Orbital Period Calculator

This orbital period calculator uses Kepler’s third law, T = 2π·√(a³ / GM), with the orbit radius and the central body. Example: the ISS at 408 km altitude needs about 92.6 minutes for one orbit of the Earth.

Example: Central body: Earth · What do you want to calculate: Period from the orbit · The distance is: Altitude above the surface · Distance 408 · Unit: km → Result: Orbital period: 92.578 min. Source: Wikipedia – Keplersche Gesetze. Updated: .

For elliptical orbits enter the semi-major axis.
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Result

Result
Orbital period: 92.578 min
Orbit data
ItemValue
Orbital period92.578 min (5,554.7 s)
Orbit radius (semi-major axis)6,779 km
Altitude above the surface408 km
Orbital speed (circular orbit)7.6681 km/s (27,605 km/h)
Escape velocity at this distance10.844 km/s
Orbits per day15.554
Angular speed233.32 °/h
Steps
  • Gravitational parameter of the central body GM = 3.986 × 10¹⁴ m³/s².
  • Orbital period T = 2π·√(a³ / GM) with a = 6,779,000 m gives T = 5,554.7 s.
  • Orbital speed on a circular orbit v = √(GM / a) = 7,668.1 m/s.
The calculation treats the satellite’s mass as negligible, with no air drag and no disturbance from other bodies. For ellipses the period uses the semi-major axis, while the speed applies to a circular orbit only. GM values from IAU and JPL.

How it is calculated

Kepler’s third law links the orbital period of a body to the semi-major axis of its orbit: T² / a³ is the same for all bodies around the same centre. With the gravitational parameter GM of the central body this becomes T = 2π·√(a³ / GM).

How it is calculated

The orbit radius a is measured from the centre of the central body, so for an altitude it is radius plus altitude. The ISS flies at about 408 km, so a is 6,779 km and T comes out at roughly 92.6 minutes at 7.67 km/s. A geostationary satellite must orbit once per sidereal day (23 h 56 min), which needs a ≈ 42,164 km, an altitude of 35,786 km. The Moon at 384,400 km gets about 27.5 days; the real value is a little lower because its own mass counts too. Earth around the Sun needs one year at 1 AU.

Backwards and with a custom body

In the second mode you set the period you want and get the orbit radius and altitude needed. With “Custom body” you can calculate for any mass, such as asteroids or exoplanets.

Frequently asked questions

How do you calculate the orbital period of a satellite?

With T = 2π·√(a³ / GM). For Earth GM = 3.986·10¹⁴ m³/s², and a is Earth’s radius plus the altitude. At 408 km that gives about 92.6 minutes.

How high is a geostationary satellite?

About 35,786 km above the equator, an orbit radius of roughly 42,164 km. One orbit takes exactly one sidereal day, so the satellite seems to stand still in the sky.

Why does the orbital period not depend on the satellite’s mass?

Because the satellite’s mass cancels between gravitational force and inertia. Only the mass of the central body (GM) and the orbit radius determine the period.

Sources and legal basis

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