Geostationary Satellite

A geostationary satellite is an Earth satellite that appears completely stationary in the sky when viewed from any fixed point on the ground -- an enormously useful property for communication and weather-monitoring satellites, since a ground antenna aimed at it never needs to be re-pointed to keep tracking it. This appearance of standing still requires three conditions to hold simultaneously:

1. The orbit must lie exactly in the equatorial plane of the Earth (an orbit tilted at any angle to the equator would still carry the satellite visibly north and south across the sky over the course of each day, even if its period matched the Earth's rotation).
2. The satellite must orbit in the same sense as the Earth's own rotation -- eastward.
3. Its orbital period must exactly equal the Earth's own rotation period (very close to $24\ \text{h}$; more precisely, one sidereal day, about $23\ \text{h}\ 56\ \text{min}$, though $24\ \text{h}$ is used as the standard approximation in problems).

With all three conditions met, the satellite completes exactly one orbit in exactly the time the Earth itself takes to complete one rotation, so it stays permanently above the same fixed point on the equator, appearing motionless from the ground (see the accompanying figure).

Finding the orbital radius. Since a geostationary orbit is circular, the period relation derived in Section 7.11, $T = 2\pi\sqrt{r^3/GM}$, applies directly. Squaring and solving for $r$,

Taking $M = 6 \times 10^{24}\ \text{kg}$, $G = 6.674 \times 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$ (so $GM \approx 4.00 \times 10^{14}\ \text{m}^3/\text{s}^2$) and $T = 24\ \text{h} = 86400\ \text{s}$,

Subtracting the Earth's own radius, $R \approx 6400\ \text{km}$, gives the satellite's height above the surface,

$h = r - R \approx 42{,}000\ \text{km} - 6400\ \text{km} \approx 3.6\times 10^{4}\ \text{km}$ …