Imagine the Universe!
Imagine Home  |   Ask an Astrophysicist  |  
Ask an Astrophysicist

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 geo-stationary 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 period2/(4 x pi2) )(1/3)

where G is Newton's constant of gravity (6.61x10-11 m3kg-1s-2), M is the mass of the Earth (5.93x1024 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.


Jonathan Keohane and Gail Rohrbach
-- for Imagine the Universe!

Questions on this topic are no longer responded to by the "Ask an Astrophysicist" service. See for help on other astronomy Q&A services.

Previous question
Main topic
Next question

Imagine the Universe is a service of the High Energy Astrophysics Science Archive Research Center (HEASARC), Dr. Alan Smale (Director), within the Astrophysics Science Division (ASD) at NASA's Goddard Space Flight Center.

The Imagine Team
Acting Project Leader: Dr. Barbara Mattson
All material on this site has been created and updated between 1997-2012.

DVD Table of Contents
Educator's Index