Orbital period of a circular orbit

Time for one circuit of a small body in a circular orbit around a much larger mass, in a Newtonian gravitational field.

Ref: FPH-51
Rev: v1
Field: Foundations: Physics
Result: T [s]

T = 2 π \sqrt{\frac{r^{3}}{G M}}

For a small body in a circular orbit the gravitational pull supplies the centripetal force, which gives T = 2π√(r³/GM), where G = 6.674 × 10⁻¹¹ N m²/kg², M is the mass of the central body and r the distance from its centre (not the height above the surface). The period does not depend on the mass of the orbiting body. It contains Kepler’s third law: T² is proportional to r³.

Time for one circuit of a small body in a circular orbit around a much larger mass, in a Newtonian gravitational field. The radius is measured from the centre of the central body.

Calculator

Solve for

TOrbital periodComputed

MMass of the central body

1e201e31

rOrbital radius from the centre

100,0001e9

Orbital speed, 2πr/T7,672 m/s
Orbits per day15.58
Period in minutes92.42 min

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Orbital period (s) current value 5,545 s

rOrbital radius from the centreKey driver

9.959,950,000

MMass of the central body

4.291,360,000

At 6,771 km from the centre of the Earth the period is 92.42 minutes (a low Earth orbit), so the satellite circles the planet about 15.6 times a day at 7.67 km/s.
Doubling the radius multiplies the period by 2^1.5 = 2.83; a geostationary orbit, with a period of one sidereal day, sits at about 42,164 km from the centre.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (s)	Local step	Change in result
r Orbital radius from the centre	100,000 to 1e9 m	9.952	9,952,000	9,952,000	+10%	+15.4 %
M Mass of the central body	1e20 to 1e31 kg	4.285	1,355,000	1,355,000	+10%	−4.65 %

Sweep of Orbital radius from the centre

Orbital radius from the centre (m)	Orbital period (s)
100,000	9.952
316,200	55.97
1,000,000	314.7
3,162,000	1,770
1e7	9,952
3.162e7	55,970
1e8	314,700
3.162e8	1,770,000
1e9	9,952,000

Sweep one variable

Vary

Orbital period (s)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Orbital period

Result · s

What the equation gives you. Everything else feeds it.

Mass of the central body

kg · typical 1e20 to 1e31

The mass of the body being orbited, in kg. For the Earth it is 5.972 × 10²⁴ kg. Shortens the period with the inverse square root of M: a stronger pull needs less time per circuit. Raising it by 1×10²⁴ (to 6.972×10²⁴) moves the result by −7.45 %, from 5,545 to 5,132 s.

Orbital radius from the centre

m · typical 100,000 to 1e9

The orbital radius measured from the centre of the central body, in m. For a low Earth orbit add the altitude to the Earth’s radius of 6,371 km. Lengthens the period with the 1.5 power of r: a higher orbit is both longer and slower. Raising it by 500,000 m (to 7,271,000 m) moves the result by +11.3 %, from 5,545 to 6,170 s.

Worked example with the values above

Cube the radius: r³ = 6,771,000³ = 3.104×10²⁰ m³; and GM = 6.6743×10⁻¹¹ × 5.972×10²⁴ = 3.986×10¹⁴ m³/s².
Divide and take the square root: √(3.104×10²⁰ / 3.986×10¹⁴) = 882.5 s.
T = 2π × 882.5 = 5,545 s, which is 92.42 minutes.

Assumptions and limits

A circular orbit about a much larger central mass, with only Newtonian gravity acting.
The orbiting body is small compared with the central mass; for comparable masses use the total mass.

Common mistakes

Using the altitude above the surface instead of the distance from the centre.
Using the orbiting body’s mass in place of the central mass.
Mixing km with m: the radius must be in metres when G is in SI.

Statistics and uncertainty

Because T depends on r^1.5, a 1 % error in the radius gives about a 1.5 % error in the period, while the same error in the mass gives only 0.5 %. The radius is therefore the quantity to know best; remember it includes the body’s own radius.

Where it is used

Setting the period of a satellite orbit or a planetary year.
Finding the central mass from an observed orbit.

Maths you need

Newton’s law of gravitation: the pull falls with the square of distance
Centripetal force: the inward force that keeps a body on a circle

Related formulas

Practise this formula