# Problem 9-76 Geosynchronous orbit - Part 12

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**

\(F = GmM/r^2 \\ F = mv^2/r \\ v^2 = GM/r \\ v = d/t = 2\pi r/T, \; \text{where T is the time to make one revolution} \\ v^2 = \frac{GM}{r} = \frac{4\pi ^2r^2}{T^2} \\ r^3 = \frac{GMT^2}{4 \pi^2} \\r^3 = \frac{GMT^2}{4 \pi^2} = \frac{(6.67 \times 10^{-11})(5.98 \times 10 ^{24})(24\; hr \times \frac{3600 \;s}{1\;hr})}{4 \pi ^2} \\ r = \sqrt[3]{7.57 \times 10^{22}} = 4.23 \times 10 ^7 \; m \\ \text{height} = 4.23 \times 10^7\;m - 6.368 \times 10^6\; m = 3.6 \times 10^7 \;m \; \text{ (answer to part (a))}\)

\(v = 2\pi r/T = 2\pi(4.23 \times 10^7 )/(24 \times 3600) = 3.1 \times 10^3\; m/s \quad \text{(answer to part (b))}\)

**You have completed this problem.**