Kepler’s Third Law Calculator

How it works

1. 1Enter the central mass M in kilograms (default is the Sun, 1.989×10³⁰ kg) and the semi-major axis a in astronomical units (AU).
2. 2The calculator converts a to metres (1 AU = 1.496×10¹¹ m) then evaluates T = 2π·√(a³ ÷ GM).
3. 3Results show the period in years, days, and seconds, plus the Kepler constant T² ÷ a³ (≈ 1 for solar orbits with T in years and a in AU).

Use cases

• Verify a planet’s period matches its semi-major axis — Earth at 1 AU gives T ≈ 1 year.
• Estimate the period of an exoplanet around a star of known mass.
• Demonstrate Kepler’s third law: T² ÷ a³ ≈ 1 for all solar-system planets.

Frequently asked questions

What does Kepler’s third law state?

The square of the orbital period is proportional to the cube of the semi-major axis: T² ∝ a³. The exact SI relation is T = 2π·√(a³ ÷ GM), where G ≈ 6.674×10⁻¹¹ N·m²/kg² and M is the central mass.

Why is T² ÷ a³ ≈ 1 for solar-system planets?

When T is in years and a in AU, the constant equals 1 for anything orbiting the Sun, because the AU and the year are defined by Earth’s orbit. Since T² ∝ a³ is universal, every solar-orbit body gives the same ratio.

Can I verify this with a real example?

Earth: a = 1 AU → T ≈ 1.000 year. Mars: a = 1.524 AU → T = √(1.524³) ≈ 1.881 years, matching the ~687-day Martian year. Jupiter: a = 5.204 AU → T ≈ 11.87 years.