Time Period of Satellite Formula: T = 2π√(r³/GM) Explained

The time period of satellite formula is $T = 2\pi\sqrt{\frac{r^3}{GM}}$ , where r is the radius of the circular orbit measured from the planet's centre and M is the planet's mass. With r = R + h for a satellite at height h, it reads $T = 2\pi\sqrt{\frac{(R + h)^3}{GM}}$ .

Only two things set the period: how big the orbit is and how massive the planet is. The satellite's own mass plays no part. Squaring gives $T^2 = \frac{4\pi^2}{GM}r^3$ , which is Kepler's third law for circular orbits: the square of the period is proportional to the cube of the orbit radius. Read this with the orbital velocity formula, since the two come from the same balance of forces.

Quick facts

Formula
$T = 2\pi\sqrt{\frac{r^3}{GM}}$
Near surface
$T = 2\pi\sqrt{\frac{R}{g}} \approx 84$ min
Kepler form
$T^2 \propto r^3$
From speed
$T = \frac{2\pi r}{v_o}$
SI unit
second (s), often converted to minutes or hours
Geostationary
T = 24 h at r ≈ 4.22 × 10⁷ m

Where the time period of satellite formula comes from

Orbital speed for a circular path: $v_o = \sqrt{\frac{GM}{r}}$ .
One lap covers the circumference $2\pi r$ , so $T = \frac{2\pi r}{v_o}$ .
Substitute: $T = 2\pi r\sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}$ .
Using $GM = gR^2$ : $T = \frac{2\pi}{R}\sqrt{\frac{(R+h)^3}{g}}$ . For h much less than R, $T \approx 2\pi\sqrt{\frac{R}{g}}$ .

The last form is the same shape as the period of a pendulum, which is a neat link students sometimes meet in objective questions.

Worked examples

Example 1: the shortest Earth orbit

Find the period of a satellite grazing Earth's surface, starting from g at ground level and Earth's radius.

$T = 2\pi\sqrt{\frac{6.37 \times 10^6}{9.8}} = 2\pi \times 806 \approx 5.07 \times 10^3\ \text{s}$

That is about 84 minutes, the fastest any satellite can circle Earth.

Example 2: orbit 400 km high

Find the period at h = 400 km.

r = 6.77 × 10⁶ m, so r³ ≈ 3.10 × 10²⁰ m³, and GM ≈ 3.98 × 10¹⁴ m³/s².

$T = 2\pi\sqrt{\frac{3.10 \times 10^{20}}{3.98 \times 10^{14}}} = 2\pi \times 882 \approx 5.54 \times 10^3\ \text{s}$

About 92 minutes, so roughly 15 orbits a day.

Example 3: geostationary check

Show that r = 4.22 × 10⁷ m gives a period of about one day.

r³ ≈ 7.52 × 10²² m³. $T = 2\pi\sqrt{\frac{7.52 \times 10^{22}}{3.98 \times 10^{14}}} \approx 2\pi \times 1.37 \times 10^4 \approx 8.63 \times 10^4\ \text{s}$ .

Dividing by 3600 gives about 24.0 h. The calculator above solves for r if you enter T = 86400 s and Earth's mass.

Kepler's third law and weighing a planet

Rearranged, the formula gives the mass of the central body: $M = \frac{4\pi^2 r^3}{GT^2}$ . This is how astronomers find the mass of a planet from its moon's orbit.

Example 4: Earth's mass from the Moon

The Moon orbits at 3.84 × 10⁸ m with T = 27.3 days. Estimate Earth's mass.

27.3 days is 27.3 × 24 × 3600 s, which is about 2.36 × 10⁶ s. $M = \frac{39.48 \times (3.84 \times 10^8)^3}{6.674 \times 10^{-11} \times (2.36 \times 10^6)^2} \approx 6.0 \times 10^{24}\ \text{kg}$ .

Close to the accepted 5.97 × 10²⁴ kg.

Ratio questions are common: if r grows by a factor k, T grows by $k^{3/2}$ . Quadruple the radius and the period rises eight times. The general period formula page covers T for waves and oscillations.

Geostationary orbits

A geostationary satellite circles above the equator, in the same direction as Earth's spin, with a period equal to one rotation of Earth. It therefore stays over one spot, which suits television and weather work. Only one radius gives this period, about 42 200 km from Earth's centre, or roughly 35 800 km above the surface.

Strictly, Earth turns once in about 23.93 h relative to the stars, so the exact geostationary radius is slightly smaller, about 4.216 × 10⁷ m. School problems use 24 h.

Errors to avoid

Using the height h instead of r = R + h.
Forgetting to cube r, or cubing it before converting km to m.
Thinking a heavier satellite has a different period. Its mass cancels.
Leaving the answer in seconds when the question asks for hours.

For the force behind it all, see the gravitational force hub, and for leaving orbit altogether, the escape velocity page.

Frequently asked questions

What is the time period of a satellite formula?

T = 2π√(r³ / GM), with r measured from the middle of the planet, so r = R + h. Just above the ground T ≈ 2π√(R / g), close to 84 minutes for Earth.

Does the period of a satellite depend on its mass?

No. Only the orbit radius and the planet's mass matter, so a small and a large satellite at the same height share one period.

How does the period change with height?

T is proportional to r to the power 3/2, so higher satellites take much longer. A geostationary satellite at about 36 000 km height takes 24 h.

How are the period and the orbital speed linked?

T = 2πr / vₒ. The satellite covers one circumference at its orbital speed in one period.