Orbital Period (Kepler) Calculator
The orbital period is the time a body takes to complete one full trip around the object it orbits, and Kepler's third law lets you compute it from just two numbers. In its Newtonian form the period depends on the size of the orbit, set by the semi-major axis, and the mass of the central body that does the pulling. Larger orbits take dramatically longer because the period grows with the three-halves power of the semi-major axis, while a heavier central mass tightens its grip and shortens the period. This calculator takes the semi-major axis in meters and the central mass in kilograms, holds the gravitational constant G fixed at 6.674e-11 N m^2/kg^2, and returns the period in both seconds and days. The defaults describe Earth circling the Sun: a semi-major axis of 1.496e11 meters, one astronomical unit, around a central mass of 1.989e30 kilograms. As a check on the physics, those numbers return very close to one Earth year, which is exactly what they should. The same formula works for moons, satellites and planets. Every figure here is computed deterministically from the standard physics formula shown below, with a worked example that reconciles exactly to the calculator so you can follow each step.
Kepler's third law gives the orbital period as two pi times the square root of the semi-major axis cubed divided by G times the central mass: T = 2 pi sqrt(a^3 / (G M)). With a semi-major axis of 1.496e11 m around a mass of 1.989e30 kg and G fixed at 6.674e-11, the period is 31,554,896.93 s, which is 365.22 days, one Earth year.
Orbital period formula
T = 2 pi sqrt(a^3 / (G M))
T = orbital period in seconds
a = semi-major axis in meters
M = central mass in kilograms
G = 6.674e-11 N m^2/kg^2 (fixed constant)
The gravitational constant G is held fixed at 6.674e-11 N m^2/kg^2. Cube the semi-major axis, divide by the product of G and the central mass, take the square root, and multiply by two pi. The result is in seconds; divide by 86,400 to convert to days.
Worked example
Earth orbits the Sun with a semi-major axis of 1.496e11 meters around a central mass of 1.989e30 kilograms, with G fixed at 6.674e-11 N m^2/kg^2.
- a^3 = (1.496e11)^3 = 3.348e33
- G x M = 6.674e-11 x 1.989e30 = 1.3275e20
- a^3 / (G M) = 3.348e33 / 1.3275e20 = 2.5221e13
- sqrt(2.5221e13) = 5,022,082.5
- T = 2 pi x 5,022,082.5 = 31,554,896.93 s
- 31,554,896.93 / 86,400 = 365.22 days
The period is 31,554,896.93 seconds, which is 365.22 days, one Earth year as expected. These are the calculator's default inputs, so the result above matches the widget exactly.
Reference orbits
| Orbit | Semi-major axis (m) | Central mass (kg) |
|---|---|---|
| Earth around the Sun | 1.496e11 | 1.989e30 |
| Moon around the Earth | 3.844e8 | 5.972e24 |
| Mars around the Sun | 2.279e11 | 1.989e30 |
Values are standard published figures for the semi-major axis and central mass. Use them to check the calculator against known orbits.
Orbital period calculator: frequently asked questions
What is Kepler's third law?
Kepler's third law relates the orbital period of a body to the size of its orbit. In its Newtonian form the period T equals 2 pi times the square root of the semi-major axis cubed divided by the product of the gravitational constant G and the central mass M. The square of the period is proportional to the cube of the semi-major axis, which is why widely separated orbits take far longer to complete.
What value of G does this calculator use?
It uses the gravitational constant G fixed at 6.674e-11 N m^2/kg^2, the standard accepted value. G is treated as a constant inside the calculation, so you only enter the semi-major axis and the central mass. The result is the time for one full orbit, reported in both seconds and days.
What is the semi-major axis?
The semi-major axis is half the longest diameter of an elliptical orbit. For a circular orbit it is simply the radius. It is entered in meters. For Earth's orbit around the Sun the semi-major axis is about 1.496e11 meters, which is one astronomical unit, the default value in this calculator.
Why does the Earth default return one year?
With a semi-major axis of 1.496e11 meters and a central mass of 1.989e30 kilograms, which is the mass of the Sun, the formula returns roughly 31,554,896.93 seconds. Dividing by 86,400 seconds per day gives about 365.22 days, one Earth year. The small difference from 365 reflects the actual length of the sidereal year.
Does this work for moons and satellites?
Yes. The same formula governs any two-body orbit where one mass dominates. Enter the semi-major axis of the orbit and the mass of the central body, whether that is a planet for a moon, the Earth for a satellite, or a star for a planet. The period scales with the three-halves power of the orbit size.
Official sources
- Planetary fact sheets and orbital mechanics: US National Aeronautics and Space Administration (NASA). As at 25 June 2026.
Reviewed by the CalculatorHub team, edited by James Graham, 25 June 2026. See our methodology. This is general information, not financial, tax, legal or investment advice.