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Kepler's Third Law Orbital Period Calculator

Solve for orbital period, semi-major axis, or central body mass via Kepler's third law, T = 2π√(a³ ÷ GM), for a body orbiting a much larger central mass.

Orbital period
365.2105 days
Orbital period (years)
0.9999
Semi-major axis
149,600,000,000 m
Central body mass
1,989,000,000,000,000,000,000,000,000,000 kg

How it works

Kepler's third law, in its full Newtonian form, relates the time a body takes to complete one orbit to how far out it orbits and how massive the central body is: T = 2π√(a³ ÷ GM), where a is the orbit's semi-major axis, M is the central body's mass, and G is the gravitational constant (6.6743×10⁻¹¹ m³kg⁻¹s⁻², the same CODATA 2022 value this site's escape-velocity calculator uses). This assumes the orbiting body's own mass is negligible next to the central body's — true for a planet around a star or a satellite around a planet, not for two comparably-sized bodies orbiting each other.

Given any two of {period, semi-major axis, central mass}, the calculator rearranges that same equation to solve for whichever one you leave as the target: a = ∛(GMT² ÷ 4π²) for the axis, or M = 4π²a³ ÷ GT² for the mass.

Period is entered and shown in days regardless of the metric/imperial toggle, since a day count means the same thing in either system — only the distance and mass fields change with the toggle. The default values are Earth's own real orbit (1 AU semi-major axis around the Sun's actual mass), which the formula correctly returns as almost exactly a 365.25-day year.

FAQ

Why does the orbiting body's own mass not matter?

It does, technically — the exact two-body form uses the *combined* mass of both objects. But for anything astronomically interesting (a planet around a star, a moon around a planet, a satellite around Earth), the orbiting body is so much less massive than what it orbits that its contribution is negligible, which is why this calculator — like the standard textbook statement of Kepler's third law — treats the central mass alone as M.

Can I use this for a satellite orbiting Earth?

Yes — set the central mass to Earth's mass (about 5.972×10²⁴ kg) and the semi-major axis to the satellite's orbital radius (Earth's radius, about 6,371 km, plus its altitude above the surface). A satellite at low Earth orbit altitude comes out to roughly a 90-minute period, matching the well-known figure for the International Space Station.

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