Circular Orbit Period

Circular orbital period: T = 2π·√(r³/GM)

Write Kepler's third law the Newtonian way and you get T² = (4π²/GM)·r³. Here r is the orbital radius measured from the center of the primary (not its surface), and GM is the standard gravitational parameter. For Earth, GM ≈ 3.986×10¹⁴ m³/s². The ISS sits at r ≈ 6,781 km, which works out to T ≈ 5,550 s, or 92.5 min. Geostationary satellites (GEO) ride at r ≈ 42,164 km with T = 86,164 s ≈ 23h 56min 4s, exactly one sidereal day. Go out to the Moon at r ≈ 384,400 km and T climbs to about 27.3 days. Mercury, at r ≈ 58 Gm around the Sun, needs 88 days.

Applications

Mission design leans on this constantly. So do satellite constellations (Iridium/Starlink in LEO at ~550 km with T ≈ 95 min; GPS in MEO at 20,200 km with T ≈ 12 h), the rendezvous timing for capsules creeping up on the ISS, and planetary work like the Ingenuity helicopter syncing to Mars' 24.6 h rotation.

FAQ

Why is GEO at exactly 42,164 km? That's the radius where T comes out to one sidereal day, which lets the satellite hang above a fixed point on the equator.

Does the satellite's mass matter? No. When a small body orbits a much larger one, T depends only on r and the GM of the primary.

What about elliptical orbits? Kepler's third law still works. Just swap r for the semi-major axis a: T² = (4π²/GM)·a³.