Orbital Period Calculator

Kepler's Third Law in practice

Johannes Kepler discovered that a body's orbital period depends only on its semi-major axis and the mass of the body it orbits - not on the orbiting object's own mass, shape, or orbital eccentricity in the simplified two-body case.

Worked example

For a satellite at a 6,780 km semi-major axis around Earth (mu = 398,600 km3/s2), close to the International Space Station's actual orbit:

T = 2π x sqrt(6780³ / 398600) = 92.6 minutes

Frequently asked questions

Why does this closely match the ISS's real orbital period? The ISS orbits at roughly 400 km altitude, which combined with Earth's 6,371 km radius gives a semi-major axis close to 6,780 km - the ISS actually completes an orbit in about 92-93 minutes, matching this calculation closely.

Does this work for elliptical orbits? Yes, as long as you use the semi-major axis (the average of the closest and farthest orbital distances), not just one specific radius - this is what makes Kepler's Third Law apply equally to circular and elliptical orbits.