Last updated: 2026-06-15

Orbital Period

The orbital period is how long one body takes to make a full loop around another — a year for Earth around the Sun, 27.3 days for the Moon around Earth, 24 hours for a geostationary satellite. Kepler found that the period depends only on the size of the orbit and the mass at its centre, not on the orbiting object itself. This calculator uses his third law to turn a semi-major axis and a central mass into a period, and also reports the speed the orbit is travelled at.

Where the formula comes from

For a circular orbit, gravity supplies exactly the centripetal force needed to keep turning: $\dfrac{GMm}{a^2} = \dfrac{mv^2}{a}$. The orbiting mass $m$ cancels, leaving $v = \sqrt{GM/a}$. Since one lap of circumference $2\pi a$ takes $T = 2\pi a / v$, substituting the speed gives Kepler's third law $T = 2\pi\sqrt{a^3/GM}$. The same relation holds for elliptical orbits when $a$ is the semi-major axis.

Formula

Squaring the period gives the familiar proportion $T^2 \propto a^3$: the square of the period grows with the cube of the distance.

Worked example

For Earth around the Sun, $a = 1.496\times10^{11}\ \text{m}$ and $M = 1.989\times10^{30}\ \text{kg}$:

The orbital speed at that distance is $v = \sqrt{GM/a} \approx 29\,800\ \text{m/s}$, about 29.8 km/s.

How distance stretches the period

Because $T \propto a^{3/2}$, the period grows faster than the distance but slower than the distance squared. An orbit four times wider takes eight times longer. That single rule explains the whole Solar System: Mercury circles the Sun in 88 days, Earth in a year, Jupiter in nearly 12 years, and Neptune in 165 years. Move a satellite higher and its day-length grows; the geostationary belt is simply the altitude where the period matches Earth's 24-hour spin.

Limitations

The formula assumes a small body orbiting a single, much larger mass — a "two-body" approximation. It ignores the orbiting object's own gravitational pull (negligible for a satellite, not for two comparable stars), the tug of other bodies, atmospheric drag on low orbits, and relativistic effects close to very massive objects. For an elliptical orbit it gives the period correctly from the semi-major axis, but the speed shown is the circular-orbit value and is only the average for an ellipse.