Orbital period calculator
Orbital period from the central body mass and the orbit radius.
Fill in the fields and the result will appear here automatically.
Find period and speed for a circular orbit around a spherical central body. Radius is measured from the centre: a 400 km altitude above the illustrative 6371 km Earth requires r=6771 km. Satellite mass is negligible compared with the central body. Orbits per day uses 86,400 seconds, not a sidereal day or a pass schedule.
How it works
Formula and logic
T = 2π√(r³/GM) with G = 6.6743·10⁻¹¹, and the orbital speed is v = √(GM/r).
Example
A 6771 km orbit around Earth takes 5545 seconds — about an hour and a half.
Fields and units
- Central body mass — 10²⁴ kg
- Orbit radius — km
How to use
- — Central body mass in units of 10²⁴ kg: Earth is 5.972, the Sun 1,989,100.
- — The orbit radius is measured from the CENTRE of the body, not the surface: 400 km above Earth is 6771 km.
- — Near 42,164 km from Earth’s centre the period is close to a sidereal day, about 23 h 56 min. A geostationary orbit must also be circular, equatorial and prograde. Radius alone does not let this calculator verify those conditions.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: circular orbit and period
- Limitation
- Circular orbit outside a spherical central body, negligible satellite mass, G=6.6743·10⁻¹¹. Eccentricity, inclination, perturbations and atmosphere are not entered. For comparable masses, a two-body period depends on the sum of both masses.
FAQ
Why is the radius measured from the centre?
Because gravity depends on the distance to the centre of mass. A satellite 400 km up sits at a radius of 6371 + 400 = 6771 km, and substituting the altitude alone would be wrong by a factor of several.
How does this differ from centripetal force?
That one takes the period from a given speed. Here the speed is derived from the central mass, so knowing what you orbit and at what radius is enough.
Why is geostationary orbit at 42,164 km?
Near 42,164 km from Earth’s centre the period is close to a sidereal day, about 23 h 56 min. A geostationary orbit must also be circular, equatorial and prograde. Radius alone does not let this calculator verify those conditions.
Does the satellite mass matter?
Omitting satellite mass is a small-body approximation. In the two-body model T=2π√(a³/[G(M+m)]), with a the semimajor axis of the relative orbit. This page calculates the circular case with m≪M.