Orbital Period Calculator
Orbital Period23.935 h
In Seconds86,166.7 s
In Days0.9970 days
The Orbital Period Calculator applies Kepler third law to find how long a satellite, moon, or planet takes to complete one orbit around a central body. Enter the mass of the central body in kilograms and the orbital radius in metres, and the tool returns the period in seconds, hours, and days. It is ideal for satellite mission planning, astronomy coursework, and sanity-checking orbital mechanics problems.
Formula
T = 2π·√(r³ / (G·M))
- T
- Orbital period (s)
- r
- Orbital radius / semi-major axis (m)
- G
- Gravitational constant, 6.674e-11 N·m²/kg²
- M
- Mass of the central body (kg)
How it works
- Enter the central body mass in kilograms (scientific notation such as 5.972e24 is accepted) and the orbital radius — the semi-major axis — in metres.
- The calculator evaluates T = 2π·√(r³/(GM)) using the gravitational constant G = 6.674e-11, then converts the period into hours and days.
- Because the period scales with r to the 3/2 power, an orbit twice as large takes about 2.83 times longer to complete.
Worked example
A geostationary satellite orbiting Earth at a radius of 42,164 km.
- Use M = 5.972e24 kg and r = 4.2164e7 m.
- T = 2π × √((4.2164e7)³ ÷ (6.674e-11 × 5.972e24)).
- T ≈ 86,166.7 s ≈ 23.935 hours ≈ 0.997 days.
About 23.935 hours, one sidereal day, as expected for geostationary orbit.
Frequently asked questions
- What is Kepler third law?
- It states that the square of an orbital period is proportional to the cube of the orbital radius. This calculator uses the Newtonian form T = 2π·√(r³/GM), which adds the central mass explicitly.
- Should I use the radius or the altitude?
- Use the orbital radius measured from the centre of the central body, not the altitude above its surface. For a low Earth orbit, add the planet radius of about 6,371 km to the altitude.
- Does the satellite mass affect the period?
- No. For a satellite far lighter than its central body, the period depends only on the central mass and the orbital radius, so a small probe and a large station at the same radius share the same period.
- Why is a geostationary orbit about 24 hours?
- A geostationary satellite must orbit once per sidereal day so it stays above a fixed point on the equator. That requirement sets its radius to roughly 42,164 km and its period to about 23.93 hours.