🛰️ Circular Orbital Velocity

Written and maintained by the PhDino author · Last reviewed 21 September 2026 · Checked against 3 independent reference calculations · how PhDino checks its numbers

How fast an object must travel to maintain a stable circular orbit at a given altitude.

An object in a stable circular orbit is continuously falling toward the planet it orbits — but moving sideways fast enough that the surface curves away beneath it at the same rate it falls, so it never actually gets closer to the ground. The exact speed needed for this balance depends only on the central body's mass and the orbital radius, not on the orbiting object's own mass.

Higher orbits require lower orbital velocity, not higher — a satellite farther from a planet's center feels a weaker gravitational pull, so it needs to move slower (though it still travels a longer circumference) to keep that same falling-but-missing balance.

Orbital period follows directly from velocity and orbital circumference, and this same relationship (Kepler's third law, expressed here in its full physical form) is why geostationary satellites sit at one specific, well-known altitude: it's the unique altitude where the orbital period exactly matches Earth's rotation.

Key formula

v = √(μ ÷ r)
T = 2π × √(r³ ÷ μ)

Variables

μ

standard gravitational parameter of the central body (μ = GM)

r

orbital radius, measured from the central body's center — not altitude above the surface

Notes & limitations

• This assumes a perfectly circular orbit — a real orbit is generally elliptical to some degree, in which case velocity actually varies along the orbit (fastest at the closest point, slowest at the farthest), and this formula only gives the single-speed circular special case.
• Real low orbits around Earth also experience atmospheric drag at low enough altitude, which this formula (pure gravity and geometry) doesn't account for — it's why very low orbits decay over time and eventually need reboosting or reenter.