35,786km for geo-stationary, which is probably what you meant.
2006-09-25 13:56:51
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answer #1
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answered by midwestbruin 3
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From Kepler's law,
T^2=4*pi^2/GM * R^3
G = 6.6 * 10^-11 (Gravitation constant)
M = 6 * 10^24 (Mass of earth)
T=24 hr = 86400 seconds.
Plug in and solve for R, you get R=42000 km.
Now that is the distance of the satellite from earth's center.
So height above earth is 42000 km - earth's radius(6400 km approx) = 35600 km approx.
2006-09-26 11:10:58
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answer #2
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answered by astrokid 4
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The orbital period of a satellite in a circular orbit is given by:
TP = 2 * PI * ((R)^(3/2)) / ((mu^(1/2))
TP = period of orbit
where PI = 3.14159....
R = radius of the orbit
mu = (constant of gravitation) * (mass of the primary)
For earth, the mass * G = mu = 3.986012*10^5 km^3/s^2
For geosynchronous orbits, you want the orbital period to match the earth's rotation period.
Plug it all in and solve for "R".
2006-09-25 22:58:35
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answer #3
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answered by Anonymous
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I always thought it was 22400 miles up. comm satelllites generate a 600 msec round trip delay in end-to-end and back again comms.
So 186000 miles * 0.6 sec = 112000 miles -then divide that by 4 and get 28000. Okay, I am in the right ball park.
oops, sorry I didn't see your details section where you asked for the formulae. looks like you got a couple of good answers that way though.
2006-09-27 15:14:28
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answer #4
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answered by Anonymous
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Theoretically, a satellite can be geosynchronous at any altitude. It just has to match the rotation of the earth. The higher the altitude, the faster is has to go to stay geosynchronous.
2006-09-25 20:03:08
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answer #5
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answered by the handyman 2
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At about 42000Km altitude.
2006-09-25 20:05:23
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answer #6
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answered by NaughtyBoy 3
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yah its 36k km
2006-09-25 21:30:48
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answer #7
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answered by Anonymous
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22,000 is the approximate miles up
2006-09-25 23:43:20
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answer #8
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answered by orion_1812@yahoo.com 6
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idk....lol sorry
2006-09-25 20:21:09
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answer #9
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answered by Anonymous
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