Boneyard Tools

Circular Orbital Velocity Explained

How v = sqrt(GM/r) comes from balancing gravity and circular motion, and what it tells you about satellites and moons.

Gravity as the string on a sling

A body moving in a circle is constantly accelerating toward the centre, and that centripetal acceleration has to come from somewhere. In orbit, gravity plays the role of the string on a sling: it pulls the satellite inward just hard enough to bend its straight-line motion into a closed loop. Set the gravitational pull GMm/r^2 equal to the centripetal requirement mv^2/r, and the satellite mass m cancels from both sides. That single cancellation is why a bolt and a space station at the same altitude orbit at exactly the same speed.

Deriving v = sqrt(GM/r)

Starting from GMm/r^2 = mv^2/r, multiply both sides by r and divide by m to get GM/r = v^2, so v = sqrt(GM/r). The speed depends only on the central mass M and the orbit radius r, with the gravitational constant G tying the units together. Because r sits under the square root, moving to a higher orbit lowers the speed: a satellite twice as far out orbits about 1.41 times slower. This calculator evaluates that expression directly and then derives the period from the circumference.

Altitude versus orbit radius

A frequent mistake is to plug in the altitude above the surface instead of the radius from the centre. The formula needs the full distance to the centre of mass, so for an Earth satellite you add Earth's mean radius of about 6371 km to the altitude. A 400 km altitude orbit therefore uses r near 6771 km, which yields about 7.67232 km/s. Forgetting the planet's radius throws the answer off badly for low orbits, though the error shrinks for distant ones like the Moon at 384,400 km.

From speed to period and altitude choices

Once the speed is known, the period is just how long it takes to travel the circumference, T = 2 x pi x r / v. A low Earth orbit laps the planet in about 5545 seconds, close to 92 minutes, while a geostationary orbit at 42,164 km takes about 86,167 seconds, close to one sidereal day, which is exactly why such satellites appear to hover over a fixed spot. Comparing the periods of different radii shows the trade every mission makes: low orbits are fast and cheap to reach but sweep past the ground quickly, while high orbits are slow, expensive to reach and linger.

Frequently asked questions

Why do geostationary satellites stay over one point?

At a radius of about 42,164 km the orbital period is roughly 86,167 seconds, matching Earth's rotation, so the satellite keeps pace with the ground below and appears fixed.

How does orbital velocity relate to escape velocity?

Escape velocity is sqrt(2) times the circular orbital velocity at the same radius, about 1.414 times faster, since escaping needs enough energy to reach zero speed at infinity.