Last updated: 2026-07-01

Orbital Velocity

Orbital velocity is the speed an object must travel to stay on a circular orbit around a planet or star — fast enough that its fall toward the body curves into a continuous loop instead of an impact. A satellite in low Earth orbit moves at about 7.7 km/s; a geostationary satellite, far higher up, needs only about 3.1 km/s. This calculator finds the circular orbital velocity from the central body's mass and the orbit radius, and also reports how long one full orbit takes.

Where the formula comes from

A circular orbit is a balance between two effects. Gravity pulls the orbiting object inward with force $GMm/r^2$, while circular motion at speed $v$ demands a centripetal force $mv^2/r$ directed toward the centre. An orbit is stable exactly when gravity supplies that demand:

The orbiting object's mass $m$ appears on both sides and cancels, leaving $v^2 = GM/r$. That is why orbital speed depends only on the central body and the radius — never on what is orbiting.

Formula

Note that $r$ is measured from the centre of the central body, not from its surface — so for an orbit around Earth you add the satellite's altitude to Earth's radius of about 6371 km.

Worked example

For a satellite in low Earth orbit at 400 km altitude, the radius is $r = 6371 + 400 = 6771\ \text{km} = 6.771\times10^{6}\ \text{m}$, with $M = 5.972\times10^{24}\ \text{kg}$:

The orbital period is the circumference divided by this speed:

So a satellite skimming low Earth orbit laps the planet roughly every hour and a half.

Higher orbits are slower

Because $v = \sqrt{GM/r}$, orbital speed drops as the radius grows. Climb to geostationary altitude — $r \approx 42\,164\ \text{km}$ — and the orbital velocity falls to about 3.1 km/s, with a period of almost exactly 24 hours. That is the point of a geostationary orbit: matching Earth's rotation so the satellite hovers over one spot on the equator. The Moon, far further still at $r \approx 384\,400\ \text{km}$, orbits at only about 1 km/s and takes 27 days.

Orbital versus escape velocity

Orbital velocity is closely related to escape velocity. Escaping a body entirely takes $v_{esc} = \sqrt{2GM/r}$, exactly $\sqrt{2}$ times the circular orbital velocity at the same radius. So a spacecraft already in low orbit at 7.7 km/s needs only about 41% more speed — roughly 10.9 km/s — to leave Earth's gravity for good.

Limitations

This is the ideal velocity for a perfectly circular orbit around a single spherical body, ignoring atmospheric drag, the gravity of other bodies, and any thrust. Real orbits are usually slightly elliptical, in which case the speed varies along the orbit and $r$ is replaced by the semi-major axis $a$ for the period. Near the surface, atmospheric drag is significant, which is why crewed and satellite orbits sit above roughly 200 km, where the air is thin enough to coast for long stretches.