Answer
The altitude of a geosynchronous orbit is $~~35,880~km~~$ above the Earth's surface.
Work Step by Step
Note that the period of the orbit is $T = 24~h$
We can express the period in units of seconds:
$T = (24)(3600~s) = 86,400~s$
We can use Kepler's Third Law to find the radius $r$ of the satellite's motion:
$T^2 = (\frac{4\pi^2}{GM})~r^3$
$r^3 = \frac{GM~T^2}{4\pi^2}$
$r = \sqrt[3] {\frac{GM~T^2}{4\pi^2}}$
$r = \sqrt[3] {\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)~(86,400~s)^2}{4\pi^2}}$
$r = 42,250~km$
We can find the altitude $h$ above the Earth's surface:
$h = (42,250~km)-(6370~km) = 35,880~km$
The altitude of a geosynchronous orbit is $~~35,880~km~~$ above the Earth's surface.
