The Question
(Submitted April 08, 1997)
Could you tell me the minimum height of that a satellite can be at in order
to remain in a geostationary position? I have been told
it is as low as 400 km and as high as 40,000 km. Thanks.
The Answer
Thank you for your question. A geostationary orbit is one that
appears to stay above one point on the Earth. This means that it has a
period of almost a day.
Please forgive us a digression about the length of a day. We usually
think of a day as having 24 hours: the amount of time it takes for the
Sun to go from highest in the sky (noon) on one day, to highest in the
sky (due south in the USA) on the next. In fact, the Earth makes one
complete rotation on its axis in slightly less than 24 hours (23 hours
+ 56 min., to be exact). The reason noon tomorrow is not 23 hours and
56 min. after noon today is that the Earth has moved a little bit in
its orbit around the Sun during this time. Because of this, the Sun is
now in a slightly different direction from the center of the Earth, and
the Earth has to turn slightly more than one rotation to bring the Sun
to due south again. (This, of course, takes 4 minutes longer, making
the familiar 24 hour day.)
Now back to the actual orbit of a satellite around the Earth: Since
geostationary satellites remain over the same point on the Earth, their
orbits must have a period equal to the Earth's rotation on its axis = 23h56m.
They also must go around the equator (or else they would appear to move
North and South throughout the day), and go in a circular orbit (or else
they would appear to move East and West throughout the day).
Now from these constraints, we can calculate the one specific height above
the Earth where a geostationary satellite has to go. If we put it too high,
the satellite would move too slow. If we put it too low, it moves too fast.
This distance from the center of the Earth is given by:
R = (G x M x period^{2}/(4 x pi^{2}) )^{(1/3)}
where G is Newton's constant of gravity (6.61x10^{11}
m^{3}kg^{1}s^{2}), M is the mass of
the Earth (5.93x10^{24} kg), and period is 23h56m = 86160 s.
If you subtract the radius of the Earth from this answer, you get
the height above the Earth for a geostationary satellite:
We calculate 35,000 km
Therefore, whoever told you 40,000 km was correct.
Sincerely,
Jonathan Keohane and Gail Rohrbach
 for Imagine the Universe!
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