Orbital Period

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

Worked example: Geostationary: r = 42 164 km around Earth → T = 86 164.8 s — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter

G = 6.6743 \times 10^{-11}\ \text{m}^{3}\text{/(kg}{\cdot}\text{s}^{2}\text{)}

Newtonian constant of gravitation

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The year of a satellite →

Grade 12Grade 12 Physics

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Orbital Period explained

Kepler noticed in 1619 that the square of a planet's year grows with the cube of its distance from the Sun; Newton later showed why — gravity's inverse-square pull makes T² = 4π²r³/(GM). A worked example: a geostationary satellite must orbit once per sidereal day, T = 86 164 s, so $r = (GMT^2/4\pi^2)^{1/3}$ = $(6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 86\,164^2 / 39.48)^{1/3}$ ≈ 4.22 × 10⁷ m — the 35 800 km altitude where every TV broadcast satellite parks.

The rearrangement for M is one of astronomy's sharpest tools: watch anything orbit, time it, measure the orbit's size, and the central mass falls out. The Moon's 27.3-day circuit at 3.84 × 10⁸ m weighs the Earth; Jupiter's moons weigh Jupiter; and the 16-year orbit of the star S2 around Sagittarius A* revealed a central mass of four million Suns packed into a region smaller than our solar system — a black hole.

Orbital Period formula

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

Where

$T$ = Orbital period (s)
$r$ = Orbital radius (m)
$M$ = Central mass (kg)

Missing one of these? Work it out first, then come back

Orbital period — Speed in Circular Motion (v = 2πr/T), Angular Velocity from Period
Orbital radius — Orbital Velocity, Centripetal Acceleration (a = v²/r)
Central mass — Orbital Velocity, Escape Velocity

Part of this formula set

Gravitation and orbital mechanics

8 formulasNewton's inverse-square law and what falls out of it in orbit — field strength, orbital speed and period, Kepler's third law, and escape velocity.