Calculate how long one orbit takes with Kepler’s third law, T = 2π√(a³/GM). Pick the Sun, Earth or another planet, enter the orbit size as a semi-major axis or an altitude above the surface, and get the period, orbital speed and the full working.
Last updated:
Central body
kg
Scientific notation is fine, e.g. 5.97e24
kg
Leave empty for a small satellite; enter it for two massive bodies (e.g. the Moon)
Orbit size
Enter the orbit as
Enter your values to see the result.
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How to use the Orbital Period Calculator
Choose the central body, or Custom and type its mass in kg.
Choose how to give the orbit: semi-major axis, or altitude above the surface for a circular orbit.
Enter the distance and pick its unit.
Add the orbiting mass only if it is significant (a moon or planet).
Read the period, speed and working, and see how it changes with distance in the table.
The orbital period formula
T = 2π √( a³ / G(M + m) )
T is the period in seconds, a the semi-major axis in metres, G the gravitational constant and M + m the combined mass of the two bodies. Rearranged, the semi-major axis for a wanted period is a = ∛( G(M + m) T² / 4π² ). The circular orbital speed at distance a is v = √( G(M + m) / a ), and the escape speed is √2 times that.
Reference orbits
Orbit
Semi-major axis
Period
International Space Station
~6,791 km (420 km altitude)
~92.8 min
GPS satellites
26,560 km
~11 h 58 min (half a sidereal day)
Geostationary orbit
42,164 km
23 h 56 min
The Moon around Earth
384,400 km
27.32 days (sidereal)
Mercury around the Sun
0.387 AU
88.0 days
Earth around the Sun
1.000 AU
365.26 days
Mars around the Sun
1.524 AU
687 days (1.88 yr)
Jupiter around the Sun
5.203 AU
11.86 years
Neptune around the Sun
30.07 AU
164.8 years
Standard gravitational parameters used
Body
GM (m³/s²)
Mean radius
Sun
1.32712440018 × 10²⁰
695,700 km
Earth
3.986004418 × 10¹⁴
6,371 km
Moon
4.9048695 × 10¹²
1,737.4 km
Venus
3.24858592 × 10¹⁴
6,051.8 km
Mars
4.282837 × 10¹³
3,389.5 km
Jupiter
1.26686534 × 10¹⁷
69,911 km
Saturn
3.7931187 × 10¹⁶
58,232 km
Frequently asked questions
What is Kepler's third law?
The square of the orbital period is proportional to the cube of the semi-major axis: T² = 4π²a³ / G(M + m). For planets around the Sun, T in years and a in AU gives the simple form T² ≈ a³.
How long does the ISS take to orbit Earth?
At about 420 km altitude the semi-major axis is about 6,791 km, giving a period of roughly 92.8 minutes, about 15.5 orbits per day.
What altitude gives a 24-hour orbit?
A sidereal day (23 h 56 min) needs a semi-major axis of 42,164 km, about 35,786 km above the equator. That is the geostationary orbit used by many communication satellites.
Does the satellite's mass change the period?
Only slightly, through the M + m term. For artificial satellites m is negligible. For the Moon (about 1.2% of Earth's mass) or Jupiter around the Sun it changes the result by a small but measurable amount.
Does the shape of the orbit matter?
No. For an elliptical orbit the period depends only on the semi-major axis (half the longest diameter), not on the eccentricity. The speed shown is the circular-orbit speed at that distance.
What value of G is used?
G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² (CODATA 2018). For the preset bodies the calculator uses their measured standard gravitational parameters GM, which are known far more precisely than G or M separately.
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