About orbital periods and Kepler's third law
The orbital period is the time a body takes to complete one full revolution around another: a year for Earth around the Sun, about 27.3 days for the Moon around Earth, roughly 90 minutes for the International Space Station around Earth. Johannes Kepler discovered in 1619 that these periods are not arbitrary; they follow a precise law tying the period to the size of the orbit. Isaac Newton later derived that law from his theory of gravity, turning an empirical pattern into an exact equation you can solve with two numbers: the size of the orbit and the mass at its centre.
The remarkable feature of the law is what it leaves out. The period does not depend on the orbiting body's own mass or on how elongated the orbit is, only on the semi-major axis and the central mass. That is why a tiny CubeSat and a massive space station orbit at the same altitude in the same time, and why every planet's year is fixed by its distance from the Sun alone.
How the orbital period formula works
Newton's form of Kepler's third law gives the period directly from the semi-major axis and the central mass.
T = 2 pi x sqrt( a^3 / (G x M) ) T = orbital period in seconds a = semi-major axis in metres M = mass of the central body in kilograms G = 6.674 x 10^-11 m^3 / (kg x s^2) (gravitational constant)
- Semi-major axis (a) = half the longest diameter of the orbit; for a circle it is just the radius.
- Central mass (M) = the mass being orbited (the Sun, a planet), in kilograms.
- G = the fixed gravitational constant, which sets the units of the answer.
- T squared is proportional to a cubed, so doubling the orbit size multiplies the period by 2 to the power 1.5, about 2.83.
Worked example
Compute Earth's orbital period around the Sun. Use a = 1.496 x 10 to the 11 metres (1 AU) and M = 1.989 x 10 to the 30 kilograms.
- Cube the axis: a^3 = (1.496e11)^3 = 3.35 x 10^33 cubic metres.
- Compute G times M: 6.674e-11 x 1.989e30 = 1.327 x 10^20.
- Divide and take the square root: sqrt(3.35e33 / 1.327e20) = sqrt(2.524e13) = 5.02 x 10^6 seconds for the factor inside.
- Multiply by 2 pi: T = 2 pi x 5.02e6 = 3.156 x 10^7 seconds.
- Convert: 3.156e7 seconds / 86,400 = 365.3 days, almost exactly one year.
Orbital periods in the Solar System
Each planet's year follows directly from its semi-major axis and the Sun's mass.
| Body | Semi-major axis (AU) | Orbital period |
|---|---|---|
| ISS (around Earth) | ~6,790 km radius | ~92 minutes |
| Geostationary satellite | ~42,164 km radius | 23 h 56 min |
| Moon (around Earth) | ~384,400 km radius | 27.3 days |
| Mercury | 0.39 | 88 days |
| Earth | 1.00 | 1.00 year |
| Mars | 1.52 | 1.88 years |
| Jupiter | 5.20 | 11.86 years |
| Neptune | 30.07 | 164.8 years |
Common pitfalls
- Mixing units. The gravitational constant is in SI units, so the axis must be in metres and the mass in kilograms. Plugging in kilometres or astronomical units without converting throws the answer off by huge factors.
- Using radius instead of semi-major axis. For elliptical orbits use the semi-major axis (the average of closest and farthest distance), not the instantaneous radius.
- Adding the orbiting body's mass when it is negligible. The strict law uses (M plus m). For satellites and planets m is tiny and ignored, but for binary stars you must include both masses.
- Confusing the central mass. A moon's period depends on its planet's mass, not the Sun's. Use the mass of the body actually being orbited.
- Sidereal versus solar day. A geostationary orbit matches the sidereal day (23 h 56 min), not the 24-hour solar day, a four-minute difference that matters for precise satellite placement.
Frequently asked questions
What is Kepler's third law?
Kepler's third law states that the square of a body's orbital period is proportional to the cube of the semi-major axis of its orbit. In its Newtonian form the period is T = 2 pi times the square root of a cubed divided by G times M, where a is the semi-major axis, M is the mass of the central body, and G is the gravitational constant 6.674 x 10 to the minus 11. The law applies to planets around the Sun, moons around planets, and satellites around Earth alike.
Does the orbiting body's mass affect the period?
For the common case where the orbiting body is far lighter than the central body, no: a feather and a cannonball at the same altitude orbit Earth in the same time. The simple formula uses only the central mass M. Strictly, the period depends on the combined mass (M plus m), so for binary stars or a planet comparable in mass to its star you must use the total mass. For satellites and planets, m is negligible and the approximation is excellent.
What units should I use in the formula?
Use SI units to match the gravitational constant: the semi-major axis a in metres and the central mass M in kilograms give a period in seconds. Earth's orbit uses a = 1.496 x 10 to the 11 metres and the Sun's mass M = 1.989 x 10 to the 30 kilograms. The calculator then converts the result into days and years for readability. Mixing kilometres or astronomical units without converting is the most common source of wrong answers.
Why is a geostationary orbit at about 35,786 km?
A geostationary satellite must have an orbital period of exactly one sidereal day (about 23 hours 56 minutes) so it appears fixed above one spot on the equator. Solving Kepler's third law for the radius that gives that period around Earth yields an orbital radius of about 42,164 km from Earth's centre, which is roughly 35,786 km above the surface. Any higher and it would lag; any lower and it would race ahead.
Does the formula work for elliptical orbits?
Yes. Kepler's third law uses the semi-major axis, not the radius, so it works for any elliptical orbit, not just circular ones. The semi-major axis is half the longest diameter of the ellipse, equal to the average of the closest (periapsis) and farthest (apoapsis) distances. Two orbits with the same semi-major axis have the same period even if one is nearly circular and the other highly elongated.