Kepler's third law: T² is proportional to r³

Gravitational Fields & Orbits — AQA A-Level Physics

Key Definition

Kepler's third law — For planets or satellites in a circular orbit about the same central body, the square of the orbital period is proportional to the cube of the orbital radius.

Deriving T² = 4π²r³ / GM

The orbital speed is the circumference divided by the periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s).:

v = \frac{2\pi r}{T}

Substitute into v² = GM/r:

\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}

Expand and rearrange:

T^2 = \frac{4\pi^2 r^3}{GM}

$T$: orbital periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (s)
$r$: orbital radius (m)
$G$: Newton's gravitational constant
$M$: mass of the central body (kg)

T² ∝ r³. A log-log plot of T against r gives a straight line with gradient 3/2.
This derivation is commonly examined. You must be able to reproduce it from scratch.

Examiner Tips and Tricks

Many gravitation calculations depend on circular motion equations.
Make sure you can combine v = 2πr/T, $v^{2} = GM/r$, and $F = mv^{2}/r fluently$.