# Problem 9-76 Geosynchronous orbit - Part 5 - 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

$F = GmM/r^2$

No. That is not the expression for centripetal force. If you check the units you will see that is so.

$F = mv^2r | \text{Units of} \; F =[\text{mass}][\text{acceleration}] = M·LT^{-2} \\ \text{Units of} \; mv^2r | = M·L^2T^{-2}·L = ML^3T^{-2}$

These are not the same so the equation cannot be right.

Try again.
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