Problem 9-76 Geosynchronous orbit - Part 5 - A

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.

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

Correct.

You have two expressions for the same thing; what should you do now?

(A)   There must be a mistake there cannot be two different expressions for the same thing.

(B)   Set them both equal to zero and solve for \(v\) and \(M.\)

(C)   Set them equal to each other and solve for \(v^2.\)