Orbital Period Calculator (Kepler’s Third Law)
Compute how long an orbit takes from the central mass and the semi-major axis using Kepler’s third law.
How it works
T = 2π · √(a³ ÷ (G·M)) · Kepler’s third law: T² ∝ a³
Kepler’s third law says the square of an orbital period grows with the cube of the semi-major axis. Newton showed the exact constant depends on the central mass: T = 2π√(a³/GM). Double the orbital distance and the year gets about 2.83 (=2^1.5) times longer, which is why the outer planets crawl through their slow, decades-long orbits.
Worked example
For Earth around the Sun, a = 1 AU = 1.496×10¹¹ m and M = 1.989×10³⁰ kg: T = 2π√((1.496×10¹¹)³ ÷ (6.674×10⁻¹¹ × 1.989×10³⁰)) ≈ 3.156×10⁷ s ≈ 365.3 days.
Sources: Kepler’s laws — Wikipedia
Frequently asked questions
What is the semi-major axis?
It is half the longest diameter of the elliptical orbit — the average of the closest and farthest distances from the central body. For a circular orbit it is simply the radius.
Does the orbiting body’s mass affect the period?
For everyday cases where the central body is far heavier (a planet around the Sun, a satellite around Earth), no — the small mass cancels out. It only matters for comparable masses, such as binary stars.
Why is the period independent of orbital eccentricity?
Kepler’s third law uses the semi-major axis, not the shape. A circular and a highly elliptical orbit with the same semi-major axis have exactly the same period.
Related calculators
This calculator is for educational purposes. Double-check important results before acting on them.