Geostationary satellite rotates

Concept — For a satellite moving in a circular orbit of radius r about the Earth, gravity supplies exactly the centripetal force: GMm/r2 = mv2/r. The satellite's mass m cancels, leaving v = √(GM/r) and an orbital period T = 2π√(r3/GM). For a circular Earth orbit, the period therefore depends on one quantity only — its orbital radius. A geostationary satellite is defined by circular, equatorial, prograde motion that keeps it permanently above the same ground point; its period must equal one sidereal day.

1. Impose the defining condition: the orbital period must equal one sidereal day, T = 23 h 56 min 4 s = 86164 s.

2. Invert the period relation for the radius: r = (GM·T2/4π2)1/3, with GM = 3.986 × 1014 m3 s-2.

3. Evaluate the expression: r ≈ 4.2164 × 107 m, that is about 42164 km measured from the centre of the Earth.

4. Subtract the Earth's equatorial radius, about 6378 km, to reach the height above the surface: about 35786 km.

Cross-check — Because m cancelled at the very first step, this height is the same for every satellite: a 500 kg and a 5000 kg spacecraft share the identical geostationary orbit, so no height that varies with mass can exist. And since every orbital plane must pass through the Earth's centre, no orbit can hold a satellite permanently over a Pole; only an equatorial orbit keeps the ground track pinned to a single point.

The defining period selects exactly one radius, so a geostationary satellite revolves at a fixed height of about 35786 km above the equator — a value independent of its mass, and one that cannot be realised over the Poles.