An orbit is not a balance of forces. Gravity alone supplies the centripetal force, and that one fact fixes a satellite's speed, its period, and the special geostationary radius.
Gravity supplies the centripetal force for an orbit, GMm / r² = mv² / r, giving v = √(GM / r) and T = 2π√(r³ / GM). The field strength of a point mass is g = GM / r², nearly constant close to a surface.
A satellite in a circular orbit has only one force on it: the planet's gravity, and that single force is the centripetal force. Setting them equal fixes the speed and period at every radius. Drag the orbit in the simulation and find the radius where the period is exactly 24 hours.
For a point mass the field strength is g = GM / r². Equating gravity to the centripetal force, GMm / r² = mv² / r, gives v = √(GM / r) and T = 2π√(r³ / GM), so a higher orbit is slower with a longer period. A geostationary satellite sits at the one radius (about 4.2 × 10⁷ m) where T is 24 hours, in the equatorial plane. Close to a planet's surface r barely changes over a few kilometres, so g is very nearly constant there.
Four quick checks on orbits, the geostationary condition and field strength. Each correct answer earns XP and lights this skill on your star map.
For a satellite in a stable circular orbit, the centripetal force is provided by:
Setting GMm / r² equal to mv² / r shows that the orbital speed is:
A geostationary satellite must have an orbit that is:
Compared with sea level, the field strength g = GM / r² at the top of a mountain is:
An orbit is not a balance between gravity and an outward force; there is no outward force. Gravity alone is the centripetal force, so GMm / r² = mv² / r gives v = √(GM/r) and T² proportional to r³: a higher orbit is slower with a longer period. Geostationary is one specific radius where T = 24 h in the equatorial plane, not any 24 h orbit.
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