Problem 9-76 Geosynchronous orbit - Part 7 - C

Most telecommunications satellites are in geosynchronous orbits above the earth, that is, they have periods of \(24 \;h.\) As a result, since the earth turns on its axis once in \(24\; h\) and each satellite goes around the earth once in \(24\; h,\) any individual satellite stays positioned above a particular point on the earth. (a) How far above the earth's surface must a geosynchronous satellite be? The earth's mass and average radius are \(5.98 \times 10^{24}\; kg\) and \(6368 \;km.\) (b) What is the satellite's speed?

Accumulated Solution

Diagram of earth and satellite.

\(F = GmM/r^2 \\ F = mv^2/r \\ v^2 = GM/r\)

Correct.

This speed is given by:

(A)   \(v = d/t = 2\pi r/T,\) where \(T\) is the time to make one revolution

(B)   \(v = dt = 2\pi rT\)

(C)   \(v = t/d = T/2\pi r\)