Last updated: 2026-07-13

Kepler's Third Law

Kepler's third law ties together the three things that describe an orbit: how far a body sits from the object it circles, how massive that central object is, and how long one lap takes. Written in Newton's form it reads $T^2 = 4\pi^2 a^3/(GM)$ — the square of the period grows with the cube of the distance. This calculator rearranges that single relationship three ways, so you can find the orbital period from a distance and a mass, recover the orbital distance from a known period, or even weigh the central body from an orbit you have measured.

The formula and its three forms

Here $a$ is the semi-major axis (the average orbital distance, and simply the radius for a circular orbit), $M$ is the mass of the body being orbited, and $G = 6.674\times10^{-11}\ \text{N·m}^2/\text{kg}^2$ is the gravitational constant. Notice that the mass of the orbiting body never appears — a dust grain and a moon at the same distance from the same star take exactly the same time to go around.

The solar-system shortcut

The reason Kepler's third law is so easy to use for planets is a lucky choice of units. Measure distance in astronomical units (1 AU is the Earth–Sun distance), mass in solar masses (1 M☉ is the mass of the Sun), and time in years, and the whole clump $4\pi^2/(GM)$ collapses to $1/M$. The law becomes:

with $T$ in years, $a$ in AU and $M$ in solar masses. That is the form this calculator uses internally, which is why its defaults — Earth's orbit — return a clean answer.

Worked example: Earth around the Sun

For Earth, the semi-major axis is $a = 1\ \text{AU}$ and the central mass is $M = 1\ \text{M}_\odot$:

The same numbers in full SI units — $a = 1.496\times10^{11}\ \text{m}$, $M = 1.989\times10^{30}\ \text{kg}$ — run through $T = 2\pi\sqrt{a^3/GM}$ to give about $3.156\times10^{7}\ \text{s}$, which is one year to within a small fraction of a percent. The astronomical-unit form just hides the arithmetic.

Worked example: the outer planets

The a³ scaling is why the outer solar system moves so slowly. Jupiter sits at about $a = 5.20\ \text{AU}$, so

matching Jupiter's real 11.86-year orbit. Neptune, at roughly 30 AU, needs $\sqrt{30^3} \approx 164$ years to circle the Sun once — a single Neptunian year is longer than the planet has been known to science.

Weighing a star

Turned around, Kepler's third law is one of the most important tools in astronomy: it lets you measure a mass you can never put on a scale. If you can watch a companion object orbit — a planet around a star, a star around a black hole — and measure its period and semi-major axis, then $M = 4\pi^2 a^3/(GT^2)$ gives the mass of whatever it is circling. This is how the mass of the Sun, of exoplanet host stars, and of the supermassive black hole at the centre of the Milky Way were all determined.

Assumptions and limits

The law as used here assumes one body far more massive than the other, so the central object stays effectively fixed while the smaller one orbits it. For two comparable masses — a binary star, for instance — the $M$ in the formula is the sum of the two masses, and both bodies orbit their common centre of mass. The result is also the ideal two-body value: it ignores the tug of other planets, the slight flattening of real bodies, and relativistic effects, all of which cause small, slow drifts in true orbits over long spans of time.