Kepler’s third law
In 1619 Johannes Kepler found that the square of a planet’s orbital period is proportional to the cube of its distance from the Sun. Newton later explained why. For a small satellite in a circular orbit:
T = 2π × √(a³ ÷ GM)
- a = radius of the orbit = radius of the central body + altitude
- GM = standard gravitational parameter of the central body
The orbital speed is v = √(GM ÷ a).
Values used
| Body | GM (km³/s²) | Mean radius (km) |
|---|---|---|
| Earth | 398,600 | 6,371 |
| Moon | 4,905 | 1,737 |
| Mars | 42,828 | 3,390 |
| Jupiter | 126,686,534 | 69,911 |
| Sun | 132,712,440,018 | 696,000 |
Worked example: the International Space Station
- a = 6,371 + 420 = 6,791 km.
- T = 2π × √(6,791³ ÷ 398,600) ≈ 5,569 s = 92.8 minutes.
- v = √(398,600 ÷ 6,791) ≈ 7.66 km/s (about 17,100 mph).
- The station completes about 15.5 orbits per day, which is why astronauts see 16 sunrises a day.
Common Earth orbits
| Orbit | Altitude | Period |
|---|---|---|
| Low Earth orbit (ISS) | 420 km | 93 min |
| Starlink | 550 km | 96 min |
| GPS | 20,200 km | 11 h 58 min |
| Geostationary | 35,786 km | 23 h 56 min |
| Moon | 384,400 km (center to center) | 27.3 days |
A geostationary satellite has a period of one sidereal day, so it stays above the same point on the equator. Satellite TV dishes point at it without moving.
Why lower is faster
A satellite in a lower orbit is pulled harder and must move faster to keep falling around the planet rather than into it. That is why the ISS laps a geostationary satellite many times a day. To catch a spacecraft ahead of you in the same orbit, you actually slow down briefly, drop to a lower and faster orbit, then climb back.
Limits
The calculator assumes a circular orbit around a spherical body and ignores atmospheric drag and the mass of the satellite. For elliptical orbits, use the semi-major axis as a: the period depends only on it. Very low Earth orbits, below about 300 km, decay quickly because of drag.