Orbital Period Calculator
Enter a semi-major axis and the central mass to get the orbital period in seconds and days — and see why wider orbits take far longer to complete.
Seconds and days at once
Enter the semi-major axis and the central mass and the calculator returns the orbital period in seconds (T = 2π·√(a³/GM)) and converted to days together.
Use SI units
Semi-major axis in metres and central mass in kilograms give the period in seconds — the days figure is simply that divided by 86,400.
What is the orbital period?
The time for one revolution
The orbital period is the time a body takes to complete one full revolution around the object it orbits. This orbital period calculator turns two measurements — the semi-major axis in metres and the central mass in kilograms — into the period in seconds, alongside the same figure expressed in days. It is governed by Kepler's third law, the relationship that ties an orbit's size to its duration, and it depends only on the orbit's size and the central mass, not on the mass of the orbiting body itself when that mass is small. It is the number behind a satellite's revisit time, a planet's year, and the altitude needed for a geostationary orbit.
Enter a semi-major axis in metres and a central mass in kilograms to get the orbital period in seconds and in days instantly.
The orbital period is two pi times the square root of the cube of the semi-major axis divided by the product of the gravitational constant and the central mass.
T = 2π × √(a³ / (G × M))Here G is the gravitational constant, 6.6743 × 10⁻¹¹ m³·kg⁻¹·s⁻². The semi-major axis is cubed, so distance dominates the result: the period grows with the 3/2 power of the orbit size. Use metres and kilograms and the period comes back in seconds, which the calculator also divides by 86,400 to show in days.
Suppose the Earth orbits the Sun at a semi-major axis of 1.496 × 10¹¹ m, with the Sun's mass at 1.989 × 10³⁰ kg.
Cube the semi-major axis
(1.496e11)³ ≈ 3.348 × 10³³ — the orbit size raised to the third power.
Divide by G × M
Divide by 6.6743e-11 × 1.989e30 ≈ 1.328 × 10²⁰ to get the ratio under the root.
Take the root and multiply by 2π
2π × √(ratio) ≈ 3.155 × 10⁷ s, which divided by 86,400 is about 365.2 days — one Earth year.
The formula is exact for an ideal two-body system, but a few practical points are worth keeping in mind.
Two-body ideal and consistent units
This calculator uses the standard two-body form of Kepler's third law, which assumes the orbiting body is much lighter than the central mass and ignores the pull of other bodies, drag, and relativistic effects. For a binary where the masses are comparable, replace M with the combined mass. Keep your units consistent — metres for the semi-major axis and kilograms for the central mass — or the seconds will be wrong.