What this calculator does
Kepler noticed that the further a planet is from the Sun, the longer its year, and that the relationship follows a precise rule: the square of the period is proportional to the cube of the orbit size. He found it in observational data decades before Newton explained why.
The modern form adds the mass being orbited, which is what turns Kepler's proportionality into an equation that works for any system: planets round a star, moons round a planet, or satellites round Earth.
The formula
The period is 2π times the square root of the semi-major axis cubed divided by the gravitational constant times the central mass. Distances are converted from kilometres to metres before the calculation.
| Term | Meaning |
|---|---|
| Semi-major axis (a) | Half the longest diameter of the elliptical orbit. For a circular orbit it is simply the radius. |
| Orbital period (T) | The time for one complete orbit. |
| Central mass | The mass being orbited. For planets round the Sun this is the solar mass, 1.989 × 10³⁰ kg. |
The inputs explained
| Field | What to enter |
|---|---|
| Semi-major axis (km) | The semi-major axis in kilometres. Earth's orbit averages 149,597,870.7 km, which is one astronomical unit. |
| Central mass (kg) | The mass of the central body in kilograms. |
When to use it
Finding a planet's year
Given an orbit size and the Sun's mass, the orbital period follows directly, which is how the length of any planet's year is calculated.
Weighing a star or planet
Run backwards, an observed period and orbit size give the central mass, which is how the masses of distant bodies are actually measured.
Planning a satellite orbit
The same relationship fixes the altitude needed for any desired period around Earth.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
How long is a year at different distances from the Sun?
Orbits at the distances of Mercury, Earth and Jupiter.
| Semi-major axis | Orbital period | Average orbital speed |
|---|---|---|
| 57.9 million km | 87.958 days | 47.878 km/s |
| 149.6 million km | 365.210 days | 29.789 km/s |
| 778.5 million km | 4,335.533 days | 13.058 km/s |
Questions
Why is period proportional to distance to the power of 1.5?
Because a wider orbit is both longer to travel and slower to travel at. The circumference grows in proportion to radius while the speed falls with the square root of radius, and combining the two gives the three-halves power.
Does the orbiting body's mass matter?
Only slightly, and this calculator ignores it. Strictly the equation uses the combined mass, which matters for binary stars but is negligible when a planet orbits a star that is vastly heavier.
Does it work for elliptical orbits?
Yes, which is the remarkable part. The semi-major axis alone determines the period regardless of how elongated the ellipse is, so two very differently shaped orbits with the same semi-major axis share a period.
How is this used to find exoplanet masses?
By observing a planet's period and orbital distance and running the relationship backwards to solve for the star's mass, or by watching the star wobble to infer the planet's. It is the standard tool of observational astronomy.
For circular orbits specifically, see the orbital velocity calculator. For the gravitational force underlying all of it, see the Newton's law of gravitation calculator.