Orbital Period Calculator: Kepler's Third Law T = 2π√(a³/GM)
Kepler's Third Law, published by Johannes Kepler in 1619 and later derived from Newton's law of gravitation, establishes a precise relationship between the size of an orbit and the time it takes to complete: T = 2 pi sqrt(a^3 / (G M)), where T is the orbital period, a is the semi-major axis (half the longest diameter of the ellipse), G is the gravitational constant (6.67430 x 10^-11 N m^2 kg^-2, NIST CODATA 2018), and M is the mass of the central body. This single formula governs the orbits of planets around the Sun, moons around planets, and satellites around Earth. It is the basis for calculating geostationary orbit altitude, GPS satellite positioning, and the masses of distant stars from the observed periods of their planets. This calculator computes orbital period in seconds, hours, days, and years, and also shows the circular orbital velocity, for any semi-major axis and central body you choose. Select from common bodies (Sun, Earth, Moon) or enter a custom central mass.
Orbital period: --
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Kepler's Third Law
T = 2π × √(a³ / (G × M))
where T is the orbital period in seconds, a is the semi-major axis in metres, G = 6.67430 x 10^-11 N m^2 kg^-2 (NIST CODATA 2018), and M is the mass of the central body in kg.
Circular orbital velocity
v = √(G × M / a)
For a circular orbit, the orbital velocity is constant. For elliptical orbits, velocity varies: fastest at periapsis (closest approach), slowest at apoapsis (furthest point).
Unit conversions used
- 1 km = 1,000 m
- 1 AU = 149,597,870,700 m (IAU 2012 definition)
- 1 year = 365.25 days (Julian year)
- 1 day = 86,400 s
Reference orbital periods
| Body | Orbiting | Semi-major axis | Period |
|---|---|---|---|
| Moon | Earth | 384,400 km | 27.32 days |
| ISS | Earth | ~6,771 km | ~92.5 min |
| GPS satellite | Earth | 26,560 km | ~12 hours |
| Geostationary | Earth | 42,164 km | 23 h 56 min |
| Earth | Sun | 1.000 AU | 365.25 days |
| Mars | Sun | 1.524 AU | 686.97 days |
| Jupiter | Sun | 5.203 AU | 11.86 years |
Sources: NASA JPL Solar System Dynamics (2024).
Frequently asked questions
What is Kepler's Third Law?
Kepler's Third Law, published in 1619, states that the square of an orbit's period is proportional to the cube of its semi-major axis: T^2 proportional to a^3. Newton later showed the proportionality constant is 4 pi^2 / (G M), where M is the mass of the central body. The full formula is T = 2 pi sqrt(a^3 / (G M)). This law applies to all orbiting bodies: planets around the Sun, moons around planets, and artificial satellites around Earth.
What is a geostationary orbit and how is the altitude found?
A geostationary orbit (GEO) is the altitude at which a satellite orbits Earth with a period of exactly 24 hours (one sidereal day, 86,164 seconds), so it appears stationary above a fixed point on the equator. Rearranging Kepler's third law for semi-major axis: a = (G M T^2 / (4 pi^2))^(1/3). Using Earth's mass and a sidereal day gives a = approximately 42,164 km from Earth's centre, or about 35,786 km above the surface.
What is the Moon's orbital period and why does it differ from the calendar month?
The Moon's sidereal orbital period (relative to fixed stars) is approximately 27.32 days, consistent with Kepler's Third Law given the Moon's semi-major axis of about 384,400 km. The synodic month (new Moon to new Moon) is approximately 29.53 days because the Earth-Moon system also orbits the Sun, so the Moon must travel slightly further than one full orbit to realign with the Sun as seen from Earth.
Why does the ISS orbit Earth faster than the Moon?
According to Kepler's Third Law, orbital period increases with the cube root of the semi-major axis. The ISS orbits at about 400 km altitude (a approximately 6,771 km from Earth's centre) giving a period of about 92 minutes. The Moon orbits at 384,400 km from Earth's centre, giving a period of 27.32 days. Smaller orbit, shorter period: lower orbits are faster both in orbital speed and in period.
Does Kepler's Third Law apply to elliptical orbits?
Yes. For elliptical orbits, a in the formula T = 2 pi sqrt(a^3 / (G M)) refers to the semi-major axis, which is half the longest diameter of the ellipse. The period depends only on the semi-major axis, not the eccentricity. Two orbits with the same semi-major axis but very different eccentricities will have exactly the same period. This is a consequence of conservation of energy in an inverse-square force field.