Kepler's Third Law Calculator
Orbital period from semi-major axis (or the reverse) around the Sun, Earth, Moon, Jupiter or any central mass — with orbital speed, geostationary check and the mass of the primary from an observed orbit.
How it works
T² = 4π²·a³ ÷ (G·M) → T = 2π√(a³/GM), a = ∛(GM·T²/4π²); v = √(GM/a)
Kepler found empirically that the square of a planet’s period is proportional to the cube of its distance; Newton showed the constant is 4π²/GM, so the law measures the mass of whatever is being orbited — that is how we weigh the Sun, planets and galaxies. For a circular orbit a is the radius; for an ellipse it is the semi-major axis, and the period does not depend on eccentricity at all. Geostationary satellites sit where T equals a sidereal day: 42,164 km from Earth’s centre.
Worked example
Earth, a = 42,164 km: T = 2π√(7.5 × 10²² ÷ 3.986 × 10¹⁴) = 86,164 s = 23.93 h, orbiting at 3.07 km/s at 35,793 km altitude — the geostationary belt.
Frequently asked questions
Does the satellite’s mass matter?
Only if it is comparable to the primary; strictly M is the sum of both masses.
Why is the ISS period 93 minutes?
At 6,791 km from Earth’s centre (420 km altitude) T = 5,560 s — low orbits are fast.
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This calculator is for educational purposes. Double-check important results before acting on them.