perturbed orbits

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: If the satelite is perturbed a 100 or more meters by the moon's gravity all the GPS satelites would be sending signals perturbed by different amounts. How is this managed by the system?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The law of gravity is well known. We can calculate the exact amount by which any body changes the position of any other body. This has been done for planets perturbing one another for centuries, and is the reason why "epicycles" are no longer needed. The success of this method is the reason why the law of gravity is held in such high esteem and considered "universal". It works.

And it works for predicting the exact deviation of each GPS satellite also. The Sun's gravity produces an effect similar to the Moon's. The flattening of the Earth's shape produces an even larger effect. But all these are now a routine part of the calculation of the orbit of any body in space. The Air Force Monitor Stations inform the satellites about their own orbits, and the satellite's true position in space is part of the message each satellite sends to GPS receivers on the ground. -|Tom|-

It makes sense that all the points you you made need to be factored into a signal for the GPS to work right and I'm not at all puzzled by this revelation. I'm puzzled by the total distance a satelite moves because of the moon's gravity and by how difficult it is to find any data about this. I guess the calculation of the motion gives correct results or else it would have been replaced. By the way, how is that calculated? I tryed but didn't get anything like 100 or so meters

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Jim]: I'm puzzled by the total distance a satelite moves because of the moon's gravity and by how difficult it is to find any data about this.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

The "satellite" problem is discussed in most books on celestial mechanics. My guess is that you have avoided reading the right books because the required math can be intimidating.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>how is that calculated? I tryed but didn't get anything like 100 or so meters.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>

I just made a quick estimate. A proper calculation would use the formulas for the variations of orbital elements. My guess is that Larry's estimate was done in this more rigorous way, and is therefore better than mine. I assumed you were just looking for the right order of magnitude.

For my "quick and dirty" estimate, I started with the acceleration of gravity at Earth's surface, 10 m/s^2. (That is 32 feet/second/second in English units, which might be more familiar to some.)

Then I divided by 81 to get the equivalent acceleration (still at one Earth-radius) if Earth's mass were replaced by the Moon's mass. (I'm now at an acceleration of 0.123 m/s^2.)

But the Moon is 60 times farther away from Earth and satellite than is Earth's surface. So we need to reduce by another factor of 60^2 = 3600, which leaves us with 0.000034 m/s^2.

Finally, we must recognize that the direct force of the Moon affects satellite and Earth in a very similar way. To the extent that it pulls both equally, we see no change in the satellite's orbit. So we are interested in the differential force of the Moon on satellite and Earth. Because the GPS satellite orbits are 4 Earth radii high and the Moon is 60, the factor we need for the differential force is 4/60 or 1/15. This leaves us with 0.000 002 3 m/s^2.

The last step is to change the acceleration estimate into a maximum distance change. A GPS satellite needs 6856 seconds to move through one radian along its orbit. So we multiply the acceleration by 6856^2, which yields our estimate: 110 meters. This is good only to about a factor of two even though I carried two significant digits in the calculation. -|Tom|-

Hello Jim,

Check your e-mail.

LB

I think I see an error but let me look this over thanks.

<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I think I see an error but let me look this over thanks.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
An error ? ! ! Well - I suppose it had to happen sooner or later.

Actually, there IS an error. The orbital period of the GPS satellites is (very close to) 12 hours. For half of each orbit the satellite is closer to the Moon than is Earth, and for the other half it is farther away than is Earth.

I divided the close half of the orbit into 12 sample periods and used 3600 seconds for the duration of that period. But I should have used 1800 seconds (half an hour instead of a full hour).

Here is the factor of 2 error I suspected when comparing my results to Dr. Van Flandern's. My calculations now give an answer of 64.1 (meters).

(My gpa is safe once again.)


Regards,
LB