Orbital Period Calculator — Kepler's Third Law
Enter the semi-major axis of an orbit and the mass of the central body to find the orbital period using Kepler's Third Law. Works for planets orbiting stars, moons orbiting planets, and any other gravitationally bound two-body system.
Axis unit
Central body
Time for one complete orbit around the central body
- 1
Semi-major axis in metres
1 × 149,600,000,000 = 149,600,000,000 - 2
G × M
6.674 × 10⁻¹¹ × 1,989,000,000,000,000,000,000,000,000,000 = 132,745,859,999,999,980,000 - 3
a³ ÷ (G × M)
149,600,000,000³ ÷ 132,745,859,999,999,980,000 = 25,221,667,447,858.64 - 4
Square root
√25,221,667,447,858.64 = 5,022,117.825√(a³/GM) in seconds per radian. - 5
Period T = 2π × √(...) in days
2π × 5,022,117.825 ÷ 86,400 = 365.22
How does this calculator work?
Orbital period follows Kepler's Third Law: T = 2π √(a³ / GM), where a is the semi-major axis and M is the central body's mass. A larger orbit always means a longer period (T² ∝ a³). Earth's 1 AU orbit around the Sun takes 365.25 days; double the distance and the period grows by a factor of 2^1.5 ≈ 2.83.
Formula
How this is calculated
Kepler's Third Law states that the square of a body's orbital period is proportional to the cube of its semi-major axis: T² ∝ a³. In full SI form, T = 2π √(a³ / (G M)), where a is the semi-major axis in metres, M is the mass of the central body in kilograms, and G = 6.674 × 10⁻¹¹ N m² kg⁻² is the gravitational constant. The result T is in seconds and is then converted to days and years for readability.
The formula assumes a two-body system where the orbiting mass m is negligibly small compared to M (so the barycentre sits at the centre of M). For Earth orbiting the Sun this assumption is excellent — Earth's mass is about 0.0003% of the Sun's. For binary star systems of similar mass the full reduced-mass correction is needed. The orbit is also assumed to be a closed ellipse; hyperbolic or parabolic trajectories (escape trajectories) do not have a period.
The orbital speed shown is the mean speed averaged around a circular orbit of radius a. Real elliptical orbits move faster at periapsis and slower at apoapsis (Kepler's Second Law), so the displayed speed is an approximation.
Frequently asked questions
One year is defined as Earth's orbital period. With a semi-major axis of 1 AU (1.496 × 10¹¹ m) and the Sun's mass (1.989 × 10³⁰ kg), Kepler's Third Law gives T ≈ 3.156 × 10⁷ seconds, which is 365.25 days — the basis of the Gregorian calendar.
Yes. For a satellite orbiting Earth, set the central body to Earth (5.972 × 10²⁴ kg) and enter the orbital radius (Earth's radius ≈ 6371 km plus altitude). The International Space Station orbits at about 408 km altitude, giving a period of roughly 92 minutes.
The semi-major axis a is half the length of the longest diameter of the ellipse. For a circular orbit it equals the radius. For Earth's slightly elliptical orbit around the Sun, a = 1 AU, midway between the perihelion (closest) and aphelion (farthest) distances.
Also known as
TG we-Calculate Editorial Team. (2026). Orbital Period Calculator — Kepler's Third Law [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/orbital-period-calculator
TG we-Calculate Editorial Team. "Orbital Period Calculator — Kepler's Third Law." TG we-Calculate. 2026. https://we-calculate.com/calculator/orbital-period-calculator.
TG we-Calculate Editorial Team, "Orbital Period Calculator — Kepler's Third Law," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/orbital-period-calculator
@misc{wecalculate_orbital_period_calculator, title = {Orbital Period Calculator — Kepler's Third Law}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/orbital-period-calculator}}, year = {2026}, note = {TG we-Calculate} }
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