Problem 9-76 Geosynchronous orbit - Part 6 - 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.

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

\(v^2  = GM/r\)

However, at this stage \(v\) is unknown

\(v\) is the speed:

(A)   at which the satellite would hit the ground if released from a distance \(r,\)

(B)   unknown, because we dont know the launch speed.

(C)   the speed required to make one orbit in \(24\; hr.\)