Orbital Period Calculator
Find the time for one orbit from the semi-major axis and central mass with Kepler’s third law.
That is 3.1555e+7 seconds. G = 6.674×10⁻¹¹ N·m²/kg²; enter a in metres and M in kilograms (e-notation works, e.g. 1.989e30).
The orbital period is T = 2π√(a³ ÷ GM), where a is the semi-major axis, M the central mass, and G = 6.674×10⁻¹¹ N·m²/kg². For Earth (a = 1.496×10¹¹ m, M = 1.989×10³⁰ kg) this gives T ≈ 3.156×10⁷ s ≈ 365 days — one year.
What the orbital period tells you
The orbital period T is the time a body takes to complete one full loop around a much heavier central body. In the two-body limit — where the central mass M is far larger than the orbiting mass m — the period depends only on the semi-major axis a (the average orbital radius) and M. This is Newton’s form of Kepler’s third law, and it works for planets, moons, and artificial satellites alike.
a = semi-major axis (m), M = central mass (kg), G = 6.674×10⁻¹¹ N·m²/kg², T in seconds
Worked example
How long does Earth take to orbit the Sun? Use a = 1.496×10¹¹ m (one astronomical unit) and M = 1.989×10³⁰ kg (the Sun’s mass).
- 1 Write Kepler’s third law. T = 2π√(a³ ÷ GM), with G = 6.674×10⁻¹¹ N·m²/kg².
- 2 Cube the semi-major axis. a³ = (1.496×10¹¹)³ ≈ 3.348×10³³ m³.
- 3 Compute GM. GM = 6.674×10⁻¹¹ × 1.989×10³⁰ ≈ 1.3275×10²⁰ m³/s².
- 4 Divide, take the square root, multiply by 2π. T = 2π√(3.348×10³³ ÷ 1.3275×10²⁰) ≈ 3.156×10⁷ s ≈ 365.3 days.
Orbital periods from Kepler’s third law
Periods computed from T = 2π√(a³ ÷ GM) with the listed semi-major axis and central mass.
| Orbit | Semi-major axis a | Period T |
|---|---|---|
| Earth around the Sun | 1.496×10¹¹ m | ≈ 365 days |
| Moon around Earth | 3.844×10⁸ m | ≈ 27.5 days |
| ISS around Earth | 6.78×10⁶ m | ≈ 93 minutes |
Kepler’s third law and why the orbiting mass drops out
T² ∝ a³. Squaring the formula gives T² = 4π²a³ ÷ GM, so for a fixed central body the square of the period is proportional to the cube of the semi-major axis. A larger orbit always means a longer year, and the relationship is steep — quadrupling the radius roughly octuples the period.
The satellite’s own mass doesn’t appear. Because M ≫ m, the orbiting mass m cancels out of the equation of motion. A feather and a space station at the same altitude share the same period. This only breaks down when the two masses are comparable, where the exact form uses G(M + m) instead of GM.