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Circular Orbital Speed Calculator

Enter the orbital radius and the central body's GM to find the circular orbital speed.

Result

7,668.6 m/s
7.669 km/s
92.6 min

How it is calculated

For a circular orbit, gravity alone supplies the centripetal force, which gives v = √(GM ÷ r). GM is the standard gravitational parameter of the central body in m³/s², and r is the distance from the body's centre in metres. The tool takes the radius in kilometres and multiplies by 1,000 before using it. It then finds the orbital period from T = 2πr ÷ v, converted to minutes. Speed is shown in m/s with one decimal, in km/s with three decimals, and the period in minutes. Both radius and GM must be above zero. The formula assumes a perfectly circular orbit around a point mass and ignores atmosphere, other bodies and the planet's shape, and the radius is not checked against the body's own size.

Worked example

An orbit 6,778 km from Earth's centre, with Earth's GM entered as 3.986e14. Convert the radius: r = 6,778,000 m. Divide: 3.986e14 ÷ 6,778,000 ≈ 58,808,000. Take the square root to get a speed of 7,668.6 m/s, which is 7.669 km/s. The period is 2π × 6,778,000 ÷ 7,668.6 ≈ 5,553 s, which the tool shows as 92.6 min. Raising the radius to 42,164 km with the same GM drops the speed to about 3,074.7 m/s and stretches the period to 1,436.1 min.

Try this example →

Common mistakes

  • Entering the altitude above the surface instead of the distance from the centre. For a 400 km altitude around Earth you must add Earth's radius, giving about 6,778 km.
  • Entering GM in km³/s², for example 398,600, instead of m³/s². The tool then returns about 0.2 m/s, which is clearly wrong.
  • Typing the radius in metres in the kilometre field, which makes the speed far too low.
  • Entering the gravitational constant G or the body's mass in kilograms in the GM field. The field needs the product GM, in m³/s².
  • Using the circular speed for an elliptical or escape trajectory. This value is only valid for a circle at that exact radius.

About this calculator

For a satellite in a stable circular orbit, gravity must supply exactly the centripetal force needed to keep it moving in a circle, and solving that balance gives v = √(GM/r), where GM is the standard gravitational parameter of the central body (Earth's is about 3.986×10¹⁴ m³/s²) and r is the orbital radius measured from the body's center, not its surface.

This equation is basic astrodynamics: it shows why low orbits require higher speeds than high ones (v shrinks as r grows), why the International Space Station at roughly 400 km altitude travels near 7.66 km/s, and why geostationary satellites at about 35,786 km altitude move much slower — around 3.07 km/s — while still completing one orbit every 24 hours because of the much larger circle they trace.

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