About the Orbital Period Calculator
This orbital period calculator applies Newton’s form of Kepler’s third law, T = 2π√(a³/G(M+m)), to find how long an object takes to complete one orbit. Choose the Sun, Earth, the Moon, Mars, Jupiter, Saturn or a custom central mass, enter the orbit’s semi-major axis in kilometres or astronomical units, and get the period in hours, days and years along with the average orbital speed. Switch modes to work backwards from a known period to the orbit size — for example, the altitude of a geostationary satellite.
Astronomy students use it to verify Kepler’s law against real planets, satellite enthusiasts use it to relate altitude and period, and exoplanet fans can estimate how far a planet sits from its star given its year length. The secondary-mass field lets you include the orbiting body’s own mass, which matters for moons and binary stars.
For an elliptical orbit the period depends only on the semi-major axis (half the longest diameter), not on the eccentricity. For a satellite around a planet, remember the semi-major axis is measured from the planet’s centre, so add the planet’s radius to the altitude.
How to use the orbital period calculator
- 1Choose whether to solve for the period or the semi-major axis.
- 2Select the central body, or enter a custom mass in solar masses, Earth masses or kg.
- 3Enter the semi-major axis (or the period) and pick its unit.
- 4Optionally add the orbiting body’s mass for moons or binary systems.
- 5Read the period in hours, days and years, the orbit size and the mean speed.
Formula and method
Johannes Kepler found that the square of a planet’s orbital period is proportional to the cube of its average distance from the Sun. Newton’s law of gravitation supplies the constant: T² = 4π²a³ / G(M + m), where a is the semi-major axis, M the central mass and m the orbiting body’s mass. For most satellites and planets m is negligible, so it defaults to zero.
The calculator uses gravitational parameters GM from NASA fact sheets (more precise than G and M separately) and G = 6.6743 × 10⁻¹¹ m³·kg⁻¹·s⁻² for custom masses. Rearranging gives the semi-major axis from a period. Mean speed is the circumference of a circle of radius a divided by T, which is exact for circular orbits and an average for mildly elliptical ones. Years are Julian years of 365.25 days.
- T
- Orbital period
- a
- Semi-major axis (from the centre of mass)
- G
- Gravitational constant
- M
- Central body mass
- m
- Orbiting body mass (often negligible)
Worked examples
Earth around the Sun (1 AU)
T = 2π√((1.496 × 10¹¹)³ ÷ 1.327 × 10²⁰) ≈ 31,558,000 s = 365.257 days — Earth’s sidereal year — at an average speed of about 29.8 km/s.
Jupiter at 5.2038 AU
With a = 5.2038 AU, T = 5.2038^1.5 years ≈ 11.87 years, matching Jupiter’s observed 11.86-year orbit to within 0.1%. About two-thirds of that small gap is Jupiter’s own mass — enter 317.8 Earth masses as the orbiting mass to include it.
Satellite 400 km above Earth (a = 6,771 km)
Around Earth, GM = 398,600.44 km³/s². T = 2π√(6,771³ ÷ 398,600.44) ≈ 5,545 s, about 92.4 minutes — the period of the International Space Station.
Geostationary orbit from one sidereal day
a = ∛(398,600.44 × 86,164.09² ÷ 4π²) ≈ 42,164 km from Earth’s centre, or about 35,793 km above the mean surface (35,786 km above the equator).
Frequently asked questions
What is Kepler’s third law?+
It states that the square of an orbital period is proportional to the cube of the semi-major axis: T² ∝ a³. For planets around the Sun measured in years and AU, T² = a³ almost exactly, so Mars at 1.524 AU takes 1.881 years.
Does the orbiting body’s mass matter?+
Only slightly in most cases. The formula uses the combined mass M + m. For a satellite around Earth or a planet around the Sun it is negligible, but for the Moon (1.2% of Earth’s mass) or binary stars it noticeably shortens the period.
What is the semi-major axis?+
It is half the longest diameter of an elliptical orbit — effectively the average distance between the two bodies. For a circular orbit it equals the radius. For satellites it is measured from the planet’s centre, not its surface.
Why does a geostationary satellite orbit at 35,786 km?+
Because that is the only altitude where the orbital period equals one sidereal day (23 h 56 min 4 s). Kepler’s law fixes the semi-major axis at about 42,164 km from Earth’s centre, which is 35,786 km above the equator.
Does eccentricity change the orbital period?+
No. Two orbits with the same semi-major axis have the same period whether they are circular or highly elliptical. Eccentricity changes how the speed varies around the orbit, not the time for one lap.