Kepler's Third Law Calculator
Relate an orbit's size to its period with T² = 4π²a³/GM. Pick the central body (Sun, Earth, Mars, or Jupiter), then either enter the semi-major axis (AU, million km, or km) to get the period, or enter the period (years, days, or hours) to get the orbit size — plus the mean orbital speed.
Example: with Solve for Period T from orbit size a · Central body Sun · Semi-major axis a (if solving for T) 1 · Distance unit AU (astronomical units) · Orbital period T (if solving for a) 27.32 → Result: T = 365.3 days.
- Same value, other units1 yr = 365.3 days = 31,558,196 s
- Mean orbital speed29.78 km/s (66,626 mph)
Computed by the calculator below using its default values. Change any input to see your own numbers.
T² = 4π²a³/GM — Kepler's harmonic law (1619) with Newton's gravitational constant supplying the proportionality. For the Sun in AU and years it collapses to T² = a³.
The law that sized the solar system
Kepler noticed in 1619 that the square of a planet's period is proportional to the cube of its mean distance from the Sun: T² ∝ a³. Newton later showed why — for a small body circling mass M, gravity supplies the centripetal pull, and the algebra collapses to T² = 4π²a³/GM. The stunning consequence is that the period depends only on the central mass and the orbit's semi-major axis. A feather and a battleship at the same altitude orbit in the same time.
Around the Sun, with distance in AU and time in years, the constant becomes exactly 1: T = a^1.5. Jupiter at 5.20 AU takes √(5.20³) ≈ 11.9 years, no calculator required — though this one keeps the units honest.
Reading orbits from periods
The law runs both ways, which is how astronomers weigh things. Measure a moon's period and distance and you get the planet's mass; measure an exoplanet's period and you get its orbital distance. Try it here: the Moon's 27.32-day sidereal month around Earth returns about 383,000 km, within half a percent of the true 384,400 km — the small gap exists because the Moon is massive enough that Earth-plus-Moon, not Earth alone, sets the pace. GPS satellites make a tidy exercise too: a 12-hour period puts them near 26,600 km from Earth's center.
How it’s calculated
T = 2π√(a³/μ) and a = (μT²/4π²)^(1/3), using standard gravitational parameters μ = GM (more precise than G×M separately): Sun 1.32712440018×10²⁰, Earth 3.986004418×10¹⁴, Mars 4.282837×10¹³, Jupiter 1.26686534×10¹⁷ m³/s² (JPL/IAU values). 1 AU = 1.495978707×10¹¹ m (exact); 1 year = 365.25 days = 31,557,600 s. Mean orbital speed = 2πa/T.
Assumes the orbiting body's mass is negligible and reports the circular-equivalent mean speed — for a two-body system of comparable masses, μ should be G(M+m), and eccentric orbits vary in speed along the path.
Kepler check: the planets
| Planet | a (AU) | Period |
|---|---|---|
| Mercury | 0.387 | 88.0 days |
| Venus | 0.723 | 224.7 days |
| Earth | 1.000 | 365.3 days |
| Mars | 1.524 | 687 days |
| Jupiter | 5.204 | 11.87 years |
| Saturn | 9.537 | 29.45 years |
| Neptune | 30.07 | 164.9 years |
Computed with T = 2π√(a³/GM☉) from JPL mean orbital elements; matches published sidereal periods to within rounding.
Common mistakes
- Using the orbital radius above the surface for satellites — a is measured from the central body's center, so add Earth's 6,371 km radius to altitude.
- Mixing unit systems: the tidy T² = a³ shortcut only works for the Sun with AU and years; everything else needs the full 4π²/GM constant.
- Applying the Sun's constant to moons — the central mass is whatever the body orbits, so lunar problems use Earth's GM.
- Entering the synodic period (e.g., 29.5 days full-moon-to-full-moon) instead of the sidereal one (27.32 days) — Kepler's law wants the true orbital period.
Frequently asked questions
What is Kepler's third law formula?
T² = 4π²a³/(GM): the square of the orbital period equals 4π² times the semi-major axis cubed, divided by the gravitational parameter of the central body. Around the Sun in AU and years it simplifies to T² = a³.
Does the mass of the satellite matter?
Not unless it rivals the central body. The satellite's mass cancels out of the derivation, which is why a CubeSat and the ISS at the same altitude share the same period. For the Moon (1.2% of Earth's mass) the approximation is off by about 0.6%.
What does this law say about the ISS?
With a ≈ 6,371 + 420 = 6,791 km around Earth, T = 2π√(a³/GM) ≈ 5,580 s — about 93 minutes, which is why astronauts see 16 sunrises a day.
Semi-major axis or radius — which do I enter?
For a circular orbit they are the same. For an ellipse, use the semi-major axis: half the longest diameter. Kepler's law holds exactly with that value regardless of eccentricity.
Why do the tool's planet periods differ slightly from almanac values?
Real planets tug on each other and orbits are ellipses defined by mean elements, so two-body Kepler math lands within a fraction of a percent, not exactly on, the published sidereal periods.