Physics Laws & Theorems
Intermediate

Kepler's Third Law

The square of the orbital period is proportional to the cube of the semi-major axis.

Formula

T2∝a3T^2 \propto a^3

Variables

TPeriod
aSemi-major axis

Example

Neptune, 30× farther than Earth, takes 30^1.5 ≈ 165 years to orbit the Sun.

How it works

Kepler’s Third Law says that objects orbiting the same central body move more slowly when their orbits are larger: if the semi-major axis increases, the period grows much faster because the orbital path is bigger and the orbital speed is lower. For bodies orbiting the same star or planet, the ratio T^2/a^3 stays constant, which comes from Newton’s version of gravity and the balance between gravitational pull and orbital motion. In the Solar System, this lets us compare or predict orbital periods using only the size of the orbit.

When to use it

Use this law when comparing planets, moons, or satellites that orbit the same central mass, or when estimating an unknown orbital period from the orbit size. It is especially useful in astronomy and astrodynamics for circular and elliptical orbits around one dominant body.

Worked example, step by step

Example: A dwarf planet orbits the Sun with a semi-major axis of 4 AU. Find its orbital period in Earth years. Step 1: Use the convenient Solar System form: T^2 = a^3 Step 2: Substitute a = 4 AU: T^2 = 4^3 = 64 Step 3: Take the square root: T = sqrt(64) = 8 Step 4: Final answer: the orbital period is 8 years

Common mistakes to avoid

  • Using the formula for objects orbiting different central bodies without adjusting for the central mass
  • Forgetting that a must be the semi-major axis, not the average distance or the closest approach
  • Squaring or cubing the wrong quantity, such as writing T^3 ∝ a^2 instead of T^2 ∝ a^3

Did You Know?

Kepler's harmonic law ties a planet's year to its distance — the key Newton used to deduce gravity.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026