# 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

$F = GmM/r^2 \\ F = mv^2/r$

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 $v^2.$