Equate Gravity to Centripetal Force: Everything Follows
For a satellite in a circular orbit, gravity supplies the centripetal force. That single equation gives the speed, the period and the energies, and it shows that none of the orbital motion depends on the satellite's own mass.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Speed and period
G M m / r² = m v² / r gives v = sqrt(GM/r) = R sqrt(g/r), so a satellite just above the surface has v0 = sqrt(gR) and one at a height R/2 (r = 1.5 R) has v = v0 sqrt(2/3).
For two satellites at 4R and R, the speeds are in the ratio 1 : 2, so if one moves at 3v the other moves at 6v. Two satellites in the same orbit, whatever their masses, have the same speed and the same period. If the force law were 1/r^m, then v² / r would be proportional to 1/r^m, so T proportional to r^((m + 1)/2).
Energies
Compare satellites by (m / r): for masses 3 : 1 at radii r and 4r the ratio of total energies is (3/r) : (1/4r) = 12 : 1. For two identical satellites at heights R and 7R, the orbit radii are 2R and 8R, so the kinetic, potential and total energies are all in the ratio 4 : 1; a claim that the total energy ratio is 5 is wrong.
Geostationary satellites
A geostationary (relay) satellite has a period equal to the Earth's rotation period (24 h) and orbits above the equator in the Earth's rotation direction. From r³ = GMT² / (4π²) its height above the surface is h = (GMT² / 4π²)^(1/3) - R, and since the mass of the satellite cancels, replacing it with a satellite of double mass keeps the same orbit radius, a ratio of 1 : 1.
When gravity stops
If gravity suddenly disappeared, the satellite would fly off in a straight line along the tangent with the speed v it had at that instant: nothing changes its velocity, so it neither falls nor stays in the orbit.
OlympIQ exam lens
- ✓v = sqrt(GM/r), T = 2π sqrt(r³/GM); neither depends on the satellite's mass
- ✓KE = GMm/(2r), PE = -GMm/r, E = -GMm/(2r); compare satellites by m/r
- ✓Geostationary: T = 24 h, equatorial orbit, h = (GMT²/4π²)^(1/3) - R
- ✓Orbit radius is height plus the Earth's radius, not the height alone
- ✓If gravity vanishes the satellite moves off tangentially at speed v
Do not use the height above the surface as the orbit radius. Orbital formulas need the distance from the centre of the Earth, r = R + h, so a satellite at height R orbits at radius 2R.
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