Result

Kepler's third law

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Statement

The square of an orbital period is proportional to the cube of the orbit's semi-major axis a.

Symbols and units

$T$	orbital period	s
$a$	semi-major axis	m
$G$	gravitational constant	N·m²/kg²
$M$	central mass	kg

When it applies

Any orbit (circular or elliptical, with a as the semi-major axis) around a fixed central mass M.

Don't use when

The simple T² ∝ a³ form assumes the orbiting body's mass is negligible compared with the central mass — a two-body system of comparable masses needs the total mass in place of M.

Derivation

For a circular orbit, the period is the circumference divided by the orbital speed:

T = \frac{2 π r}{v} = 2 π r \frac{r}{GM}

Squaring gives the period-radius relation (a replaces r for a general elliptical orbit):

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Worked examples

Period of a geostationary satellite

Solve for a given T = 86164 s (one sidereal day) around Earth:

a = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} \approx 4.22 \times 1 0^{7} m

About 35,800 km above the surface — the standard geostationary altitude, derived from nothing but Newton's law of gravitation.

Watch out for

The constant 4π²/GM depends only on the central mass M — every satellite around the same body obeys the same T²-a³ ratio, regardless of the satellite's own mass.

Connects to

Circular orbital speed Newton's law of gravitation