v = √(GM/r) — Circular Orbital Velocity

Circular orbital velocity — how fast satellites must travel — with a worked ISS-altitude example.

v = \sqrt{\frac{GM}{r}}

Orbital speed equals the square root of G times central mass over orbital radius.

What it means

Orbit is continuous falling: the speed $\sqrt{GM/r}$ makes gravity bend the path into a circle instead of pulling the satellite down. Higher orbits are slower (larger $r$ ), and the satellite's own mass is irrelevant — a feather and a space station orbit identically.

Variables

Symbol	Name	Unit	Meaning
$v$	Orbital velocity	m/s	Speed along the circular orbit
$G$	Gravitational constant	N·m²/kg²	Universal gravitational constant
$M$	Central mass	kg	Mass of the planet or star
$r$	Orbital radius	m	Orbit radius from the center

Assumptions

Circular orbit
Central mass M dominates (satellite mass negligible)
Only gravity acts

Derivation

Set centripetal force equal to gravity: $mv^2/r = GMm/r^2$ . The satellite mass $m$ cancels, leaving $v^2 = GM/r$ — the orbit speed is independent of the satellite's mass.

Worked example

Problem

What speed keeps a satellite in circular orbit 400 km above Earth's surface?

Given:

GM = 3.986×10¹⁴ m³/s² (Earth)
r = 6.771×10⁶ m (Earth radius + 400 km)

Write the formula: $v = \sqrt{GM/r}$
Substitute: $v = \sqrt{3.986\times10^{14} / 6.771\times10^6}$
Calculate: $\sqrt{5.887\times10^7} \approx 7673\ \mathrm{m/s} \approx 7.67\ \mathrm{km/s}$

≈ 7.67 km/s

The ISS must travel 7.67 km/s — about 27,600 km/h — to stay in orbit. Any slower and gravity wins; this is why reaching orbit is about going sideways fast, not just going up.