch Earth is far from sun, moon and all other objects. I.e. There is no outside agent to act. The angular momentum of a closed system such as this is conserved, with no torques from the outside the system acting NO MATTER HOW YOU RE-ARRANGE THE PARTS OF THE SYSTEM. No matter what forces one part of the system applies to another of the system to achieve the re-arrangement or how complex these forcres are, etc.

If you do not believe the standard "no change in angular momentum unless system is acted upon by torque applied by outside agent" law, please state your different requirement or conditions under which angular moment is conserved.

I think BillyT and D H would be in agreement if the method of launch for the satellite had been defined as an equatorial earth-mounted cannon, pointed east, aimed toward the horizon at some angle. Certainly, in this case, there would be some slowing of the earth's period of rotation, however miniscule, (according to both D H and BillyT's hypotheses).

If we allow the cannon to recoil back on wheels, intuitively it seems like the the earth's rotation would not be affected as much, but in theory it should still be affected by the same amount because of conservation of (angular) momentum laws. In a way, this seems analogous to fuel and the atmosphere in the original experiment. It is not easy (for me, anyway) to visualize all of the angular momentum being transferred back to the earth via the fuel particles settling out of the atmosphere.

Ok...it's been years since my classical mechanics class, but I believe DH is correct regarding launching a rocket from the surface of a rotating sphere. However, I believe you could easily fix the problem by modifying the question so that you are launching your satellites with some sort of cannon/railgun/whatever.

Edit:woops, Neddy Bate beat me to it.
Imagine a figure-skater spinning on her tows with her arms outstretched, holding a metal ball. Her angular velocity won't change if she simply releases the ball, even if the ball later accelerates under its own power; the ball no longer has anything to do with her angular velocity once it's released. On the other hand, if she actively tosses the ball away from her tangential to her direction of rotation, she would indeed slow down slightly.

...It is not easy (for me, anyway) to visualize all of the angular momentum being transferred back to the earth via the fuel particles settling out of the atmosphere.It is good that you can not because it is not. The momentum of the satellite in geostationary orbit has increased from the value it had while sitting on the launch pad at the center of mass of the rocket. Ergo the angular momentum of all other parts of the system has decreased by an equal amount.*
----------------------------------------
*To be absolutely correct in this, one should catch all the falling rocket parts and Al2O3 dust on a platrom with altitiude h above the Earth, so that there has been exactly zero change in the in radial distantance of this mass (at height h both before and after satellite is in orbit at radius R.) Then the first paragraph's decrease of Ae is exactly compensated by the increase of the angular momentum in the satellite.

...However, I believe you could easily fix the problem by modifying the question so that you are launching your satellites with some sort of cannon/railgun/whatever. ...Perhaps that would make it easier for you to see /understand, but it makes absolutely no difference what forces are used to place the satellite into orbit,* so long as satellite is the only thing to permanently leave the earth and the mass distribution of all else remains the same. (No change in the moment of inertia of anything but the satellite. -See footnote of the post I just made replying to Ned.)
-------------------------------------
* To be extreme (and have some fun): Assume rocket starts out below the sea and air pressure in a polaris sub's launch tube sends it to the surface. There it ignites, but seem to be sputting. Fortunately a cowboy in the "photograph helicopter" throws a rope arround the sputtering rocket. The force of sputtering rocket plus the lift of the helicoper get it to 10,000 feet, but it is wildly swinging and the sputtering exhaust burns thru the rope, so without the lift of the helicopter, it begins to fall back towards the sea, but fortunately at 1000 feet, enough fuel has been burnt so that even the sputteing fuel rocket can keep its self in level flight for 3 minutes. Then the sputtering stops and full rocket thrust begins. The guidance system get it headed upwards. At "flame out" it is finally going fast enough to coast to thru the last part of the atmosphere and coast into elliptic orbit. It's increditable luck continues as it is hit by another satellite and after this collision is in perfect circular geostationary orbit at R from the center of the Earth!!!! (The "other satellite" no doubt has its angular moment altered, as did the atmosphere, as did the sub, as did the sea, as did the helicoper, the rope, etc. but all these are in the term Ae, so these details as to how Ae is divided up do not matter.)

If I may steal from DH:

All these details do not make "one iota" of difference. What I have shown mathematically by considering only the initial and final states remain exactly unchanged!

Ain't "conservation of momentum" wonderful (and powerful)! ;)

Billy T, you are wrong, AGAIN. The rocket does not 'steal' angular momentum from the Earth to gain orbit. It is powered by its chemical engines. No angular momentum is lost from the Earth, no work is done by the Earth. No, I am not a physicist, but I knew this before D H gave his excellent analysis. That is the main reason I didn't vote in your gedankin, you had no 'zero' option.

Angular momentum is most certainly not conserved between the "before launch" and "after in orbit" states. The spacecraft throws itself into orbit by firing its rockets. Those rockets change the energy of the system; the conservation laws do not apply.
:bugeye:

So what if the (kinetic) energy of the system is changed?
Angular momentum (and linear momentum, of course) must still be conserved.

Imagine a figure-skater spinning on her toes with her arms outstretched, holding a metal ball. Her angular velocity won't change if she simply releases the ball, even if the ball later accelerates under its own power; the ball no longer has anything to do with her angular velocity once it's released.
Right. Unless whatever the ball uses as reaction mass is tied to the skater, or later comes to rest on the skater.

Eg if the ball accelerates in a straight line by ejecting ball bearings behind it which are caught by the skater, the skater will slow down, right?

Wow. I'm learning more about angular momentum than I expected. It's interesting that I can find no examples of how mass loss effects (or dosen't effect) the angular momentum of a rotating body. So, if I am spinning in a chair holding a weight, and I drop the weight (no torques applied) my rotation rate will remain unchanged. So, if we do our experiment on the moon (no atmospher), and the rocket launches normal to the surface, the answer to the poll is exactly zero change. Yes?
Yes, if we ignore gravitational interaction :)

Of course, I don't want to go there... because then the answer would depend on the relative angular position of the rockets, and to get really pedantic we might have to consider the interaction between the two rockets as well, which puts us in the realm of a three-body problem. :eek:

Edit:woops, Neddy Bate beat me to it.
Imagine a figure-skater spinning on her tows with her arms outstretched, holding a metal ball. Her angular velocity won't change if she simply releases the ball, even if the ball later accelerates under its own power; the ball no longer has anything to do with her angular velocity once it's released. On the other hand, if she actively tosses the ball away from her tangential to her direction of rotation, she would indeed slow down slightly.

I think this captures the problem with the anology of the skater releasing a weight. Every action has to have an equal and oppisite reation. If there were no gravity or atmosphere maybe the anology would be correct - but it isn't is it. A better anology would be a girl made out of steel throwing magnets into orbit around her. There may as well be a string between the magnet and the girl - both pulling on eachother. You would have to take into consideration conservation of angular momentum.

Billy T, you are wrong, AGAIN. The rocket does not 'steal' angular momentum from the Earth to gain orbit. ....Never said it did "Steal", only that the total system angular momentum is constant, so if one part increases (the satellite) the remained must decrease. You are welcome to your opinions, even if contradicted by mathematical analysis, as they are and have been.

Never said it did "Steal", only that the total system angular momentum is constant, so if one part increases (the satellite) the remained must decrease. You are welcome to your opinions, even if contradicted by mathematical analysis, as they are and have been.

Billy T, when I said the satellite does not 'steal' any angular momentum from the Earth, that is exactly what I meant. The satellite has exactly the same angular momentum sitting on the launch pad as it does in orbit, geostationary or otherwise. Angular momentum is conserved in both the satellite and in the Earth's rotation. It does not increase or decrease for EITHER body. The KINETIC energy of the satellite does increase, the conversion of gravitational energy (potential energy) to kinetic energy.

I don't think many people understand angular momentum or the conservation of it, and I don't think I want to argue or explain it. I'll just ask how is an orbiting body different then an ice skater that abducts her or his arms? An orbiting body in geosync. orbit is "tethered" by gravity as much as the skaters arms are tethered by bones and ligaments.
http://en.wikipedia.org/wiki/Angular_momentum


The satellite has exactly the same angular momentum sitting on the launch pad as it does in orbit, geostationary or otherwise

Oh, really. How would you like to be hit by boxer with 3 foot arms as opposed to a boxer with 1 foot arms (depending on type of punch thrown)?
An object traveling around the planet has much more energy for the very same reason.

Get 2 pieces of strings, one 6in long and one 3 feet long. Tie a small rock on the ends. Which do you think would have more energy if you swung them around your body as you were turning in a circle?

Now after you've thought about this - do you really think a satellite has the same amount of energy on the surface of the earth as it does in space?

If you are still not convinced, I have one more experiment for you:
Step 1. Take the rocks with strings on them and make one 2" long and one 12" long.
Step 2. Swing the one that is 2" long as hard as you can with your dominent hand and hit your self square on the top of the head with it. Measure the bump.
Step 3. Repeat step 2 only use the one 12" long.
Compare the size of the bumps and if bump number 1 is smaller then bump number 2 then you don't understand the conservation of angular momentum correctly. If bump number 1 is bigger then you may submit your evidence (the bumps on your head) to this forum by way of digital pictures, and if you're convincing enough, we may help you to rewrite the law of angular momentum!!

After flipping a coin, having destroyed my weejee board in a trash fire long ago, I have decided to guess that an object in orbit around the Earth has greater angular momentum than it has if it is sitting still on the surface of the Earth.

After the thread is over, I might want to calculate it. Does anyone know who to ask for the definition and mathematical equation of angular momentum?

...The satellite has exactly the same angular momentum sitting on the launch pad as it does in orbit, geostationary or otherwise....Perhaps you do not know the definition of angular momentum? For a point mass (as our satellite has been assumed to be) the magnitude of the angular momentum about a point "X" is the product of following three things:

(1) Mass at the point, M.
(2) Distance, D, along the vector line, L, from the point X to the mass point.
(3) Speed, S, perpendicular to the "X" to "mass point" line.

Thus the magnitude of the angular momentum about any point "X" is MDS. In our case of interest, "X" is the center of the Earth, but please note that the angular momentum is linear in the value of D, which would be different if the angular momentum were being claculated about some other point. For example, the angular momentum (neglecting quantum "spin") about the point mass itself is zero because then D = 0.

Both "D" and "S" are much greater when the satellite is in orbit than when it was on the launch pad. Thus you comment is incorrect.

Note, however, that angular momentum is a vector. How that vector is calculated is discused below:

There are two different well-defined "multiplication operations" for vectors. One called the "cross product" and the other the "dot product." Put the word "vector" in front of these names and Google to read more.

Speed, S, is not a vector, but velocity, V, is.
(A very common convention, used here, is to make vectors bold.)

The above names for these product come from their standard notation. I.e. for the two vectors given above:

The dot product is written: L . V where the "dot" should be a little higher, but impossible to do here. L*V looks better, but does not make the origin of the name as clear. The result of the dot product is always a scaler. I will not explain it further, because it is of no use for this problem, but hope you will "google" to learn what it is, when it is useful, etc. I just note it is very useful, sometimes essential, in calculating the work done or changes in potential energy as something is moved in a force field etc. (Just to be complete, I should have "conservative" in front of "force field" as if force field is not "conservative," then the simple concepts of "potential" do not exist, but that is an entirely different lecture.)

The cross product is written: L x V.
The result of the cross product is always another vector, which is perpendicular to the plane formed by the two original vectors. (The side of that plane on which it extends requires too many words to describe here, but is well defined by the "right hand rule.")

In this vector notation, the definition of angular momentum, A, of point mass about X is:

A = M(L x V)

The equation, given above, (A = MDS) for the magnitude only follows from this definition. (There is no other definition, only this one. - One should never give two different definitions for the same thing, unless you have a solid proof that the two definition are in fact identical, but only "look" different. How the A = MDS equation follows from the definition is illustrated near the end of this lecture.)

If you think, even slightly, what the effect of L is, you will understand that the satellite's angular momentum is much larger when the satellite is far from the center of the Earth. Also it is much larger because the "perpendicular speed", S, which in this circular orbit case is the magnitude of the velocity vector, is much larger also. (On the equatorial lanch pad the satellite travels approximatel 25,000 miles around with the Earth in 24 hours, but much much farther in 24 hours when it is in the geostationary orbit.)

Note speed is always the magnitude of the velocity, but only at four points in an elliptical orbit would the velocity be perpendicular to L. That is, at all other points in an elliptical orbit the following is true:
|L x V| / |L| < S where the vertical bars indicate magnitude. (I think you can understand why this is true, if you try.)

For the perfectly circular orbit, however, the following is always true:
|L x V| / |L| = S.

You should learn both what the definition of angular momentum is and how to use this powerful, compact, vector notation. I hope this helped you, but expect it will not; however, I like to teach, even though I am now retired, and think at least some will learn from this effort, even if you do not.

To make another example of what you can do with conservtion of angular momentum, consider a comet at apogee and perigee using:
A = M(L x V). Note this is true at ALL points on the eliptical orbit. Not only true, but A is constant in both magnitude and direction! In fact this is not restricted to objects in orbit, but follows from the definition for all objects at all times, in any type of motion or stationary! (It is the vector form of the definition of angular momentum - so must be true "by definition.") For "extended objects" you apply this to each "point" and integrate over the volume of the object to get the total angular momentum of the object about the point "X."

Apogee and perigee are two of the 4 points in an elliptical orbit where:
|L x V| / |L| = S
so
|A| = M|(L x V)| = A = M(|L|)S, but we have called |L|, "D" above, so for the comet at apogee and perigee (plus two other points):
A = MDS as before, but note this is a scaler equation about magnitudes, not a vector one.

Thus the ratio of the comet speed at apogee (point most distant from sun) to perigee (closest point on orbit to sun) is:
(Va / Vp) = (Sa / Sp ) = |Lp| / |La| = Dp / Da, where the subscripts refer to the two respective points.

Or in words: At apogee and perigee, the speed ratio is the inverse of the distance ratio. This comes directly from the definiton of angular momentum given above at the start of this lecture, and fact it is concerved.

My "reducto-ad-adsurdum" proof that the satellite's Ab is not same as Af, where subscript "b" is "before" and "f" is "after", also comes from this definition and conservation of A.

OPINIONS to the contray, do not count for much against math, logic (algebra) and definitions.

Hay! That looks about right. Thanks again, BillyT!

Maybe 2inq' will get it.

I have a confession to make. I already knew about angalur momuntem. I was trying to be sacrostic.

Hay! That looks about right. Thanks again, BillyT! ...MY "lecture post" just prior to your request was not made in response to you intentionally as I was writing, editing and still modifying it while you posted, but it seems to be just what you were asking for. Please look at it again, especially the last parts of the "lecture" as more has been added. BTW, you are welcome.

... main reason I didn't vote in your gedankin, you had no 'zero' option.I regret, now, that I did not include two additional response choices. I will try to extend and correct the pole results informally to permit the following "omitted choices":

Choice "Neg" is:

Satellite "B" is actually lower than "A" so distance B is above A is negative.

Choice "0" is:

Both geostationary satellites are at the same distance from center of the Earth, because the launce of B does not make "one iota" change in length of day / Earth spin rate, etc.

I thank you for not voting and now record your vote (and that of DH) for Choice "0". - Thus DO NOT VOTE AGAIN. However, you can help me if you can see who has voted for which group still and will post a table of this information. That information was available to me,prior to my voting, so I suspect it is still available to you. Thanks for the effort, if you can inform us all via a post. It will aid me in correcting for my error of omission. (The two new choice should have been there from the start. - I erroneously assumed no one would want them, but infact they are just as popular as my guess, and two others!)

Perhaps some of the 10 (especially concerned that DH may be one) who have voted for the official "less than E-4M" choice because that band does include both of the new choices, will inform us that that is what they did. For example if DH voted for that band, then the current "10 COUNT" of that band is really only 9. IF YOU REALLY WANTED ONE OF THE TWO NEW CHOICES AND VOTED FOR ONE OF THE LISTED BANDS, please inform which and which of the two new answer choices you prefer and I will correct the final vote manually.

I may again post the following summary, if new information is available in new posts made by people, like you and DH, who really wanted one of the two new anwswer choices above:

SUMMARY of Voting corrected for new choices:
Neg.......0
"0"........2
< E-4.....9*
< E-2.....1
< 1M .....2
< 100M...1
< 1000M .0
> 1000M..2
------------
*May be 8 or less. If it is 8 or less, total of the two new choices will be more than 2. Has been reduce from the offical 10 count to correct for DH's expressed wishes. (See my quote of his post, made soon after this one.)

... Why isn't "less than 0" one of the answers? It should be.
...
Thus I vote for a tiny, tiny decrease in the geosynchronous altitude.I have provided this "Neg. choice" also now, but based on several of your posts ("not one iota difference", etc.) I recorded your vote, along with that of 2inqusitive's, in the new "0" choice group. If you really want it to be recorded in the "Neg" choice group, I will move it to that group, which currently has no votes because it is not completely clear which of he two new groups you wanted. (You, and You alone, at preasent, can move your vote from "0" to "Neg" group if you like.)*

Anyone else, like DH who voted for one of the "official choice" groups only because the two new choices were not provided when you voted, PLEASE TELL HOW YOU VOTED, in new post, so the results can be correct to reflect the voter's actually wishes at time of voting
----------------------------------------------
*(No changing votes is permited by the structure of the sciforum poll, so I will usually try to enforce this also in the "corrected summary" accounts.)

OK, here is a short take on my thoughts as to why angular momentum does not increase in the orbiting body, along with a link from NASA. I will also see if I can locate a more comprehensive explaination.

To begin with, when the rocket is setting on the launch pad, it is not at 'rest'. It is spinning with the rotation of the Earth. From the frame of a distant observer, one NOT located on Earth's surface, the angular momentum of the rocket does not change when boosted into orbit. Angular momentum is conserved in the rocket. The talk by Tortise about getting hit by longer arms was not about angular momentum, but about kinetic energy. Here is one cut & paste concerning angular momentum:

"angular momentum
A quantity obtained by multiplying the mass of an orbiting body by its velocity and the radius of its orbit. According to the conservation laws of physics, the angular momentum of any orbiting body must remain constant at all points in the orbit, i.e., it cannot be created or destroyed. If the orbit is elliptical the radius will vary. Since the mass is constant, the velocity changes. Thus planets in elliptical orbits travel faster at perihelion and more slowly at aphelion. A spinning body also possesses spin angular momentum."
http://imagine.gsfc.nasa.gov/docs/dict_ad.html

I will attempt to dig up a link I have read earlier concerning angular momentum and satellites, that explains it more throughly, when I have the time to do so.

Note: edited for link

, the angular momentum of any orbiting body must remain constant at all points in the orbit, i.e., it cannot be created or destroyed. If the orbit is elliptical the radius will vary. Since the mass is constant, the velocity changes. Thus planets in elliptical orbits travel faster at perihelion and more slowly at aphelion

Ya great! Just one thing: The satelite is not naturally falling into a geo sync. orbit. We are putting it into a higher energy state.

Here is what your thinking: If the earth was spinning fast enough then stuff could fly off and be in a geo sync. orbit (wrong)

Stuff that may fly off a spinning body would not have the same angular momentum and couldn't settle into a geo sync. orbit. It wouldn't have enough energy. And to be more specific, it would have enough angular momentum. You should try to understand that it simply has a higher velocity - forget about the fact that it is in orbit for a sec. It has a much higher velocity, and it happens to be captured in an orbit.

The effect postulated by this post should be observable. By Kepler's Third Law, the period of an orbit T is proportional to the square root of the cube of the semi-major axis length. Thus a one hundred meter change in the semi-major axis of a geocentric orbit would correspond to about a 0.3 second change in the length of a day (delta T = 3/2 T/a delta a).

We can measure most physical constants to at most a handful of decimal places. Time is an exception to this rule: We know that the length of a day is increasing by 1.6 milliseconds per century. With 2.3E⁰⁹ seconds in a century, this means we know the length of a day to 13 decimal places or more. With this kind of accuracy, the effect postulated by this post should be observable even if the change in the size of a geosynchronous orbit was as small as 1.5 millimeters.

Good grief, Tortise. No, that is not what I am thinking. Satellites in orbit are not things attached to spinning bodies. You are talking centrifugal force now. Do you have any idea what angular momentum IS? It is NOT the KINETIC energy and it is not centrifugal force. What do you mean 'IF the Earth was spinning fast enough...'? I am speaking of real world physics, not some imaginary gedankin. Question. You seem to believe angular momentum places the satellite in orbit, something 'stolen' from Earth's equatorial angular momentum. Not true. Do you believe a satellite in low Earth ORBIT has less or more angular momentum than a satellite in geostationary ORBIT?

Ya, 2inq - it sounds like you have a real grasp on the concepts sorry for trying to dumb it down for you. It really didn't seem like you understood very well. But if you're comfortable with your level of understanding, maybe that's all that matters.


Satellites in orbit are not things attached to spinning bodies. You are talking centrifugal force now. Do you have any idea what angular momentum IS? It is NOT the KINETIC energy and it is not centrifugal force. What do you mean 'IF the Earth was spinning fast enough...'?


Satellites are attached to the earth by way of gravity. As much as a car on earth is attached by way of gravity. Both would float away if it were not for gravity.

It's obvious to many people on this forum that you're a very smart person, but you may not be willing to admit when you are wrong - so you keep arguing hopeless arguments.

Ha! Tortise you are an ignorant school kid. It is obvious you don't know what you are talking about. You keep flirting around with all kinds of nonsense concerning angular momentum. Now, explain to me what gravity has to do with angular momentum.

In many cases gravity is what makes angular momentum possible. But that is as far as I'm going to go into it now with you because it doesn't appear that you are looking for the truth. You are looking to be right. This is the flaw in your thinking.

The effect postulated by this post should be observable. By Kepler's Third Law, the period of an orbit T is proportional to the square root of the cube of the semi-major axis length. Thus a one hundred meter change in the semi-major axis of a geocentric orbit would correspond to about a 0.3 second change in the length of a day (delta T = 3/2 T/a delta a).

We can measure most physical constants to at most a handful of decimal places. Time is an exception to this rule: We know that the length of a day is increasing by 1.6 milliseconds per century.
That's the mean length of a solar day. The actual length of each day varies chaotically from one day to the next on the order of milliseconds. Any minute change induced by satellite launches would be swallowed up in the noise of other mass shifts (think atmosphere and ocean).


With 2.3E⁰⁹ seconds in a century, this means we know the length of a day to 13 decimal places or more.
It doesn't follow. You're assuming a steady change in the length of the day, which is not so. The change in Earth's rotation rate is a general trend, not a monotonic decline.
I think you got the maths wrong as well. Even if the length of each day varied metronomically, knowing the variation in day length to within 0.1 milliseconds over a century would lead to knowing the length of a day to within about 10^-9 seconds, not 10^-13.


For interest:
The Earth's actual rotation is monitored by the International Earth Rotation and Reference Systems Service (http://www.iers.org/), using the International VLBI Service for Geodesy and Astrometry (http://ivscc.gsfc.nasa.gov/).

To begin with, when the rocket is setting on the launch pad, it is not at 'rest'. It is spinning with the rotation of the Earth. From the frame of a distant observer, one NOT located on Earth's surface, the angular momentum of the rocket does not change when boosted into orbit.
Put some numbers in, 2inq. Assume a 1000kg rocket.
What's the angular momentum of the rocket sitting on the ground on the equator? (R = 6400km, V = 460m/s)
What's it's angular momentum in geostationary orbit? (R = 42000km, V = 3000m/s)

Here's an even easier one:
A rocket is launched from the North Pole. into a polar orbit. What's its angular momentum before launch? After?

Your quote applies to satellites in freefall. Their angular momentum does not change while their rockets are not firing.

Tortise, it is clear to everyone why you are not willing to go farther. You are in above your head. You never did answer my question if you believe a satellite in geostationary orbit has less or more angular momentum than an identical satellite in low Earth orbit. Gravity is a central force, a force which causes a torque. Gravity does not change the angular momentum of a satellite, but gravitational potential energy can be converted into kinetic energy. Angular momentum of a satellite in orbit does not change due to low or high orbits. The mass of the satellite stays the same and the speed of the satellite decreases when it is moved into a higher orbit. It takes kinetic energy to change the satellite's orbital altitude, not a change in angular momentum. The angular momentum of an ORBITING satellite never changes.

Another interesting factor in Earth instantaneous rotational velocity is traffic. Rush hour traffic represents significant action/reaction transmitted into changing the pavement's angular momentum. Where the rubber meets the road.

I'm having trouble remembering what the definition of angular momentum is. Doesn't it have something to do with velocity and the radius of the rotation?

And, I can't remember for sure, ;) , but doesn't the orbital velocity have something to do with the square of the radius of the orbit? Due to the inverse square rule of gravitational attraction?

Look 2inq - if you want to argue then let's argue. But at least admit when you are wrong ok? Otherwise it's like arguing with a child. If you agree to consider the facts and admit if you are wrong, then I will argue with you (does anyone like to argue with people who can never admit when they are wrong?)

You asked:
You never did answer my question if you believe a satellite in geostationary orbit has less or more angular momentum than an identical satellite in low Earth orbit

You have to be more specific then that but the general answer is that the velocites would have to be different in different orbits due to keplers law. A sat. in a low earth orbit would not be able to be geosync. because of kepler's law. The low earth sat. would have to be traveling much faster - therefore would have the greater energy if it were a stable orbit and if the sat. weighed the same.

You never did answer my question if you believe a satellite in geostationary orbit has less or more angular momentum than an identical satellite in low Earth orbit.
It's easy to run the numbers, 2inq. Why didn't you do it?

geostationary orbit: R = 42000km, V = 3000m/s. Angular momentum (100kg satellite) = 1.3x10¹¹ kg.m&sup2;/s

Low earth orbit (ISS): R = 6700km, V = 8000m/s. Angular momentum (100kg satellite) = 0.5x10¹¹ kg.m&sup2;/s

Put some numbers in, 2inq. Assume a 1000kg rocket.
What's the angular momentum of the rocket sitting on the ground on the equator? (R = 6400km, V = 460m/s)
What's it's angular momentum in geostationary orbit? (R = 42000km, V = 3000m/s)

Here's an even easier one:
A rocket is launched from the North Pole. into a polar orbit. What's its angular momentum before launch? After?

Your quote applies to satellites in freefall. Their angular momentum does not change while their rockets are not firing.

OK, Pete, what is the angular momentum of a rocket at the north pole before launch? What is the angular momentum of the same rocket after launch? Do you believe the Earth transferred angular momentum to the north pole rocket?

2inq - this is VERY entertaining - you trying to get the focus off your statements by challenging us to do these problems - VERY. Wouldn't it just be easier to admit when you're wrong? It's easier then you might think. Try it....Say it with me "I'm wrong" It's not that hard.

OK, Pete, what is the angular momentum of a rocket at the north pole before launch?
Zero.

What is the angular momentum of the same rocket after launch?
Not zero.

Do you believe the Earth transferred angular momentum to the north pole rocket?
No. Not directly, anyway.

The rocket gained angular momentum by ejecting fuel in the opposite direction. Fuel goes backward, rocket goes forward, right?

So, 2inq...
Do you believe a satellite in low Earth ORBIT has less or more angular momentum than a satellite in geostationary ORBIT?

From EXTENSIVE personal practise, I know that it is really easy.

W R O N G.

Just try it. ANYONE can do it.

That has no effect on the Earth's rotation, right? Now, Pete, do YOU believe a satellite in low Earth orbit has less or more angular momentum than a satellite in geostationary orbit? Do you still contend that the Earth looses angular momentum when a satellite is launched into orbit?

Now, Pete, do YOU believe a satellite in low Earth orbit has less or more angular momentum than a satellite in geostationary orbit?
:bugeye:
You missed the calculations, didn't you?

Here it is again:

geostationary orbit: R = 42000km, V = 3000m/s.
The angular momentum of a 100kg satellite in geostationary orbit is about 1.3x10¹³ kg.m&sup2;/s

Low earth orbit (ISS): R = 6700km, V = 8000m/s.
The angular momentum of a 100kg satellite in low earth orbit is about 0.5x10¹³ kg.m&sup2;/s

So what do you think, 2inq? Is 0.5 more or less than 1.3?

EDIT:
Does no one check my figures?
My final numbers were (consistently) out by a factor of 100!

2inq-

Do you still contend that the Earth looses angular momentum when a satellite is launched into orbit
so your logic goes something like: " Well the closer a sat. is to the earth the more energy it has, so an object on the surface would have more ang. momentum then a sat. in geo sync. orbit"

The only problem with this logic is that a sat. in a very near earth orbit would be circling the earth many times a day - something that objects on earth - on the ground never do.

So is there any more circular reasoning you would like us to straighten out for you?

Pete, the formula for angular momentum is the moment of inertia times the angular velocity. It is NOT the same as for linear momentum. Mass times angular velocity does not give angular momentum. I think you are calculating the rotational kinetic energy instead of angular momentum.

Let me ask you this. A spinning skater increases her rotational velocity by bringing her arms closer to her body. Her mass does not change and her angular momentum does not change, only the rotational velocity of her 'hands'. It is the same for satellites in orbit, the angular momentum for a satellite in low Earth orbit is the same as for an identical satellite in higher, or geostationary, orbit. Only the orbital velocity changes.

Of course there is the possibility that your just pretending to be wrong so you can test our physics skills? Is that what's going on here?


It is the same for satellites in orbit, the angular momentum for a satellite in low Earth orbit is the same as for an identical satellite in higher, or geostationary, orbit

If you mean that the total angular momentum of just the sat. then the answer is that it has more angular momentum in the lower orbit. If you mean the angular momentum of the entire system then you are right.

L=Iw is just another word for L=mvr.

In the ( L=mvr ) universe ;
L= angular momentum
m= mass
v= velocity
r= radius ( of mass from axis of rotation )

So, when we increase either radius or velocity, or, both, we increase L.

If this is too simple, if you let me know, I will drink another cold beverage and then try to write it again in a more garbled way.

her angular momentum does not change, only the rotational velocity of her 'hands'

The skater's entire period of rotation changes. Ask any skater or anyone else on this fourm. The skater's rotation slows down as she abducts her arms, and the skater's body speeds up as she adducts her arms - but the entire amount of angular momentum has stayed the same minus the friction.

Putting an object, originally sitting still on the surface of the planet, at optimally about 1,000 MPH velocity, into orbit at minimally about 5 MPS velocity, is not exactly the same as a skater telescoping their limb(s).

Both the velocity and the radius from the axis of rotation have increased significantly in one case, therefore, L=mvr, the angular momentum, has increased significantly.

Why has not angular momentum been unchanged? Would rapid combustion of rocket fuel explain such a mystery?

Why has not angular momentum been unchanged? Would rapid combustion of rocket fuel explain such a mystery

But just because rocket fuel is used, that does not mean that the gasses from that fuel didn't push against the earth. I think the only way it matters that rocket fuel was used is if the expelled gasses are expelled in space away from earth. The expelled gasses can slightly raise the atmospheric pressure that can be felt on the earth's surface. Eventually particles of gas or particles that were influenced by the expelled gas makes contact with the ground and causes friction. You're saying that angular momentum isn't conserved because rocket fuel was used. Well I think your wrong Cangas....the only part of the angular momentum that is outside the conservation of angular momentum is if the rocket had part of it's burn in space - the gases expelled away from the planet, and I don't think this is typical of sat. launchings.

You can take a horse to the water trough.......

am I wrong? I'll admit it if I am, that's just what I had thought. Every action has an equal and opp. reaction. I mean you could say that the rocket was expelling gases at a very high velocity (about the speed of sound or better) at an angle in the direction of the earth. The volume of gases is amazing. Why wouldn't this have an effect on the earth?

Hey, T:

You need to be checked out.

1+1=?

2. Right? Am I right? I mean I'll admit it if I'm wrong.....That's just what I thought....1 + 1 = 2 ... anyway with respect to the above argument: What is more likely that I who am never wrong is wrong or YOU who is sometimes wrong is wrong....Someone argue with me!! I must argue!!!!.........Must.. find... argument.... :D

That's the mean length of a solar day. The actual length of each day varies chaotically from one day to the next on the order of milliseconds. Any minute change induced by satellite launches would be swallowed up in the noise of other mass shifts (think atmosphere and ocean).

Yes, I know about the difference between a mean and "true" day. The point of my post was that if the hypothesized effect existed, the scientific community would know about it because time is the most precisely measured of all physical quantities. You mentioned the IERS. This IERS web page (http://www.iers.org/MainDisp.csl?pid=95-84) describes the sources of the variability of the earth-rotation vector:

The variability of the earth-rotation vector relative to the body of the planet or in inertial space is caused by the gravitational torque exerted by the Moon, Sun and planets, displacements of matter in different parts of the planet and other excitation mechanisms. The observed oscillations can be interpreted in terms of mantle elasticity, earth flattening, structure and properties of the core-mantle boundary, rheology of the core, underground water, oceanic variability, and atmospheric variability on time scales of weather or climate. The understanding of the coupling between the various layers of our planet is also a key aspect of this research.
They do not mention satellite launch as one of the causes of the variability.

A change in the size of a geosynchronous orbit by a meter to one hundred meters (your vote) is anything but minute. It corresponds to a 3 millisecond (one meter) to 0.3 seconds (one hundred meters) change in the length of a day.


You're assuming a steady change in the length of the day, which is not so. The change in Earth's rotation rate is a general trend, not a monotonic decline.

I assumed the hypothesized effect is a permanent change in the Earth's rotation rate, just as Billy T posed to start this thread.


I think you got the maths wrong as well. Even if the length of each day varied metronomically, knowing the variation in day length to within 0.1 milliseconds over a century would lead to knowing the length of a day to within about 10^-9 seconds, not 10^-13.

You're right. I did get the math wrong: 86400*365.25*100 = 3.2E09, not 2.3E09. I transposed the 3 and 2 on typing. Oops.

To be able to say with confidence that the length of a mean day is decreasing (on average) by 1.6 milliseconds per century, does that not mean we know the length of a mean day to better than 0.1 milliseconds per century, or 13.5 decimal places? (0.1E-03/3.2E9=3.2E-14.)


I stand by my previous statement that the angular momentum of the combined Earth+satellite system is not conserved when a spacecraft is launched into orbit. When applying the conservation laws, one must be very careful in correctly drawing the boundaries of the system in which to apply those laws. For example, drawing the boundary around the rocket and spacecraft and ignoring the exhaust leads to the erroneous result that the spacecraft cannot move. Similarly, drawing the boundary around the Earth and satellite leads to the erroneous result that the Earth's rotation rate changes.

The satellite is connected by gravitational strings to the Earth. However, it is also connected by the same gravitational strings to the Sun, Mars, Alpha Centauri, and to the most remote pulsar. The angular momentum of the entire universe is conserved when a satellite is placed into Earth orbit. However, the angular momentum of the Earth+satellite system is not.

....However, it is also connected by the same gravitational strings to the Sun, Mars, Alpha Centauri, and to the most remote pulsar. The angular momentum of the entire universe is conserved when a satellite is placed into Earth orbit. However, the angular momentum of the Earth+satellite system is not.That is true. Everyone even slightly informed in this area knows that the moon is distorting the Earth. - Making oceans tides (and even slightly in the "solid" Earth). Mainly these tides provide a "handle" for the moon to apply a torque on the Earth, which is slowing the Earth's spin rate (and causing the moon to move farther away from Earth.)* If mass of Earth were distributed exactly spherically at all times, the moon / Earth separation would not be changed, by the moon existing near it.**

It is partially because of the well known facts you mentioned (Earth's spin rate is changing by moon, etc. torques) that I postulated TWO satellites, A & B and asked for the CHANGE in their geostationary radii CAUSED BY THE LAUNCH of satellite B. Perhaps you voted without reading / understanding the question?

This thread does not care, and is not concerned with the fact that the moon is changing the spin rate of Earth, or that the angular momentum of the entire solar system is not exactly conserved. (Solar system is certainly not a "radially symmetric distribution of mass" and is acted upon by "outside torques" from near by stars etc.) Please read the question asked. It does not concern the total angular momentum of the Earth moon system, the solar system etc. but indirectly asks how the Earth's spin rate adjusts to the placing 1000Kg satellite into geostationary orbit.

Another part of why I asked for the difference in radii of satellites A & B is that this difference is surely (I still do not know answer) larger in MKS units that the change in spin rate is in MKS units, even though it may be, as most think, very small still. (The change in spin rate CAUSED BY B's launch into orbit may be such a small number that be people using calculators, instead of algebra, can not agree on exactly how small it is.)

---------------------------------------------
*We could start another pole, guessing how many years until total eclipse of the sun is impossible, but it would have a more speculative answer than this one as the long term variations in tidal torques are not very well quantified and vary as other things, like ice ages and even building dams, digging mines, weather's atmospheric pressure, actually the associated air masses in local columns of air, volcanoes, etc. all change the "handle" the moon can use to torque the Earth.)

**(May need to assume the moon is also a "radially symmetric distribution of mass", but I think not, because moon's day and orbit period are equal. I.e. I think that even if moon's mass were in a long pole-shaped object, but still in a gravity gradient stabilized orbit, it could not torque a spherical Earth.)

It is partially because of the well known facts you mentioned (Earth's spin rate is changing by moon, etc. torques) that I postulated TWO satellites, A & B and asked for the CHANGE in their geostationary radii CAUSED BY THE LAUNCH of satellite B. Perhaps you voted without reading / understanding the question?

I understand the question asked. I have asked two questions in return:
What mechanism causes the Earth's spin rate to change?
Why has the effect not been observed and discussed by those who are concerned with measuring the length of a mean day to 13.5 decimal places of accuracy?

I have not seen an answer to the second question. My answer is that the effect <strike>does not exist</strike> is immeasurably small.

Some have answered the first question by stating that the angular momentum of the Earth+satellite system must be conserved. You yourself raised the issue of conservation of momentum with your satellite de-spin example.

I raised the Sun, Moon, etc because I think the boundary of the system in which angular momentum is conserved is the entire universe. The angular momentum of the Earth+satellite system is not conserved.


Another part of why I asked for the difference in radii of satellites A & B is that this difference is surely (I still do not know answer) larger in MKS units that the change in spin rate is in MKS units, even though it may be, as most think, very small still. (The change in spin rate CAUSED BY B's launch into orbit may be such a small number that be people using calculators, instead of algebra, can not agree on exactly how small it is.)

It's easy to determine how changes in the spin rate affect geosynchronous altitude. By Kepler's Third Law, T&sup2; is proportional to a&sup3;. Thus a small change delta a in semi-major axis a is related to a small change in period T by

delta a = 2/3 a/T delta T

A geosynchronous orbit has a semi-major axis a=42164 km and a period T=86400 sec. Changing the length of a day by a small amount delta T seconds means changing the size of a geosynchronous orbit by about 2/3*42164km/86400sec delta T, or

about 1/3 km change in geosynchronous altitude for each 1 second change in the length of a day (valid only for small changes in the length of a day).

Perhaps you voted without reading / understanding the question?

To ensure that I am answering your question, my understanding of the question is as follows:
Satellite A is launched from the surface of the Earth to an orbit whose period is equal to the length of one mean Earth day; i.e., a geosynchronous orbit.
The launch of satellite A changes the Earth's rotation rate by some amount. Satellite B is later launched from the surface of the Earth to an orbit whose period is the length of a day as altered by the launch of satellite A.
What is the difference between the radii of the two orbits?


As I understand the question, you could eliminate the need for satellite B by asking instead "how much does the launch of satellite A change the Earth's rotation rate?".

Using the relation delta t = 3 sec/km delta a at geosynchronous altitude, a corresponding response list is (ignoring the factor of 3),

Less then 100 nanoseconds
Less then 10 microseconds
Less then 1 millisecond
Less than 100 milliseconds
Less than 1 second
More than 1 second

... I have asked two questions in return:
[1]What mechanism causes the Earth's spin rate to change?
[2]Why has the effect not been observed and discussed by those who are concerned with measuring the length of a mean day to 13.5 decimal places of accuracy?I think [1]'s answer is closely related to fact that a cannon recoils (accelerates backwards) when throwing (accelerating) a cannon ball forward.

That is, to throw forward the satellite (Accelerate to greatly increased speed so it can still go around the center of the Earth point, at geostationary radius instead of only at Earth radius, in essentially 24 hours), by any means, something must "recoil" or "accelerate backwards."

In the case of a rocket, rolling over eastward (instead of traveling radial outward only to fall back radially inward later) it is initially the exhaust gases that "recoil" - are accelerated "backwards."* The mass of these gases, at initial pitchover, have more rearward momentum than the fuel in the rocket tanks had, but of course the rearward moving gases mix with the air and slow, giving this rearward momentum to the atmosphere, which in tern shares it with the solid Earth, etc. - that is the basics coupling mechanism by which some of the angular momentum of total system (Earth + rocket + satellite) is transferred to increase the satellite's and decrease the share of total in the Earth + rocket, when rocket is again resting on the Earth.
-------------------------------------
*The exhaust gases are initially traveling backwards, but they need not be traveling "backards" in the Earth frame for them to be accelerated rearward later. After the rocket is traveling significantly faster than the speed of sound and has rolled over to the East, they will be going "forward" in Earth's frame, when they first part from the rocket, but not going forward as fast as their mass was a few seconds earlier when they were still "fuel."
It is really just Newton's laws. - A force is required to continue accelerating the rocket (and satellite inside) and a force is required to accelerate the fuelmass (converted to "exhaust gas", but so what?). These two forces are the "equal and opposite" forces Newton spoke of. The force acting on the fuel mass (exhaust) has a tangential component, acting on a "moment arm" greater that the radius of solid Earth. That is "produces a torque." First on the air the exhaust is mixing with, but eventally the entire Earth is slowed by this torque.

So much for Q1. Now as for Q2 -- I am tempted to ask a corresponding question in reply such as: Why has no one measured the sound of an ant scratching to dig a hole into the Earth to escape from the hurricane's roar? I can hardly believe you are serious, but I will answer as if you were:

The change in Eath's spin rate is very very small. Completely unobservable in comparison (like ant scratching compared to the hurricane's roar) to the effect of even a moderate rain falling anywhere on Earth (I think) as the lowering by rain of a water mass, which is huge compared to the fuel mass thrown out by the rocket, surely is constantly changing and bigger effect (speeding up the Earth's spin rate by amounts that vary every minute). Likewise the mass of ocean water being raised by the sun (evaporation) high into the air is slowing the Earth's spin rate much more that any satellite launch. Just the minute-by-minute variation in the variation in "ocean evaporation slowing" as clouds form and move over land etc. is much great that all the satelites that have ever been launched (I suspect.)

Make you a deal:
You design for me how to measure/record the sound db produced by 1000 ants in the "foot print" of a class 5 hurricane, despite its roar, I will measure the effect of a shuttle launch on Earth spin despite falling rain and ocean evaportion, but you demonstrate your measurement first. :D

...Since T&sup2; is proportional to a&sup3;, a small change delta a in semi-major axis a is related to a small change in period T by

delta a = 2/3 a/T delta T...I did not check you numbers because, in addition to being lazy, I think you have a mistake in your differentiation. The derivative of "a cubed" should leave an "a squared" term.

Despte this, I congratuate you on a very good and easy way to relate change is spin rate to corresponding difference in radii between satelites A & B. I might have thought of this approach also, but have not given much thought as to how I would calculate the answer.

Clearly the spin rate change caused by putting B into orbit is less than one second per day or as you say, someone would have noticed it (and probably several law suits would be in the US courts. :rolleyes: ) Thus, your approach, when done correctly, does set an upper bound on how big the radial difference can be.

I am waiting for someone to post an answer and their method after the 25th.

I think [1]'s answer is closely related to fact that a cannon recoils (accelerates backwards) when throwing (accelerating) a cannon ball forward.
Agreed. The question is, what recoils? My answer is that the response is distributed throughout the universe; i.e., it is unobservable.


I am tempted to ask a corresponding question in reply such as: Why has no one measured the sound of an ant scratching to dig a hole into the Earth to escape from the hurricane's roar?
All of your response choices are very measureable, and that is for but one satellite. We have launched thousands of satellites, most to LEO, fewer to GEO, and a small number even further. The cumulative effect, if it exists, should definitely have been noticable. Do you understand what incredible measurement accuracy is needed to enable scientists to say that the mean length of a day is changing by 1.6 milliseconds per century?


I did not check you numbers because, in addition to being lazy, I think you have a mistake in your differentiation. The derivative of "a cubed" should leave an "a squared" term.

Here's the derivation, step-by-step:
T^2 is proportional to a^3
Kepler's Third Law
a = k*T^(2/3)
From above, adding constant of proportionality
da/DT = 2/3*k*T^(-1/3)
First derivative of a wrt T
1/a da/DT = (2/3*k*T^(-1/3))/a
Divide both sides by "a"
1/a da/DT = (2/3*k*T^(-1/3))/(k*T^(2/3))
Substitute a=k*T^(2/3) on RHS
1/a da/DT = 2/3*1/T
Simplify RHS
da/DT = 2/3*a/T
Multiply both sides by a
a(T) = a(T0) + (T-T0) da(T0)/dT
1^st order Taylor series about T=T0
a(T) = a(T0) + 2/3*a(T0)/T0*(T-T0)
Apply da/DT = 2/3*a/T
delta a = 2/3*a/T*delta T
Equation for change in a as f(change in T)



I am waiting for someone to post an answer and their method after the 25th.
You have my answer and my method.

Our exchanges are passing each other in cyber space so I do not think we are changing Earth spin rate ;)
To ensure that I am answering your question, my understanding of the question is as follows:
Satellite A is launched from the surface of the Earth to an orbit whose period is equal to the length of one mean Earth day; i.e., a geosynchronous orbit.
The launch of satellite A changes the Earth's rotation rate by some amount.
Satellite B is later launched from the surface of the Earth to an orbit whose period is the length of a day as altered by the launch of satellite A.
What is the difference between the radii of the two orbits?

As I understand the question, you could eliminate the need for satellite B by asking instead "how much does the launch of satellite A change the Earth's rotation rate?". That is not exactly my question but surely has the same answer. You said you worked in this area (unless I am confusing you withsome one else) so you surely must know that a geostationary satellite is not launched into orbit to have a period equal to the pre-launch "length of one mean Earth day".

They consdier the launch a great success if the period is within a minute of 24 hours. If it is, everyone who had anything to do with the satellite or launch is half drunk with celebration (except for a few "kill joys" who don't touch the stuff.)

Satellite is now the operators / owners "baby." After the hang-over is gone, and they are back on job next day, they are asking "How did this schedule slip?" and "What are we going to do to catch up?" etc. They no longer care that the operators are firing tiny thrusters (which do change the total angular momentum of the satellite (and of course the Earth satellite system) as that exhaust mass escapes into deep space to get the period very close to "one mean Earth day." My question assumed that the launch crew was perfect and knew exactly what the effect of the launch of A would be on the length of the day so they made it that "after launch" period. What they did not know was another satellite was soon to go up and change the length of the day, causing A to be slghtly not geostationary anymore.

Launch crew for B, knew exactly the length of day after A is in correct orbit, i.e. matches the rotation rate of Earth changed a little by its own launch (I am here assuming that none of the operators fine thrustors changes the Earths rotatation, because it was a "perfect launch" and did not give the operators anything to do.) Launch crew B also knows how their own launch will change Earth's rotation rate, so they, again with perfection, leaving the operators again with nothing to do, placed B into the new longer day's geostationary orbit - a SLIGHTLY higher one than A had before A's operator start firing small thrustors, (I believe)

Question of the poll, is what is the value of this "SLIGHTLY"?

I have already noted your differentiation is in error as it has no a^2 term, so will not reply to rest of your post.

Agreed. The question is, what recoils? My answer is that the response is distributed throughout the universe; You asked me to describe the mechanism that torques the earth to change its spin rate and I did so. Now I ask the same of you. Please describe the mechanism that torques the rest of the universe. In fact, lets make it easier, just tell how (mechanism) alpha centura's angular momentum is changed to be part of the "recoil torque' you think is applied there and else where in univers but not the Earth to change its spin by the mechanism I suggested.

All of your response choices are very measureable, and that is for but one satellite. We have launched thousands of satellites, most to LEO, fewer to GEO, and a small number even further. The cumulative effect, if it exists, should definitely have been noticable.Support this OPINION with some numbers or reference. At least comare the effect to that of a big rain storm.

Do you understand what incredible measurement accuracy is needed to enable scientists to say that the mean length of a day is changing by 1.6 milliseconds per century? I think smaller effect have been estimated by calculation as this one has been. Most times when a leap second adjust is made to the year (as at end of 2005) it is to add a leap second, but not always. Nearly random variation of the day length is orders of magnitude greater, so actually observation is of the average rate impossible. One can average the changes that have been required over the period in which atomic clocks have existed but for example, if he oceans are warming and expanding the current rate of slowing may be differrent than what is calculated. This is best investigated, not by telescopes or atomic clocks, but by the ancient records of in what cities the eclipse of the sun was total. Your value, if it were a constant average, would make eclipse totality occur in different cities than where the Chinese saw them several thousand years ago. Must go now will read rest and reply later.

You said you worked in this area (unless I am confusing you withsome one else) so you surely must know that a geostationary satellite is not launched into orbit to have a period equal to the pre-launch "length of one mean Earth day".

They consdier the launch a great success if the period is within a minute of 24 hours.

Of course I know that. The orbit transfer is performed either using timed burns or using accelerometers to measure the accumulated delta-v. Timed burns are a terrible way to go; rocket thrust varies by 5-15 percent. Since accelerometers have an uncorrectable bias of 20 microG or so, even accumulated delta-v cutoff results in fairly large errors by the time the vehicle reaches GEO. The satellite is placed very precisely at some point in time. Who owns the satellite and exactly how it gets to its target orbit is a detail.


My question assumed that the launch crew was perfect ...

Exactly. Who owns the satellite and exactly how it gets to its target orbit is a detail. At some point it is placed very precisely. That is all that matters -- except for the minor issue of where GEO is, and how launch and travel to GEO changes GEO altitude -- i.e., the point of this thread.


Question of the poll, is what is the value of this "SLIGHTLY"?

My answer remains, none of the choices corresponds to "SLIGHTLY". With the cumulative effect of all of the spacecraft launched by mankind over the last 45 years, even a few micrometers change would be observable. Nanometers or picometers are SLIGHTLY, but you did not offer those as choices.


Support this OPINION with some numbers or reference.

I already did supply some numbers. I re-expressed your original list of changes to geocentric orbit radius to changes in length of day. The smallest entry (10e-4 meters) corresponds to a change in the length of a day of 300 nanoseconds; the largest (1 km) corresponds to a change of 3 seconds.

The Earth's space agencies collectively launch hundreds of satellites per year. Multiplying your list by 1000 (100 launches per year * 30 years of launches*2/5 LEO/GEO = 1200 or 1000) yields a cumulative change of 300 milliseconds to 3000 seconds. Do you not think such a change would be observable?

This is not the sound of an ant marching in a thunderstorm. Compared to the measurement accuracy of 13.5 decimal places for the length of a mean day, your list of choices corresponds to a large fleet of jets flying overhead at 100 feet.


I have already noted your differentiation is in error as it has no a^2 term, so will not reply to rest of your post.

Ok, bright boy, what is the correct relationship?

This is basic stuff! Check the units, Billy. My answer

delta a = 2/3*a/T*delta T

is correct. Units are length on both sides of the equation. An a^2 term would not yield correct units.

This is an application of a general rule for derivatives of power models, and is very useful in modeling.

Given some power relationship
y(x) = a*x^p
Differentiation with respect to "x" yields
y'(x) = p*a*x^(p-1)
Multiplying both sides by "x" yields
y'(x)*x = p*a*x^(p-1)*x = p*a*x^p = p*y(x)
or
y'(x) = p*y/x



I think smaller effect have been estimated by calculation as this one has been. Most times when a leap second adjust is made to the year (as at end of 2005) it is to add a leap second, but not always. Nearly random variation of the day length is orders of magnitude greater, so actually observation is of the average rate impossible.

The IERS does not mention your effect. They do however, have analyses such as this which shows the variations in length of day to sub-millisecond level from 1995 to 1999 (http://www.iers.org/MainDisp.csl?pid=95-99). I can see a long term drift in this plot, well explained by momentum transfer to the moon's orbital motion. I can see seasonal effects. I do not see any large jumps corresponding to satellite launches. There were a lot of satellite launches between 1995-1999. Where are the three millisecond jumps corresponding to a one meter change in geosynchronous altitude?

We are simply saying that it slows the earth, but to some extreemly exteemly small degree. You don't get somethig for nothing. Every action has an equal and opp. reaction. Why is this so hard to understand. It's a very simple concept! This is not rocke.....ummm ok it's rocket science but I sware that a 6th grader could understand the concept of every action has an equal and opp. reaction, and how people so smart could argue otherwise is way way beyond me. SHAME SHAME SHAME (finger furiously shaking)
Having said this, I would like to point out the the effect would be so small, that I don't think any correction at all would be needed for the 2nd sat.

According to many in this thread, launching the satellite towards the East to take advantage of Earth's equatorial velocity slows the rotation of the Earth, they say to conserve angular momentum. OK, launch the same weight satellite into orbit, except launch towards the West. Yes, it takes more fuel to launch that way, but according to your thoughts, launch to the West will INCREASE the Earth's rotation rate, correct? The satellite will increase in angular momentum regardless of direction of launch. Now, if angular momentum is conserved in the satellite/Earth system, the Earth will still have to decrease its angular momentum regardless of the direction of launch, if the satellite gains in its amount of angular momentum. Same for a polar launch, the satellite will gain angular momentum acording to calculations after launch. Do you still maintain this 'increase' in angular momentum by the satellite comes from a decrease in Earth's angular momentum?

Edit: removed 'geostationary' from post

We are simply saying that it slows the earth, but to some extreemly exteemly small degree. You don't get somethig for nothing. Every action has an equal and opp. reaction. Why is this so hard to understand. It's a very simple concept.

Agreed. But you have to draw the boundaries of the system correctly. Drawing them incorrectly leads to incorrect results.

The question can then be broken into three parts:
Does launch itself change the rotation rate of the Earth? That is, is there any change in the rotation rate in the few seconds it takes for the vehicle to clear the pad?
Does the motion of the free-flying vehicle from nanometers off the ground at Earth-rotation rate to LEO at 1 rev/90 minutes change the rotation rate of the Earth?
Does the orbital transfer of the free-flying vehicle from LEO (1 rev/90 minutes) to GEO (1 rev/day) change the rotation rate of the Earth?


I agree that the system boundary for the first question is indeed the Earth+satellite system. However, launch itself does not change the rotation rate of the Earth.

I disagree with the system boundary being the Earth and satellite for the third part. The same strings (gravity) that couple the Earth and satellite also couple them with everything else in the universe. The Earth+satellite are not an isolated system. Arbitrarily drawing the boundary around the Earth and satellite for the transfer from LEO to GEO leads to incorrect results.

The second part is no different than the third, except that the vehicle and the exhaust will transfer some momentum to the atmosphere (and thence to the Earth). Drawing the boundary around the Earth and satellite is incorrect once the vehicle is off the ground.

In the end, this is a bunch of arguing over how many angels can dance on the head of a pin. Playing devil's advocate here, let's assume that the angular momentum of the Earth+satellite system is conserved. The angular momentum of the Earth is 8E37 kg-m^2 * 2*pi/86400 seconds, or 6E33 kg-m^2/s. The angular momentum of a 1000 kg satellite in GEO is 1.3E12 kg-m^2/s. Using delta T/T = delta L/L or delta T = delta L/L * T yields a change of 2E-17 seconds in the length of a day. Finally, using delta a = 300 m/s * delta T at GEO altitude yields a change of 6E-15 meters for the change in altitude of a geosynchronous orbit.

The effect won't be noticeable until we have launched about half a million satellites into orbit.

According to many in this thread, launching the satellite towards the East to take advantage of Earth's equatorial velocity slows the rotation of the Earth, they say to conserve angular momentum. OK, launch the same weight satellite into orbit, except launch towards the West. Yes, it takes more fuel to launch that way, but according to your thoughts, launch to the West will INCREASE the Earth's rotation rate, correct? The satellite will increase in angular momentum regardless of direction of launch. Now, if angular momentum is conserved in the satellite/Earth system, the Earth will still have to decrease its angular momentum regardless of the direction of launch, if the satellite gains in its amount of angular momentum. Same for a polar launch, the satellite will gain angular momentum acording to calculations after launch. Do you still maintain this 'increase' in angular momentum by the satellite comes from a decrease in Earth's angular momentum?

Edit: removed 'geostationary' from postAngular momentum is a (pseudo) vector quantity. So an "increased" angular momentum in the opposite direction is essentially simply a large negative angular momentum.

-Dale

The IERS does not mention your effect. They do however, have analyses such as this which shows the variations in length of day to sub-millisecond level from 1995 to 1999 (http://www.iers.org/MainDisp.csl?pid=95-99). I can see a long term drift in this plot, well explained by momentum transfer to the moon's orbital motion. I can see seasonal effects.That is a really interesting plot. I assume that the seasonal effects you mention are the local minima near the half-year marks. Do you know what that is from?

I think the satellite-launch effects are obviously undetectable, particularly with such large natural variations. I don't think that the thread is intended to be anything other than an exercise in estimation. Assuming no other effects than the satellite launch, assuming no rocket exhaust escapes, assuming a perfect solid spherical earth, etc. can you estimate the magnitude of the effects to a couple orders of magnitude?

-Dale

Pete, the formula for angular momentum is the moment of inertia times the angular velocity. It is NOT the same as for linear momentum. Mass times angular velocity does not give angular momentum. I think you are calculating the rotational kinetic energy instead of angular momentum.

Do you read your own posts?


Here is one cut & paste concerning angular momentum:

"angular momentum
A quantity obtained by multiplying the mass of an orbiting body by its velocity and the radius of its orbit. ..." http://imagine.gsfc.nasa.gov/docs/dict_ad.html

Moment of inertia times angular velocity is the same as linear momentum times radius. See CANGAS's post for the proof if you don't believe NASA.

Anyway, it's still easy to plug in the numbers no matter what forumla you choose.

Why don't you do it?

Angular momentum is a (pseudo) vector quantity. So an "increased" angular momentum in the opposite direction is essentially simply a large negative angular momentum.

-Dale

So what is the 'increased' angular momentum in the polar launch/orbit called?

...The orbit transfer {from what the launch achieved} is performed either using timed burns or ......At some point it is placed very precisely. That is all that matters -- except for the minor issue of where GEO is, and how launch and travel to GEO changes GEO altitude -- i.e., the point of this thread. we agree on all this. I wanted people to only consider the thrusting of the rocket with the assumption that all the exhaust gases were retained on Earth I think we agree that where on Earth is not very important, but to be specific I suggested aluminum + oxygen making Al2O3 dust exhaust that fell to Earth. when worrying that some might get bothered by fact that the CoM of fuel before launch was slight higher than the Al2O3 would be after satellite is in orbit, I suggested the Al2O3 could be caught on a platform that circled the Earth at the original height of the CoM of the fuel in rocket pre launch. I did not want to have people worring about the change in the total system angular momentum that is associated with exo-atmospheric burns of tiny thruster need to correct exactly the orbit, not only for changes in length of the day but solar pressure, residual drag, magnetic interactions with Earth's field etc. Most people understood the question posed, correctly I think.


...My answer remains, none of the choices corresponds to "SLIGHTLY". With the cumulative effect of all of the spacecraft launched by mankind over the last 45 years, even a few micrometers change would be observable. Nanometers or picometers are SLIGHTLY, but you did not offer those as choices. yes i did so offer. your estimates are in the first group, one I have several times called bottom less. I think you voted for this group with the majority and may be correct. I will only list 2inqusitive as voting for the zero group as he clearly thinks it is exactly zero change. That is the displayed count of the bottomless group is correctly including your vote.

I think our disagreement has been reduced to the observability of the effect. You claim it is observable as timing is very precise. I claim it is not observable as there is much greater variation in larger know effects (signal to noise problem makes effect of even many, certainly one, satellite launch on Earth's spin rate unobservable.) You have given a number for the rate of spin decrease and from this conclude, with your analysis, that because such a small rate of decrease is known the effect of satellite launch would be observed, if I understand your position correctly. I can only note that current decade average rate of spin decrease, is not the same as the average observed over several thousand years. (This by noting where on Earth the total eclipse, whose time of occurrence even 3000 years ago is very accurately known, was actually observed - written down by Chinese scholars while my ancestors and most of those active here in English were ill-literate fools thinking the gods were angry or something like that if they happened to experience a solar eclipse.)

I have already congratulated you on your use of the a^3 ~ T^2 law. I probably would have done the same, but doubt I would have made it even simpler, A LOCALY LINEAR equation, as you did with a first order Taylor series expansion about the 24-hour day point. That was clever.

I.e. For people here who have no idea as to what you did, let me put it this way:

Effectively, DH plotted the "a vs. T" curve for a small region very close to T = 24 hours and then used the slope of that curve to estimate the effect of a change in A.

He found that a one-meter change would correspond to roughly 0.003 seconds on the length of the day. As he knows that the long-term average rate of change is much smaller, he concludes that this great (0.003s) a change is impossible, and he may well be right.

I note, however, that the three of us in the 1cm to 1M band are betting that the change is between 0.000,3 and 0.003 seconds. DH's argument has persuaded me, if his numbers are OK and they seemed to be, that our only hope to be correct is near the low end of this band. For example if 0.000,3 seconds is the change, then in 365 days we have about 0.11 second longer year for each 1000KG put into geostationary orbit that year.

2005 did require the addition of one "leap second" at end of year but if I could vote again, armed with HD's information, and assuming it correct, I would at least move down to the 0.1 to 1cm band if not join him and others in the "bottom less " majority. Anyone want to argue for a band other than the "bottomless band"? I think the E-4 to E-2M band surely has a chance still. - I only rejected it as stated earlier because it is 100 times smaller than the next greater one. It has been fun, and for me at least educational.

The effect won't be noticeable until we have launched about half a million satellites into orbit.

I think to many on this post this means that DH agrees that there is an effect. To me the question was: Is there an effect. The answer is yes. But I agree with DH in that the effect is so very small that it could go unnoticed for a very long time.

Certainly the effect is not large enough to have to correct for after launching one satellite.

Angular momentum is a (pseudo) vector quantity. ..
What do you mean by "pseudo"?

L = M R x V sure looks like a real vector to me.

I have already congratulated you on your use of the a^3 ~ T^2 law. I probably would have done the same, but doubt I would have made it even simpler, A LOCALY LINEAR equation, as you did with a first order Taylor series expansion about the 24-hour day point. That was clever.

This kind of linearization is widely used. Non-linear equations are a bear to work with. It is so much easier to work with linear equations. So, when faced with a non-linear system, a common solution is to linearize (and then of course pay attention to the errors that result).

You mentioned in an earlier post that spacecraft are overjoyed if the spacecraft is delivered even close to the target. The targetting algorithms used to bring the spacecraft from "close" to spot-on are linearized equations called the "Clohessy-Wilshire Equations" (or just CW).

I think to many on this post this means that DH agrees that there is an effect.

Did you read my post? I prefaced that series of analysis with "playing devil's advoate here ...".

What do you mean by "pseudo"?

L = M R x V sure looks like a real vector to me.

Pseudo-vectors don't transform like "true" vectors. In particular, a true vector transforms to its additive inverse on reflection (negation of the coordinate axes). A pseudo-vector stays the same on reflection.

Yes, I know about the difference between a mean and "true" day. The point of my post was that if the hypothesized effect existed, the scientific community would know about it because time is the most precisely measured of all physical quantities.
No, they wouldn't know about it because the size of the effect is orders of magnitude smaller than other unpredictable effects.


You mentioned the IERS. This IERS web page (http://www.iers.org/MainDisp.csl?pid=95-84) describes the sources of the variability of the earth-rotation vector:

They do not mention satellite launch as one of the causes of the variability.
That's right, because the effect is insignificant beside the other causes.


A change in the size of a geosynchronous orbit by a meter to one hundred meters (your vote) is anything but minute.
Oh!
My vote was a guess! A wildly wrong guess. By my calculation (which I'm not sure I trust, but it's better than a guess), the correct answer is perhaps a tenth of a millimeter. Sorry about that. :o


To be able to say with confidence that the length of a mean day is decreasing (on average) by 1.6 milliseconds per century, does that not mean we know the length of a mean day to better than 0.1 milliseconds per century, or 13.5 decimal places? (0.1E-03/3.2E9=3.2E-14.)
It means we know the mean length of a day now to within 0.1 milliseconds, and we know the mean length of a day then to within 0.1 milliseconds.

This means that we know the average daily change over the past century to 13 decimal places... but that's not much use when the actual daily change varies in the order of millseconds, is it?



I stand by my previous statement that the angular momentum of the combined Earth+satellite system is not conserved when a spacecraft is launched into orbit.
Including exhaust, right? I'm still disagreeing (but you knew that :) ).


The satellite is connected by gravitational strings to the Earth. However, it is also connected by the same gravitational strings to the Sun, Mars, Alpha Centauri, and to the most remote pulsar. The angular momentum of the entire universe is conserved when a satellite is placed into Earth orbit. However, the angular momentum of the Earth+satellite system is not.
Are you maintaining that the gain or loss in the Earth+satellite system angular momentum comes from other astronomical bodies? How, exactly?

If a satellite was launched from a planet in an otherwise empty universe, would the angular momentum of the planet+satellite be conserved?

Just when I thought I brought DH into the real world of cause and effect and a laws hundreds of years old Pete points out that he did say that he said:
The angular momentum of the entire universe is conserved when a satellite is placed into Earth orbit. However, the angular momentum of the Earth+satellite system is not.

DH is right that the satellite is part of a much larger system as is every thing else in the universe including the earth. Take venus for example or murcury
Venus is in some sort of resonence with the earth so that it shows the same face to earth whenever we are nearest to it. Murcury shows the same face to the sun all the time. But DH misses the point that in our earth satellite system the earth is already slowing so any additional slowing would not be counteracted by the sun or the moon or anything else likely. I cast my vote with Pete.

... Certainly the effect is not large enough to have to correct for after launching one satellite. I agree but note that both DH and I concur that the launch only get the satellite close - I.e. entire launch crew celibrate if they turn a satellite over to controllers with a period of 24 hours plus or minus one minute.

Point being that all geostationary satellites have, and use often, small thrustors to "station keep"* against much greater problems than the lauch of another geostationary satellite changing the Earth's spin rate.

To do this they will have probably made some exo- atmospheric final burns, that as Dale has explicitly noted should not be considered in the guessing of the answer (not that they would make any difference, except in principle they violate conservation of angular momentum of our system if any of their exhaust goes away from Earth.

I mentioned some reasons before why it is necessary to have some small thruster to keep on your assigned geostationary spot so here will just tell that once, when some one else was sick, it fell my lot to go the NASA, in Greenbelt MD I think it was, to place small magnets on a satellite to try to zero out the residual magnetic field satellite had. That is an art, not a science, and I needed help by NASa's local experienced "magnetic balancing artist."

I initially thought: OK they just measured the net field and orientations for me, so I will put and "equal and opposite" strength magnet there to cancel the measured field out. That idea does roughly work at first on killing the gross initial field, but the new magnet induces fields in most materials, even those not normally considered magnetic.

This satellite's main job was to carefully measure the magnet field of the Earth, solar wind effects on it, etc. so the field of the satellite was taking to space had to be almost killed. After about 15 much smaller magnets than my first try, had been added to the satellite with help of the "magnetic artist" the satellites own field at the end of the extended boom where the magnetometer would be was “nulled out.” I had to go to that NASA facility as they have a test room with large coils that took away the Earth's field.
-------------------------------------------------
A station in the geostation 360 degrees is quite valuable and the UN has given rights to various nations. I believe a small spot is "owned" by some UN member - a tiny island country in South Pacific, not on the tourist routes, and renting its spot to telcom companies is its major source of foreign exchange!

Are you maintaining that the gain or loss in the Earth+satellite system angular momentum comes from other astronomical bodies? How, exactly?

This raises the question, What is inertial mass, and why is it the same as gravitational mass? The most satisfactory answer that I have found is that inertial mass of some object (and hence its inertia) is the sum of the gravitational interaction between the object and all of the other objects in the universe. The combined gravitational acceleration toward every other object in the universe is what makes an object resist change to its state.

Turning the table, how exactly does a thrusting rocket change the Earth's rotation rate? Please don't answer with conservation of angular momentum, as I am of the opinion that you have drawn the boundaries of the system incorrectly.

Suppose the rocket is in LEO and performs a Hohmann transfer to GEO. Does that action change the Earth's rotation rate? How? Now suppose the rocket is orbiting Jupiter. Does firing the rocket's engines change the Earth's rotation rate in this scenario? What is the difference?


If a satellite was launched from a planet in an otherwise empty universe, would the angular momentum of the planet+satellite be conserved?

Given my answer to the first question, I don't think so. I suspect the laws of conservation of linear and angular momentum arise because the universe is so large and so uniformly distributed.

Pseudo-vectors don't transform like "true" vectors. In particular, a true vector transforms to its additive inverse on reflection (negation of the coordinate axes). A pseudo-vector stays the same on reflection.Thanks. Your knowing this and yet thinking the distant stars are significant and prevent conservation of angular momentum in the Earth + rocket + satellite system plus your knowledge about Kepler, orbits, etc. makes me willing to guess about you also:

Your role is in the math of trajectories, planet swing by effects etc. not in the physics of building satellites and testing them - right? or wrong? I.e what do or did you do wrt satellites?

I agree but note that both DH and I concur that the launch only get the satellite close - I.e. entire launch crew celibrate if they turn a satellite over to controllers with a period of 24 hours plus or minus one minute
I see what you're saying, but the point is not how accurate we are - even if we were off by 5 minutes - I think that is besides the point - a satellite off by 5 minutes still expended enormous amounts of fuel to get there and any small corrections it makes while in orbit is 1/10,000 at most the fuel it took to get into orbit. So while I can understand what you're saying, do you really think it is relevant to this discussion? The question was meant to be a thinking exercise I thought, not a litteral one.

This raises the question, What is inertial mass, and why is it the same as gravitational mass? The most satisfactory answer that I have found is that inertial mass of some object (and hence its inertia) is the sum of the gravitational interaction between the object and all of the other objects in the universe. The combined gravitational acceleration toward every other object in the universe is what makes an object resist change to its state.Was just about to trun off computer for day and have not read rest of this, but know you are repeating Mach's position do you not? will read tomorrow more.

Was just about to trun off computer for day and have not read rest of this, but know you are repeating Mach's position do you not? will read tomorrow more.

Yes, I know I am repeating Mach's principle, and I also know it doesn't fully agree with General Relativity. So I live in a world of conflicting models. They are the best ones available.

I too must turn off the computer and go study a different kind of rotational inertia. Thursday night is when my wife and I go out dancing.

So what is the 'increased' angular momentum in the polar launch/orbit called?Perpendicular.

-Dale

This raises the question, What is inertial mass, and why is it the same as gravitational mass? The most satisfactory answer that I have found is that inertial mass of some object (and hence its inertia) is the sum of the gravitational interaction between the object and all of the other objects in the universe. The combined gravitational acceleration toward every other object in the universe is what makes an object resist change to its state.
An interesting question it is... perhaps another thread?
Would you mind if we assume Newtonian mechanics and gravity in this thread?


Turning the table, how exactly does a thrusting rocket change the Earth's rotation rate?
The exhaust of the rocket applies a torque to the Earth.


Suppose the rocket is in LEO and performs a Hohmann transfer to GEO. Does that action change the Earth's rotation rate?
Yes (undetectably, of course), if the rocket's exhaust reaches Earth.


Now suppose the rocket is orbiting Jupiter. Does firing the rocket's engines change the Earth's rotation rate in this scenario?
Same answer.


What is the difference?
The exhaust from the rocket around Jupiter isn't likely to reach Earth.


Given my answer to the first question, I don't think so. I suspect the laws of conservation of linear and angular momentum arise because the universe is so large and so uniformly distributed.
What does Newton's laws predict?
If a satellite was launched from a planet in an otherwise empty universe, do Newton's laws predict that the angular momentum of the planet+satellite would be conserved?

What do you mean by "pseudo"?

L = M R x V sure looks like a real vector to me.I think for most quantities defined by a cross product you wind up with a Pseudovector (http://en.wikipedia.org/wiki/Pseudo_vector). If you invert the coordinates so that R' = -R and V' = -V then L' = L instead of -L.

-Dale

Pete:

I have bad news for you.

It was Newton who first proposed that inertia is the result of the gravitational activity of all of the mass in the universe. It was the result of his famous water bucket experiment. Not to be confused with Faraday's famous ice pail experiment.

It has also been expressed as the result of the majority of mass of the universe.

Mach belived this also, according to his public statements.

Einstein believed this also, according to his public statements. He also said that he regretted that he had failed to be able to incorporate Mach's principle ( actually Newton's principle ) into the relativities.

The concept that inertia is the result of all the mass in the universe IS NEWTONIAN PHYSICS.

Yes indeed. And was it not Newton that said:

" All forces occur in pairs, and these two forces are equal in magnitude and opposite in direction - except when a rocket is delivering a satellite in to an earth orbit".

That's my favorite quote of Newton's (wipeing eyes - blowing nose) - I just get so emotional when I read Newton.

Angular momentum can also be calculated by multiplying the square of the distance to the point of rotation (look familiar), the mass of the particle and the angular velocity.

I see what you're saying, but the point is not how accurate we are - even if we were off by 5 minutes - I think that is besides the point - a satellite off by 5 minutes still expended enormous amounts of fuel to get there and any small corrections it makes while in orbit is 1/10,000 at most the fuel it took to get into orbit. So while I can understand what you're saying, do you really think it is relevant to this discussion? The question was meant to be a thinking exercise I thought, not a litteral one.I agree. I only want all to assume for this thread, but not in actual launches, that the rocket burn within the atmosphere placed the satellite into perfect geostationary orbit without any exhaust being releast into space. At times, some here have stated that angular momentum is not conserved. DH has requently spoken of exo-atmospheric burns and interactions with the distant stars, etc. I have been trying to keep thread question focused onthe problem I intended, and most have understood, which can assume only the Earth and the two satellites, A & B, exist. Specifically there are no distant stars or exhaust gasses releasted into space in the thread's problem, however I do understand that given the view of Mach & DH about the origins of inertia etc, it makes no make no sense to pose the problem as I have. That is offset inmy view at least that the nearest star (Baryards?) can not even know about the launch of B until years after B is in orbit so to further define the problem I might say that both A & B were perfect launches (no positioning exo-atmospheric burns, the sun and all else of our solar system, save Earth does not exist and if distant stars must exist to make inertia, etc. then the "SLIGHTLY" of the thread is the radial difference shortly (well be the closest star "knows" about the launch of B) after B is in perfect geo-stationary orbit about the Earth which all but 2inqusitive believe to be spinning "SLIGHTLY" more slowly BECAUSE B has been placed into orbit than it was prior to launch of B.

Hi CANGAS,
Not all of Newton's ideas are included in what we refer to as "Newtonian Physics" (even if it's in capitals).

Assuming Newtonian mechanics and Newtonian gravity means assuming the truth of the three laws of motion and the universal law of gravitation.

CANGAS said:

I have bad news for you.

It was Newton who first proposed that inertia is the result of the gravitational activity of all of the mass in the universe. It was the result of his famous water bucket experiment. Not to be confused with Faraday's famous ice pail experiment.

It has also been expressed as the result of the majority of mass of the universe.

Mach belived this also, according to his public statements.

Einstein believed this also, according to his public statements. He also said that he regretted that he had failed to be able to incorporate Mach's principle ( actually Newton's principle ) into the relativities.

The concept that inertia is the result of all the mass in the universe IS NEWTONIAN PHYSICS.


"For around 200 years Newton's arguments in favour of absolute space were hardly challenged. One person to question Newton was George Berkeley. He claimed that the water became concave not because it was rotating with respect to absolute space but rather because it was rotating with respect to the fixed stars"

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Newton_bucket.html

even if you were right (which you were not).... what the crap does that have to do with anything? Stay on subject. We need a moderator in this thread.
So don't quote quantum physics or anything...quote some 300 year old experiment. Way to stay ahead of the curve. Next time don't hold back, quote something even older like cave paintings or something - but make sure you get it right. (And yes I appologise for being a smart ass)

To DH:

You have asked, more than once:

"[1]What mechanism causes the Earth's spin rate to change?"
AND more recently:
"Turning the table, how exactly does a thrusting rocket change the Earth's rotation rate?"

And have been answered more than once. Pete's reply is much shorter than mine. (I am always too long winded, but the details I gave earlier expand Pete's correct response to the finer details of how Earth is torqued.)
...The exhaust of the rocket applies a torque to the Earth. ...
I asked you for the mechanism by which you imagine that the distant stars / "universe" are used to conserve the angular momentum. You said that it was not conserved locally in the Earth+satellite system, etc. so how do the distant stars /universe assist in the conservation? Mechanistic details, like I gave you yesterday at 47 minutes past the hour, would be a nice reply. I said:

I think [1]'s answer is closely related to fact that a cannon recoils (accelerates backwards) when throwing (accelerating) a cannon ball forward.

That is, to throw forward the satellite (Accelerate to greatly increased speed so it can still go around the center of the Earth point, at geostationary radius instead of only at Earth radius, in essentially 24 hours), by any means, something must "recoil" or "accelerate backwards."

In the case of a rocket, rolling over eastward (instead of traveling radial outward only to fall back radially inward later) it is initially the exhaust gases that "recoil" - are accelerated "backwards."* The mass of these gases, at initial pitchover, have more rearward momentum than the fuel in the rocket tanks had, but of course the rearward moving gases mix with the air and slow, giving this rearward momentum to the atmosphere, which in tern shares it with the solid Earth, etc. - that is the basics coupling mechanism by which some of the angular momentum of total system (Earth + rocket + satellite) is transferred to increase the satellite's and decrease the share of total in the Earth + rocket, when rocket is again resting on the Earth.
-------------------------------------
*The exhaust gases are initially traveling backwards, but they need not be traveling "backards" in the Earth frame for them to be accelerated rearward later. After the rocket is traveling significantly faster than the speed of sound and has rolled over to the East, they will be going "forward" in Earth's frame, when they first part from the rocket, but not going forward as fast as their mass was a few seconds earlier when they were still "fuel."
It is really just Newton's laws. - A force is required to continue accelerating the rocket (and satellite inside) and a force is required to accelerate the fuelmass (converted to "exhaust gas", but so what?). These two forces are the "equal and opposite" forces Newton spoke of. The force acting on the fuel mass (exhaust) has a tangential component, acting on a "moment arm" greater that the radius of solid Earth. That is "produces a torque." First on the air the exhaust is mixing with, but eventally the entire Earth is slowed by this torque.

NOW YOU HAVE TWO ANSWERS - HOW ABOUT GIVING US ONE?

Hard to believe, but 18 pages and a day to go before this two-week only poll closes. Least people think my last post is attacking DH, let me again note I have learned from him more about the utility of the Taylor series than I did in college, where it was usually just a math exercise to be done on an exam etc., with little, (for me) real meaning or practical use (except to get good grade on the test). Thanks to DH, I now see /understand it is powerful, easy to us, tool for problems like this thread's. Thanks again DH.

I agree with Billy T. For me, the object of writing in these threads is to learn. That we get into arguments is besides the point and hopefully we don't take it personally. I'm really laughing half the time over these arguments and I think they should inspire learning and not be taken too seriously. And let's be honest - when we are in an argument - isn't that when we do some of our best thinking and research?

NOW YOU HAVE TWO ANSWERS - HOW ABOUT GIVING US ONE?

I have always been a bit leery regarding arguments by way of the conservation laws. They are a bit more handwaving than I like. They sweep all the details under the rug; they pay little attention to cause and effect. Most importantly, the results will be incorrect if the boundaries are drawn incorrectly. Because the details have been swept under the rug, there is no sanity check to ensure that the results are true.

It is imperative that the boundaries be drawn correctly when applying the conservation laws. Why did you choose the Earth+satellite for the boundaries? Why not the Earth, Moon, and satellite, the solar system+satellite, or the Milky Way + satellite?

That is not to say that I don't resort to the conservation laws myself in my work. They are very powerful because they sweep all the details under the rug. For example, I don't care about the details of contact dynamics. I definitely do care how a spacecraft behaves after docking or undocking is complete. I just sweep the details of docking by means of the conservation laws.

The reason I specifically asked about exo-atmospheric burns is because the exhaust from such burns is, for the most part, swept away from the Earth by solar radiation pressure. How does that exhaust exert a torque on the Earth? It never contacts the Earth. To argue that it must because of the conservation laws begs the question. I want forces and torques; I want equations of motion.

I do agree that, up to the point that the satellite leaves the atmosphere, the exhaust from the spacecraft does indeed change the angular momentum of the Earth by an immeasurable amount. Conservation of angular momentum can be applied to assess this change. All bets are off once Elvis has left the building.

DH said:

The reason I specifically asked about exo-atmospheric burns is because the exhaust from such burns is, for the most part, swept away from the Earth by solar radiation pressure. How does that exhaust exert a torque on the Earth? It never contacts the Earth. To argue that it must because of the conservation laws begs the question. I want forces and torques; I want equations of motion.

The percentage of the total burn of putting the satellite into orbit excapes the earth's gravity I think it is less then .1 % Why even argue about this?

But you may be right - for that very small amount - it is conserved in the universe and not the immediate system.

Why did you choose the Earth+satellite for the boundaries? Why not the Earth, Moon, and satellite, the solar system+satellite, or the Milky Way + satellite?Because it is easier. That's also why we required that all of the exhaust return to the earth. We are all aware that real rockets spew exhaust to the far reaches of the universe and said exhaust must be included in a realistic analysis, but we did not want to complicate the problem. It is just an estimation game.

-Dale

DH said:


The percentage of the total burn of putting the satellite into orbit excapes the earth's gravity I think it is less then .1 % Why even argue about this?

But you may be right - for that very small amount - it is conserved in the universe and not the immediate system.

Because the original question was about GEO, not LEO. Getting out of the atmosphere is but a fraction of the total problem of getting to geosynchronous orbit.

Getting out of the atmosphere is but a fraction of the total problem of getting to geosynchronous orbit.

Ya in as much as 9/10 is a fraction. But that's besides the point - we are talking about where the fuel winds up, we are not talking about logistics - or what a pain in the but it is. I submit that 999 parts of gas out of 1000 of the origional feul that the rocket started out with, remains in, or winds up in the earth's atmosphere, (subsequently causing friction) and you of all people should know that.

The reason I specifically asked about exo-atmospheric burns is because the exhaust from such burns is, for the most part, swept away from the Earth by solar radiation pressure. How does that exhaust exert a torque on the Earth?
It doesn't, of course. I don't think anyone said it did?


I do agree that, up to the point that the satellite leaves the atmosphere, the exhaust from the spacecraft does indeed change the angular momentum of the Earth by an immeasurable amount. Conservation of angular momentum can be applied to assess this change. All bets are off once Elvis has left the building.
Then we're done.
Like I said early in the thread, Billy's original post assumes that all fuel expended from the launching rocket returns to Earth.

Tortise:

You're too tense. Stop taking your meds again?

Yes I did. Thank you CANGAS. Are you upset you were wrong about Newton? I was just joking...when I said that you shouldn't hold back and just quote cave paintings - and that was if you were even right about the experiment you were talking about which you know now you weren't. Don't take it so personally it's just friendly humor.

It doesn't, of course. I don't think anyone said it did?


Then we're done.
Like I said very in the thread, Billy's original post assumes that all fuel expended from the launching rocket returns to Earth.

You all keep discussing 'the expended rocket fuel adds angular momentum to the Earth'. Here is what I don't understand. The total momentum of the rocket/fuel system is zero, due to conservation of momentum. The expended fuel has an equal and opposite reaction to the rocket, giving motion to the rocket of course. If the exaust gasses also give momentum to the Earth's rotation, wouldn't that upset the conservation of momentum in the rocket/exhaust gas system, effectly using the exhaust gas to give momentum to both the rocket and the Earth?

Tortise:

Whether I was exactly correct in quoting Newton or not, I am too lazy and too disinterested in engaging in an argument which I could only consider pointless to worry about it any further.

The origin of the flap was someone's comment or question regarding the theorized nature of inertia, and if anyone has ever proposed a better concept than that of the majority ( or totallity ) of mass in the universe being the agent, I genuinely would appreciate being able to study it.

Perhaps Tortise knows how inertia REALLY works?

If it was postulated ( where was that, anyway? ) that all the exhaust remained in planetary confine, then action and reaction, and conservation of momentum have to sum to zero. The only question remaining is the matter of whether the relocation of the satelite from stationary surface location to stable orbit has regaged Earth rotational velocity.

The problem is, did the rapture ( snatching up ) of the satelite alter the rotation of the planet?

You're right that the sum would be zero, and that's actually a very good way of looking at the problem. That's actually how scientists look at the problem - if the satellite has more angular momentum in the orbit, then that momentum must have mostly come from somewhere in order for the system to remain the same.

You all keep discussing 'the expended rocket fuel adds angular momentum to the Earth'. Here is what I don't understand. The total momentum of the rocket/fuel system is zero, due to conservation of momentum. The expended fuel has an equal and opposite reaction to the rocket, giving motion to the rocket of course. If the exaust gasses also give momentum to the Earth's rotation, wouldn't that upset the conservation of momentum in the rocket/exhaust gas system, effectly using the exhaust gas to give momentum to both the rocket and the Earth?
Hi 2inquisitive,
Don't forget that momentum is a vector... so if some system starts has zero momentum, it can split in two with each part having momentum in opposite directions. The total momentum of the system is still zero, because the opposite momentums add to zero.

If the exaust gasses also give momentum to the Earth's rotation, wouldn't that upset the conservation of momentum in the rocket/exhaust gas system, effectly using the exhaust gas to give momentum to both the rocket and the Earth?You are correct here. The conservation of momentum applies for isolated systems. By having the rocket exhaust impact the earth the rocket/exhaust system is no longer isolated and must be extended to rocket/exhaust/earth. Momentum is then no longer conserved between rocket and exhaust, but is conserved among all three.

This is exactly what D_H was talking about with "drawing the boundaries" being critical in applying conservation techniques and what his argument here has been. The problem has been set up so that all exhaust returns to earth, therefore you can consider the system to just be rocket/earth. D_H (correctly) points out that this is not entirely accurate and some exhaust will escape earth. This then would require knowledge of the momentum of escaping exhaust in order to correctly apply the conservation laws.

-Dale

2inq said:

You all keep discussing 'the expended rocket fuel adds angular momentum to the Earth'. Here is what I don't understand. The total momentum of the rocket/fuel system is zero, due to conservation of momentum. The expended fuel has an equal and opposite reaction to the rocket, giving motion to the rocket of course. If the exaust gasses also give momentum to the Earth's rotation, wouldn't that upset the conservation of momentum in the rocket/exhaust gas system, effectly using the exhaust gas to give momentum to both the rocket and the Earth?

In a sense you are right: You could think of it as accelerating the earth but in the oppisite direction that it is rotating. But the end result is still that the period of rotation has slowed.

...It is imperative that the boundaries be drawn correctly when applying the conservation laws. Why did you choose the Earth+satellite for the boundaries? Why not the Earth, Moon, and satellite, the solar system+satellite, or the Milky Way + satellite?

.....I do agree that, up to the point that the satellite leaves the atmosphere, the exhaust from the spacecraft does indeed change the angular momentum of the Earth by an immeasurable amount. Conservation of angular momentum can be applied to assess this change. All bets are off once Elvis has left the building.Answer to first paragraph question is because the thread is ignoring the realities that there are torques on the Earth from moon and other planets, sun, etc. Several times this has been explictitely stated, even to the extreme of suggesting that the Earth as only object in universe was the case considered by thread.

Unfortunately, you have not been willing to conform to the threads problem, but want to include many real effects, exo-atmospheric thrusting, moon etc. torques, etc. Several of these real effects are much larger and I believe make the actual observation of the thread's tiny effect impossible. - The unknown variations in some of these other effects, such as the seasonal change in Earth's I, is even "much larger" and thus you can not actually measure a tiny effect, if you can not correctly subtract out a bigger effects because knowledge of exactly how much bigger the larger effect is, is not acailable.

I also note that you still have not given even the slightest hint as to a possible mechanism by which the rest of the universe plays a role in torquing the Earth, or making the conservation of angular momentum associated with a man made change, (Place B into orbit in less thn a day), can even be known by objects that are light years away.

Again can you please describe how the rest of the universe enters into the change made by man in a day? You still have not attempted an answer to this question - obviously I think there is none, but I will hear you out if you have one.

Don't forget, the rocket is not 'turned' towards East until it is nearly free of the atmosphere. A rocket sent into orbit travels straight up until it reaches the thin upper atmosphere. The less time spent powering through the dense atmosphere, the better. In that situation, most of the exaust gas's effect is directed perpendicular to the Earth, not 'pushing it opposite the direction of rotation'. The rocket gas may push against the atmosphere, but the rocket body pushes against the atmosphere in the opposite direction. Aren't these two 'pushes' supposed to be equal in magnitude due to conservation of momentum in the rocket/exaust gas system?

Another point brought up by D H before is that the supposed slowing of the Earth's rotation through satellite launches is not borne out by evidence.

The 24 hours of a solar day equals 86,400 seconds. The Earth rotates once every 86,400.002 seconds today. By the way, the 'true' rotation of the Earth, the sidereal day, is almost 4 MINUTES less than 24 hours. The Earth is actually rotating faster than once every 24 hours.

Back to the 86,400.002 second figure. In 1820, the solar day was exactly 86,400 seconds long. The Earth has slowed .002 seconds per rotation since then, about 186 years. By checking ancient observations of eclipses, the Earth is estimated to slow about 2 milliseconds each century for the last few thousand years. The actual slowing of the Earth's rotation seems to be at a slightly lower rate now than in the past, not more due to recient satellite launches.

I stand by my statement that satellite launches have zero measurable effect on Earth's rotation.

Weeeelll,

Since the poll is closed and the original idea that I had was bogus anyway due to my naievete regarding conservation of angular momentum (thinking that tossing a satellite out from the earth was the same as a skater stretching his/her arms out - no torques though - duh) I will give you my answer assuming that the satellite is raised to geosynchronous orbit by a massless, infinetly rigid telescoping pole. Take that!

I find under these very realistic conditions that the angular speed of the earth decreases at a magnitude of 10^-20. Since the geostationary altitude is related by the cube root of the square of the orbital period, it changes by something on the order of 10^-20 to 10^-21 also. This is somewhere between an angstrom (10^-10meters) and the Planck length (1.6x10^-35meters).

I stand by my statement that satellite launches have zero measurable effect on Earth's rotation.

I'm with you 2inq.

Don't forget, the rocket is not 'turned' towards East until it is nearly free of the atmosphere. A rocket sent into orbit travels straight up until it reaches the thin upper atmosphere. The less time spent powering through the dense atmosphere, the better. In that situation, most of the exaust gas's effect is directed perpendicular to the Earth, not 'pushing it opposite the direction of rotation'. The rocket gas may push against the atmosphere, but the rocket body pushes against the atmosphere in the opposite direction. Aren't these two 'pushes' supposed to be equal in magnitude due to conservation of momentum in the rocket/exaust gas system?It is true that if the centroid of the rocket thrust were directed towards the center of the Earth, until the rocket was far above 99+% of the atmosphere and only then rolled over to the East, there would be even less effect upon the spin rate of the Earth. However this is not the assumption of the thread. Several times I have stated, and DaleSpan at leasst twice also, that NO EXHAUST GAS escapes from the Earth.

In actual practice, this restriction is nearly true. For safety reasons a US launch from the begins a roll over to the East so if the rocket explodes, the pieces will fall in the Atlantic ocean. Compared to the thrust of the rocket, the frictional drag of the air is negligible, so getting out of the air quickly s possible is not very important. I would not be surprised to learn that having he exhaust "press against the atmosphere" increases the force on the rear of the rocket more that the friction of air flowing by its side subtracts form the rocket thrust, but do not know if this is true. Likewise, there must be a very small effect fo the atmospheric presure difference between the "top side" of a rocket rolled over to 45 degrees compared to the "bottom side." (Air planes are held up by a reated pressure difference, but of course their wings are designed for this - a fast moving ponted cylinder (the rocket) will "surf" on this pressure difference very little, I think.) Also it cost very little to make part of the the kinetic energy being given to the rocket be associated with Eastward motion, instead only climbing against gravity. For all of these resaons, even in practice, rockets usually roll over to the East, if they are intended to go into orbit. If it is a much less capable rocket, used only to get a few minutes above most of the atmosphere, then a nearly straight up trajectory is desirable, if safety permits.

In any case, you like DH, are not accepting the thread's assumptions, but discussing some other case.

...
I stand by my statement that satellite launches have zero measurable effect on Earth's rotation.Here it is you, me, and SL (and probably several others) against DH. I.e. We all agree the effect is entirely impossible to MEASURE,but if you are still claiming that this is because it is exactly zero, I think you are standing alone.

Poll is closed, but I wonder if anyone want to answer the following closely related question (by math demonstration, preferred or guess/opinion if that temps you still):

Assume universe consists of two objects only, one of mass M and the other of mass m. They are rotating about their common center of mass with period P (although without at least a grain of sand on the other side of the universe as a "fixed star" to anchor the reference frame, we will need a bucket of water to see this rotation.) So that all will use the same symbols, the distance between them is D. I think most will now grant that the angular momentum about the center of mass is unchanging constant, L.

Guess yes or no to the following four questions:

Q1: Is the angular momentum about the center of mass of M also L?
Q2: Is it also L about any point on their trajectory where they are not?
Q3: Is it also L about any point in the plane of their trajectory?
Q4: is it also L about any point off the plane of the trajectory?

Note Q4 is asking if it is the same for ALL points in space, even that grain of sand on "other side of the universe."

If you answer "no" to any of the above, and can tell / know correct answer, wait until 1 April to do so.

If you are just guessing and say "no" to any of the four, guess if angular momentum is greater or smaller, or could be either, depending upon where the point is in case of Q2, q3, & Q4.

Lets see how many "April fools" we have. (Remember "fool," at least originally, referred to one who amuses us, sort of an honor, or at least a service.)

by Billy T:

"It is true that if the centroid of the rocket thrust were directed towards the center of the Earth, until the rocket was far above 99+% of the atmosphere and only then rolled over to the East, there would be even less effect upon the spin rate of the Earth. However this is not the assumption of the thread. Several times I have stated, and DaleSpan at leasst twice also, that NO EXHAUST GAS escapes from the Earth."
================================================== ==============

Billy T, no exhaust gas will escape the Earth unless that gas is moving away from the Earth at above escape velocity. Since that exhaust gas is directed toward the Earth even when the rocket trajectory is tilted in the upper atmosphere, it does not escape Earth's gravitational field until the rocket is in the vacuum and only then if the gas were directed away from the Earth at escape velocity relative to the Earth. A rocket does not reach escape velocity itself, if the objective is to place a satellite in ORBIT.
This has nothing to do with conservation of angular momentum in Earth's rotation, however. The exhaust gasses are still a part of Earth's mass. The only loss of mass concerning the Earth's angular momentum would be if the mass of the rocket itself escaped Earth's gravitational field (escape velocity, not orbital velocity) and entered orbit around the sun. Your gedankin specifically states ORBITAL insertion of a rocket/satellite.

Here it is you, me, and SL (and probably several others) against DH. I.e. We all agree the effect is entirely impossible to MEASURE,but if you are still claiming that this is because it is exactly zero, I think you are standing alone.

The way I see it, if the effect is impossible to measure, even after the accumilated effects of thousands of launches, the effect is zero.

You do know the Earth is getting more spherical due to the loss of angular momentum, don't you? Angular momentum is a centrifugal force, causing the Earth to be a larger diameter around the equator than around the poles. As the Earth slows in its rotation, this relationship is changing. The crust of the Earth is literally tearing apart around the equator, the Pacific and Atlantic ridges. The crust of the Earth is falling and drifting north/south around the equator, causing moltent rock to ooze through the cracks. This has been confirmed by the LAGEOS satellite. It has been measured.

....Billy T, no exhaust gas will escape the Earth unless that gas is moving away from the Earth at above escape velocity. True if the rocket were firing a machine gun or canon backwards, but not true if it is, as of course it is, a gas that is leaving the rear of the rocket. But again, before explaining this, let me note that neither the rocket nor any of the exhaust "go into orbit" Recall that to make this clear, impossible to miss understand, I said the fuel was aluminum dust and O2, which burn together to make Al2O3 dust that falls to the Earth surface,

Now about you gas in space, even though it is contrary to the thread's assumptions:

First note it will rapidly expand into a vacuum, and if in sunlight, surely ionize. Being very small particles, perhaps only atoms or molecules, these charged particles will orbit about the magnetic field lines. The solar wind pushes these field lines about quite strongly if you are far enough form Earth (a few Earth radii) to be speaking of "vacuum," instead of "residual atmosphere." (There is not clean line between these terms.)

These particles will also feel solar radiation pressure. Recently a private group paid Russians to but up their "solar sailor" space craft, which was to go (slowly) far from Earth on this light flux, never achieving anything like "escape velocity" but, escaping. Unfortunately, the launce failed some way - I forget details.

Because of solar particle and radiation fluxes plus the dynamics of Earth's magnetic field, especially the "tail" being whipped around by fluctuations in the solar wind (particle flux from) some of these gases do not need to coast away from Earth, with "escape velocity." They will be accelerated away by the sun. The sun will also keep them warm. In a thermal distribution of velocities (Some what doubtful that their collisional interactions can maintain this, so they really do not have any "temperature," but still will have some distribution of velocities.) a fraction will in fact have the escape velocity.

In short summary:
Because of these effects, and fact that even at geo-stationary distance the satellite is bound, (has significantly less than the escape velocity) I suspect that 99+% of all the gases that leave the micro thrustors used to fine position the satellite into its assigned geo-stationary position are:
(1) With less than Earth escape velocity,
yet
(2) do escape from the Earth.

Well, here is how I worked it. Since the satellite is going to be in geosynchronous orbit we can consider the satellite/earth system to be a single rigid body with the satellite attached to the earth via a massless rod. We can then find the geosynchronous orbit radius as a function of the length of the massless rod.

Starting with:
T˛ = 4π/(GM) ał
L = I ω
ω = 2π/T
I = Ie + m r˛
where T is the period, G is the universal gravitational constant, M is the mass of the earth, a is the geosynchronous orbit radius, L is the angular momentum, I is the moment of inertia of the satellite/earth, ω is the angular velocity, m is the satellite mass, r is the length of the rod, and Ie is the moment of inertia of the earth.

Combining we get:
√(ał) = √(GM)/L Ie + √(GM)/L m r˛

However, we are not actually interested in a but rather in changes in a. So differentiating both sides we get:
3/2 √(a) da = √(GM)/L m 2r dr
da = 4/3 mr/L √(GM/a) dr

Now, integrating over r from re, the earth radius, to a we get:
Δa = 2/3 m/L √(GM/a) (a˛-re˛)

Which is about
Δa = 6E-13 m

Like superluminal this is also between the Angstrom length and Planck length, but closer to Angstroms than his result.

-Dale

Well, here is how I worked it. Since the satellite is going to be in geosynchronous orbit we can consider the satellite/earth system to be a single rigid body with the satellite attached to the earth via a massless rod....
I = Ie + m r˛
..., r is the length of the rod, and Ie is the moment of inertia of the earth..
..
Now, integrating over r from re, the earth radius, to a we get:
Δd = 2/3 m/L √(GM/a) (a˛-re˛)...I'm pretty sure you have moment of inertia of the satellite wrong. Should it not be:
m (R+r)^2
?, where R is the Earth radius. Perhaps your rod goes all the way to center of the Earth?, If yes, then you have no error here.

also I am not clear why you want to integrate over r, or the limits you used {0 & r}? A few words about this would b helpful.

Your general approach up to this integration part is probably like one I would have used, if I were not so lazy. One thing you should note, like I did, is the neat way DH's Taylor series tricks work, but his approach only gives the linearized equation delta t = K delta a, where K is the local "slope" of the t vs a graph (without needing to draw the graph) for the RELATIONSHIP between t and a. As it turns out this is all we need:

delta a = 2/3 a/T delta T
A geosynchronous orbit has a semi-major axis a=42164 km and a period T=86400 sec. Changing the length of a day by a small amount delta T seconds means changing the size of a geosynchronous orbit by about 2/3*42164km/86400sec delta T, or
about 1/3 km change in geosynchronous altitude for each 1 second change in the length of a day (valid only for small changes in the length of a day). turning this around:
delta t = (3/2)*(86400/42164000)delat a = 0.0030672 delta a
Thus for the bottom of my "a band" (1 cm) to be correct, I need the day to be lengthend by 0.30672 seconds. Clearly it is not as that is 112 second in a year.
The E-4 to E-2 band bottom is 1.12 seconds per year, very very probably still to high, but certainly in the ball park of what is observed. Too bad it is mainly the moon, not NASA we must blame for this.

The point of this is that if I had known the answer, I would have made the "biggest band" = "one meter or greater" and added more bands at the bottom, such as E-8 to E-6M, & E-6 to E-4M, then we would need some calculation of delat a, not just a Relationship between a & t.

DH is not only clever with his Tayor series approach, he is also the beneficiary of my poor guessing the range for the poll. His method would not give the answer, if I had guessed better and then set up better set of choices.

I was misslead by the fact (discussed earlier) that the angular momentum in the satellite is linear in the length of the moment arm (distance from center of Earth) but the velocity is not. So I thought: You must need a cm of difference to "store" B's angular momentum in as 42,000,000M changed by only a few cm is also "dam small."

Here's how I got my result:

L = Angular momentum
I = moment of inertia
w = angular velocity
Ds = distance of satellite above the earth

L = Iw

For a solid sphere rotating about its own axis (the earth) I = 2/5 MeRe^2

For a point mass rotating at some distance from the earth’s axis (the satellite on the surface) I = MsRe^2

And for the satellite in orbit I = Ms(Re+Ds)^2

Linit = Lfinal (conservation of angular momentum)

so,

Iinit x winit = Ifinal x wfinal

I init = (2/5MeRe^2 + MsRe^2)

Ifinal = (2/5MeRe^2 + Ms(Re+Ds)^2)

wfinal/winit = Iinit / Ifinal

This is a VERY VERY small change. I had to use a high precision calculator to get:

0.99999999999999999998213...

This means that the angular speed of the earth/satellite system decreases at a magnitude of
10e-20. Since the geostationary altitude is related by the cube root of the square of the orbital period, it changes by something on the order of 10e-20 to 10e-21 also.

An angstrom is 10-10meters. Planck length is 1.6e-35 meters

I am quite dissapointed in my intuition.

Well, here is how I worked it. ...

Combining we get:
?(ał) = ?(GM)/L Ie + ?(GM)/L m r˛

However, we are not actually interested in a but rather in changes in a. So differentiating both sides we get:
3/2 ?(a) da = ?(GM)/L m 2r dr
da = 4/3 mr/L ?(GM/a) dr

You can't do that. r is an alias for a -- they are one and the same variable. What you needed to do, instead, was solve your first equation for a.


Edited to add,
Those question marks should be square root symbols. The reply-to-post mechanism kindly changed DaleSpam's math for me.

First note it will rapidly expand into a vacuum, and if in sunlight, surely ionize. Being very small particles, perhaps only atoms or molecules, these charged particles will orbit about the magnetic field lines. The solar wind pushes these field lines about quite strongly if you are far enough form Earth (a few Earth radii) to be speaking of "vacuum," instead of "residual atmosphere." (There is not clean line between these terms.)

These particles will also feel solar radiation pressure. Recently a private group paid Russians to but up their "solar sailor" space craft, which was to go (slowly) far from Earth on this light flux, never achieving anything like "escape velocity" but, escaping. Unfortunately, the launce failed some way - I forget details.

Because of solar particle and radiation fluxes plus the dynamics of Earth's magnetic field, especially the "tail" being whipped around by fluctuations in the solar wind (particle flux from) some of these gases do not need to coast away from Earth, with "escape velocity." They will be accelerated away by the sun. The sun will also keep them warm. In a thermal distribution of velocities (Some what doubtful that their collisional interactions can maintain this, so they really do not have any "temperature," but still will have some distribution of velocities.) a fraction will in fact have the escape velocity

Because the launch will most likely be in the day - (and if launched at night the earth would be blocking the sun - (there are just two narrow windows where this wouldn't be true)) the vast majority of the gas will be pushed away from the sun back into the earth's atmosphere.
As the rocket is attaining orbital velocity, the gas is not, because it is going the oppisite direction, so the exaust gasses are not sharing the rockets growing independence of the earth and it's atmosphere - making it again much more likely to fall to earth. Add to that the fact that the vast majority of the fuel is expent durring the first part of launch (the first part of launch is when the rocket weighs the most and when the most thrust must be applied to get the rocket up to speed) and in the atmosphere. The remaining rocket/satellite system achieves somewhat of a ballistic trajectory by the time it is completely out of the atmosphere. Gas that excapes the earth's influence (as a fraction of the total amount of exaust) is very rare in a launch of a satellite.

Assuming angular momentum is conserved, and ignoring the Earth's oblateness then this is quite easy using the first-order approximations,

delta T/T = - delta L/L from L = 2*pi*I/T
delta a/a = 2/3 delta T/T from a = k * T^(2/3)

Thus
delta a/a = -2/3 delta L/L

All one has to do is solve for the pre-launch angular momentum, the pre-launch geosynchronous orbit radius, and the change in angular momentum.

a = geosynchronous orbit radius = 4.2e7 m
re = 6.4e6 m
I = 8e37 kg-m^2
T = 86400 s

L = 2*pi*I/T = 2*pi*8e37kg-m^2/86400s = 6e33 kg-m^2/s
delta L = -2*pi/T*m*(a^2-re^2) = 1000 kg*2*pi/86400*((4.2e7)^2-(6.4e6)^2)m^2 = -1.25e14
delta a = 2/3*1.25e14/6e33*4.2e7 = 5.8e-13 m

and similarly,

delta T = - delta L/L T = 1.25e14/6e33*86400s = 1.8e-15 s

Even with about ten thousand satellite launches since the onset of the space age, (average of 100 per year for 40 years), the cumulative effect is far from being observable.

Does no one check my figures?

I'm pretty sure you have moment of inertia of the satellite wrong. Should it not be:
m (R+r)^2
?, where R is the Earth radius. Perhaps your rod goes all the way to center of the Earth?, If yes, then you have no error here.Yes, sorry I wasn't clear. All radii are measured from the center or axis of earth. Altitudes just confuse me.



also I am not clear why you want to integrate over r, or the limits you used {0 & r}? A few words about this would b helpful.No, I integrated from re, the earth radius, to a, the geosynchronous orbital radius. The satellite starts out on the surface of the earth, re, and ends up at the geosynchronous radius, a. That is where the limits of integration came from.

-Dale

Dale,

You didn't answer BillyT's question of why you chose that particular equation integrate, nor did you see my comment that that equation is invalid. You used the variable 'a' in one equation and 'r' in another, but they are clearly one and same variable (a is identically equal to r). dx/dx = 1, no matter what x(T) is.

I answered this previously as 6e-15 m. This is thanks in part to an unnamed poster who gave the angular momentum of a 100-kg satellite in geosynchronous orbit as 1.3e11 kg-m^2/s.
:o

No. I didn't check your figures. Actually, they were off by a lot more than a factor of 100. Try again, this time using 4.2^2 = 17.6, not 13 and using 42,000 km = 4.2e7 m.
I think 1.3e13 kg.m^2.s^-1 is correct.

Mass x Radius x Velocity = 4.2 x 3 x 10^12

L x w should be exactly the same, I think...

Mass x Radius^2 x Velocity/Radius = Mass x Radius x Velocity, right?

Who cares.

The current level of wine in my systenm argues that this is a point;less crock.

:o

I think 1.3e13 kg.m^2.s^-1 is correct.

Mass x Radius x Velocity = 4.2 x 3 x 10^12

L x w should be exactly the same, I think...

Mass x Radius^2 x Velocity/Radius = Mass x Radius x Velocity, right?

Well, uh, yes. Back to the drawing board. I forgot a factor of 2*pi/86400.

Amazingly agreeing with super', the current beer deficiency I am suffering, very soon to be remedied, enables me to clearly see that he has an extremely high probability of being correct.

You can't do that. r is an alias for a -- they are one and the same variable. What you needed to do, instead, was solve your first equation for a.They are not one and the same variable. r is the radius of the satellite (on the massless rod) and a is the geosynchronous orbital radius. Actually, a is a function of r. That is the whole point of the thread, and r varies from re to a. In any case, I did solve it for a, or at least √(ał)

-Dale

It is an hour before launch. The Enterprise is hovering on impulse somewhere beyond Mars orbit. Captain Kirk gives the order and Scotty instantaneously beams up the satelite. Nobody notices, so the launch sequence is completely executed.

Immediately after the satelite vanishes POOF from the launch pad, does the Earth rotation change? Has the vanishing into thin hyperspace of the satelite and its angular momentum caused the remnant Earth rotation to change?

It depends on the nature of the teleportation. Some theories about how this could occur have to do with reconstructing an exact replaca from a quantum model. It just depends on how the teleportation works.
If it is the purpose of the thought experiment to suppose that somehow it stops existing on earth and starts existing somewhere past mars, the answer is that the earth's rotation rate stays the same, but that since the satellite was part of the total inertia of earth, tidal forces ect. would slow the earth at a imperceptably faster rate.

In my cool scenario, the transporter works by instantaneously moving the object into a hyperspace which has no measurable effect upon our normally sensible 4-space.

Beyond quibbling, my scenario asks the question: would the hypothetical vanishing of the satelite and its mass and velocity and momentum make any difference at all to the rotation of the remnant of the Earth that was left behind?

If it is the purpose of the thought experiment to suppose that somehow it stops existing on earth and starts existing somewhere past mars, the answer is that the earth's rotation rate stays the same for that moment, but since the satellite was part of the total inertia of earth, tidal forces ect. would slow the earth at a imperceptably faster rate since the earth is already slowing due to tidal forces ect. This may not seem like much force, but over time it isn't insignificant. If a satellite on earth is going 1000. mph on the ground now, it will be going only 1/30 that speed in the very distant future (33 mph) this is not an insignificant amount of energy. 966 mph x aprox. 1000 lbs.

It is indeed the gist of my godunkem that the instantaneous vanishing of the satelite would, in my humble (?) opinion, not change the rotation of the Earth at any amount measurable by the best state of the art instruments that money can buy.

My reasoning being, since the prevanish angular momentum of the satelite is in perfect equillibrium with the prevanish angular momentum of the Earth/satelite system, the POOF disapearance of the satelite makes no measurable difference.

I am basing my posts on the thread as originally stated, and the choices originally provided, not any willy nilly change of rules or choices in the middle of the game. My measurements of rotation would be made within the time era indigenous to the thread start conditions, not some time period far off and eventually happening.

I agree with CANGAS (and with super :)). A chunk of mass instantaneously disappearing from the Earth wouldn't affect the Earth's rotation at all.

If the entire crust and mantle were teleported away, for example, the core would continue rotating at 1 rev per day.

It's interesting to consider what would happen if the teleportation process conserves momentum, angular momentum, and energy, but I think that these could all be conserved at the other end, so to speak. Ie the Enterprise might perhaps be jolted as it "catches" the satellite, without affecting the Earth at all.

To depart from science and indulge in beloved science fiction; the famous JOLT that the Enterprise so often enjoyed on our tv screens would be much more so if it could catch as good as Yogi Berra did.

I agree with Pete. I think it might be easier to believe that if teleportation were possible, it would be done in a way that would concerve our known laws, including concervation of angular momentum.

Toretise:

You do not believe in the Prime Directive of godunkems; anything is possible if postulated?

Besides, as Pete pointedly pointed out, the Enterprise could conserve momentum by catching a good jolt.

However, the point of the posts is not how teleportation might hypothetically function, but, rather, to rigorously examine the conservation of the main part of of the Earth's angular momentum if a very, very small part of it were "magically" removed.

I was agreeing with him and just making a comment.

Live long and prosper.

Long life and prosperity.

Live long and prosperkinesis

It seems to me that at least one of the conservation and/or laws of thermodynamics would have to be thrown out with teleportation. The teleportee is deep in a gravity well. Teleporting from the surface of the Earth to geosynchronous orbit represents a big increase in potential energy. Conserving energy would require a change in kinetic energy or temperature. A kinetic energy change would violate conservation of linear and/or angular momentum. A temperature change would (I think) violate one or more of the laws of thermodynamics.

The change in potential energy could be supplied by the energy required to do the teleportation even using "perfectly efficient" teleportation equipment.

Hmm, maybe the Klingons could power their own ships by teleporting bombs down the gravity well and capturing the energy with such equipment.

-Dale

D_H, BillyT

Sorry about my short replies yesterday, I didn't have enough time to give a detailed response to your points. I think the best way to explain the various variables is to describe my thought process in a little more depth.

After making the estimation when I was thinking about how to actually work this problem I realized that the beginning and the end points could both be described by a single rigid-body with changing moment of inertia by using the typical "massles rod" idea. Since everything involved in the problem is conservative it does not matter how the satellite gets to where it is going, only the beginning and end points matter for the result. So instead of thinking about launching the satellite in the traditional manner I thought about hoisting it aloft on this massless (telescoping) rod. Now, as the rod lengthens (r) the total moment of inertia increases (I = Ie + m r˛) and so by conservation of angular momentum (L) the sidereal day (T) lengthens and the geosynchronous orbital radius (a) increases. Therefore a, the geo orbital radius, is a function of r, the radius to which we have hoisted our satellite on its massless rod.

So, when I initially solved for a as a function of r the first thing I tried to do was to simply evaluate a(a)-a(re). Unfortunately this quantity turned out to be numerically equal to zero. Basically the problem is that the moment of inertial of the earth (Ie = 8.07E37 kg m˛) is so large that, as expected, it completely obscures the contribution of the satellite.

I tried several ways of directly simplifying the expression a(a)-a(re) in order to pull out Ie, but since it involves square roots and cubes of square roots it is annoyingly difficult to simplify. However, the expression for √(ał) is easy to separate out Ie. Unfortunately you cannot take differences in √(ał) to answer the question about differences in a. That is when I decided to differentiate the expression for √(ał). This removed the Ie term and can easily be solved for da. Integrating then gives the desired quantities without requiring any precision-losing steps. The only subtraction involved is a˛-re˛ which involves numbers that are different enough to be reasonably precise.

The one thing that concerns me is why my results are 7 or 8 orders of magnitude off from superluminal's. The only quantity I see that is on that order is the geosynchronous radius itself (a = 4.2E7 m). Perhaps, superluminal, your number represents the multiplicitave factor rather than the distance itself?

-Dale

Dale,

You can take differences in a^(1/3) if you do a first-order approximation. Since the change in a is very small, the first order approximation is very accurate (19 decimal places), which is many orders of magnitude smaller than other error sources.

BTW, your answer agrees with mine (you said 6e-13 m, I said 5.8e-13 m).

Your approach is very good, much more consice than mine. I didn't do a Taylor's expansion because I was thinking of a as a function of r not as a function of L. Although L doesn't change much r does. So your approach makes a first-order approximation reasonable, while I thought mine didn't.

I think such close agreement with such different methods lends some credibility to the results. I suspect that our small difference probably comes mostly from using slightly different numbers. The obvious one is that you used the 24-hour day and I used the sidereal day, a difference of about 0.3%, but most of my other numbers (e.g. re, M, G) came from Mathematica and probably similarly had a few tenths of a percent difference each from the numbers you used.

-Dale

The one thing that concerns me is why my results are 7 or 8 orders of magnitude off from superluminal's. The only quantity I see that is on that order is the geosynchronous radius itself (a = 4.2E7 m). Perhaps, superluminal, your number represents the multiplicitave factor rather than the distance itself?

-Dale

My number:

0.99999999999999999998213...

is just the change in ω. I didn't calculate the actual distance but assumed the change would be on the same order of magnitude. The GSO radius is:

r = ³√(G x Me / ω²)

Perhaps my assumption was wrong. Would one of you gents who have mathematica care to calculate it? If it does not agree with your answers, at least to within an order of magnitude, I'd be very interested in learning where I went wrong... I followed the simple method in my previous post, using the numbers in my other post.

Thanks.

P.S. This has been one of the more enjoyable exercises/threads. Good one BillyT.

My number:

0.99999999999999999998213...

is just the change in ?. I didn't calculate the actual distance but assumed the change would be on the same order of magnitude.

You started with
I init = (2/5MeRe^2 + MsRe^2)
Ifinal = (2/5MeRe^2 + Ms(Re+Ds)^2)

and used these to calculate the ratio

wfinal/winit = Iinit / Ifinal

That number is very close to one; in fact, it is one unless one resorts to extended precision. If you had calculated things a little differently you would not have needed extended precision.

In many cases, it is preferable to work with the difference between pairs of numbers that are nearly the same. This is certainly the case here.

Introduce
delta I = Ifinal - Iinit
delta w = wfinal - winit

Starting with
wfinal/winit = Iinit / Ifinal
then
delta w/winit = delta I/Ifinal =~ delta I/Iinit

Since the change in moment of inertia is very, very small, the error in the approximation is very, very small.

The MeRe^2 terms in delta I = Ifinal-Init cancel:
delta I = Ifinal - Iinit = Ms((Re+Ds)^2-Re^2)
Thus
delta w/winit = -delta I/Init = -1.8e-20

Expanding a = k * w^(-2/3) (Kepler's 3rd Law) about w=winit gives
a(wfinal) = k*winit^(-2/3) -2/3k*winit^(-5/3)*(wfinal-winit) + ...
or
delta a/a(winit) = -2/3*delta w/winit

Using your result with a(winit) = 4.2e7 meters,

delta a = 2/3*1.8e-20*4.2e7 = 5e-13

Your number differs from mine (5.8e-13) because you used 2/5MeRe^2 as the Earth's inertia. The Earth is not a uniform solid (its core is more dense than the mantle) and thus the solid body approximation is a bit high.

Interesting. Thanks DH.

My number:

0.99999999999999999998213...

is just the change in ω. I didn't calculate the actual distance but assumed the change would be on the same order of magnitude. The GSO radius is:

r = ³√(G x Me / ω²)

Perhaps my assumption was wrong. Would one of you gents who have mathematica care to calculate it? I plugged your formula above in using my numbers for G and Me. I took my number for ω and multiplied it by 1 for the beginning situation and multiplied it by your number above for the final situation. The difference between the two was 0 to standard numerical precision. I then repeated the procedure telling Mathematica that all of my input data had 100 decimal digits of precision. The result was 5E-13 m, so your result agrees with ours as well.

The precision of the result was 79 digits indicating that 21 decimal digits of precision were lost. Machine precision on my computer is 15.95 decimal digits, so I guess it is not surprising that the original answer was 0.

-Dale

PS I see D_H already did the calculation and came out with the same result.

Cool! I did it originally using a web-based arbitrary precision calculator. The precision on my PC is also 15 digits as I mentioned earlier.

It seems to me that at least one of the conservation and/or laws of thermodynamics would have to be thrown out with teleportation. The teleportee is deep in a gravity well. Teleporting from the surface of the Earth to geosynchronous orbit represents a big increase in potential energy. Conserving energy would require a change in kinetic energy or temperature. A kinetic energy change would violate conservation of linear and/or angular momentum. A temperature change would (I think) violate one or more of the laws of thermodynamics.
I posted a thread on a simlar topic a while ago, thinking about whether energy is conserved when moving things through wormholes, but no one responded. Here it is.

In the same vein as my initial thoughts in that dead-end thread, perhaps there might effectively be a very strong gravity field across a teleporter... ie pushing something through a teleporter "gate" (I'm thinking Stargate rather than Star Trek now) might be very very difficult if the gravitational potential is different on each side of the gate.

might be very very difficult if the gravitational potential is different on each side of the gate.

I might be reading this wrong - I just woke up, but isn't that the crux of the problem? I mean if you were transporting someone from the earth to the moon, the higher gravity on earth might be the problem if you were to take them by rocket or if you were to transport them without violating one of our laws? Would it be naive to assume that such a tremendous amount of would could be done for nothing?

I think it would be naive... but that's exactly what happens in the popular idea of teleportation, right?

Yes. But the idea of maping atoms and creating an exact replaca might get around this. Sinse knowing the exact position and speed/direction of atoms in imposible according to the uncertainty princible, a quantum model might have to be used. But this is just reiterating what has already been said before. Michael Chrichton made a book about a time machine that used MRI technology - MRIs map the position of hydrogen atoms. Food for thought.

I work with MRI. They do map the position of hydrogen, but the sensitivity is incredibly low. You have to have supermolar concentrations to get any signal at all. Typically only 6 atoms per million are detectable, and then it takes millions of those to generate a detectable signal.

-Dale

Hi Dale, I used to work in a related field (CT) it's nice to know that there are other people on here that have similar backgrounds to me.

Cool! I don't know if you still keep up with the field, but there is some really interesting work being done on using carbon nanotubes as a low temperature electron source for x-ray generation. Could be used to have multiple really fast-switching sources so you could do without the spinning tube and detectors arrangement.

-Dale

No, I haven't heard about that - it sounds very interesting - and if you had not said it, I would have insisted that it was BS