Kepler found the pattern in 1619 without knowing its cause: the square of an orbital period is proportional to the cube of the orbit’s size. Newton later supplied the cause — and the star’s mass in the bargain. This Kepler’s third law calculator runs the law both ways (distance to period, period to distance) for any central mass, using the solar-system units that make the classic form read T² = a³.
The Law in Solar Units
Measure distance in astronomical units, time in years, and mass in Suns, and the law becomes arithmetic you can do at a glance: T = √(a³ ÷ M). Earth calibrates it (a = 1, T = 1, M = 1). The page’s default is Mars: a = 1.524 AU → T = 1.524^1.5 ≈ 1.881 years, the Martian year of 687 days. Jupiter at 5.2 AU: 5.2^1.5 ≈ 11.9 years. Run the law the other way and a newly discovered exoplanet’s period hands you its distance immediately — a one-year period around a Sun-like star means a 1 AU orbit, in the habitable-zone conversation before any telescope resolves a thing.
The Newtonian Form Behind the Shortcut
Newton derived what Kepler fitted: T² = 4π²a³ ÷ (G × M), with the orbiting body’s mass added to M when it is not negligible. The solar-unit version hides G, 4π² and the unit conversions inside its clean numbers — set a in metres, M in kilograms and G = 6.674 × 10⁻¹¹ and the same law prices a satellite’s orbit around Earth (a geostationary orbit falls out at 42,164 km from the centre for T = one sidereal day). The two-body refinement matters for binary stars and the Earth–Moon system; for planets around stars, the planet’s mass shifts the third decimal and the simple form on this page is the working tool.
Weighing Stars With Their Planets
The law’s professional superpower runs backwards: observe a planet’s period and distance and the star’s mass follows — M = a³ ÷ T² in solar units. This is, to first order, how stars get weighed at all (visual binaries provide both quantities directly), and how the mass of the Milky Way’s central black hole was measured: stars whipping around Sagittarius A* in 16-year, 0.12-AU-scale orbits demand a central mass of about four million Suns. Any time astronomy announces the mass of something it cannot touch, Kepler’s third law is somewhere in the derivation — four centuries old and still load-bearing.
Ellipses Welcome
The “a” is the semi-major axis — half the long diameter of the elliptical orbit — not the distance at any particular moment. A comet on a screamingly eccentric orbit and a planet on a near-circle with the same semi-major axis share the same period: Halley’s comet (a ≈ 17.8 AU, period ≈ 75.3 years) checks against 17.8^1.5 ≈ 75.1. Orbits are priced by their a alone, a profound economy of nature that the law states and Newton explained — the period cannot see eccentricity at all.
Where It Applies — and Stops
Any two-body gravitational system: planets, moons (swap in Earth’s mass and kilometres for the Moon: a = 384,400 km gives T = 27.3 days through the Newtonian form), artificial satellites, binary stars, exoplanets. It stops where other forces compete: atmospheric drag on low satellites, radiation pressure on dust, third bodies perturbing close pairs, and the relativistic corrections that move Mercury’s perihelion — the famous 43 arcseconds per century that Newtonian gravity, Kepler included, could not explain and Einstein could. Knowing a law’s border is part of knowing the law.
Common Mistakes
Using diameter or perihelion distance instead of the semi-major axis. Mixing unit systems mid-formula — the clean T = √(a³/M) holds only in AU, years and solar masses; SI work needs the full Newtonian form with G. Forgetting the central mass when leaving the solar system (a one-year period means 1 AU only around a one-solar-mass star). And expecting eccentricity to matter: two orbits of equal a and wildly different shapes keep identical periods.
Related Tools
The orbital bench continues in the Physics Calculators hub — the Length Contraction Calculator takes motion to relativistic speeds, and the E = mc² Calculator prices the mass-energy those stars burn.
Frequently Asked Questions
What is Kepler’s third law?
The square of an orbital period is proportional to the cube of the semi-major axis: T² ∝ a³. In AU, years and solar masses it computes directly as T = √(a³/M) — Mars at 1.524 AU takes 1.881 years.
How do you find the semi-major axis from the period?
Invert the law: a = (T² × M)^(1/3) in solar units. A planet with a 8-year period around a Sun-mass star sits at (64)^(1/3) = 4 AU. This calculator’s reverse mode does it directly.
Does the planet’s own mass matter?
Strictly, the law uses the total mass of both bodies. For planets around stars the planet’s share is negligible (Jupiter is a thousandth of the Sun), so the star’s mass alone is the working input; binary stars need both.
Why do eccentric and circular orbits share periods?
Because the period depends only on the semi-major axis, not the orbit’s shape. Halley’s comet, on a very stretched ellipse with a ≈ 17.8 AU, returns every ~75 years — the same period a circular orbit at 17.8 AU would keep.
Can the law be used for satellites?
Yes — through the Newtonian form T² = 4π²a³/(GM) with Earth’s mass. It yields low-orbit periods near 90 minutes and places the geostationary orbit at 42,164 km from Earth’s centre.
How are stars weighed with this law?
Measure an orbiting companion’s period and semi-major axis; then M = a³/T² (solar units) gives the central mass. The four-million-solar-mass black hole at the Milky Way’s centre was weighed exactly this way, from the stars circling it.