Using Kepler’s third law, this tool finds how long one orbit takes from the central mass and the orbit’s semi-major axis. It works for satellites around a planet, moons, or planets around a star — anything dominated by a single central gravitational body.
How it works
The period comes from Kepler’s third law in Newtonian form:
T = 2π · √(a³ ÷ GM)
where a is the semi-major axis in metres, M is the central mass in kilograms, and G is the gravitational constant, 6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻². The tool converts your axis from kilometres to metres, applies the formula to get the period in seconds, then also displays it in days (÷ 86,400) and years (÷ 31,557,600). Presets fill in the mass of the Sun, Earth, Jupiter or the Moon.
Example
Earth orbiting the Sun (the defaults): M = 1.989 × 10³⁰ kg and a = 149,597,870 km (1 AU = 1.496 × 10¹¹ m).
- T = 2π · √((1.496×10¹¹)³ ÷ (6.6743×10⁻¹¹ × 1.989×10³⁰))
- T ≈ 3.156 × 10⁷ seconds ≈ 365.25 days ≈ 1 year
| Central body | Semi-major axis | Period |
|---|---|---|
| Sun | 1 AU (Earth) | ≈ 365 days |
| Sun | 5.2 AU (Jupiter) | ≈ 11.9 years |
| Earth | 6771 km (LEO, 400 km up) | ≈ 92 minutes |
All values are computed in your browser; nothing is uploaded.