Orbital Period and Speed Calculator

Orbital speed and period

The force that keeps a satellite in orbit is gravity. In a circular orbit the gravitational force equals the centripetal force: G × M × m / r² = m × v² / r. This gives

v = √(G × M / r) and T = 2 × π × √(r³ / (G × M))

M is the mass of the central body and r the radius of the orbit measured from its centre: add the radius of the planet to the height above the surface. The mass of the satellite does not affect the result.

Example: a satellite at 400 km

For the Earth r = 6371 + 400 = 6771 km. The speed is √(3.986 × 10¹⁴ / 6.771 × 10⁶) = 7673 m/s, about 27 600 km/h, and one orbit takes 92.4 minutes.

Geosynchronous orbit

A satellite whose period equals the rotation period of the Earth (23.9345 hours) stays above the same longitude. Turning the formula around gives an orbit radius of 42 164 km, which is 35 786 km above the equator.

Kepler's third law

The same relation holds for planets: T² = 4 × π² × a³ / (G × M). The square of the period is proportional to the cube of the semi-major axis. A body 1 au from the Sun has a period of 1 year, and Mars at 227.956 million km has a period of 687 days. If the orbit and the period are known, the same formula gives the mass of the central body, which is how the mass of the Earth follows from the orbit of the Moon.

Limits

The mass of the orbiting object is neglected next to the central body. Atmospheric drag and the pull of other bodies are not included. Masses and radii are NASA Planetary Fact Sheet values.