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Orbital Period Calculator

Find how long an orbit takes around any mass.

How it is calculated

The tool applies Kepler's third law in its Newtonian form: T = 2π × √(a³ / (G × M)). The semi-major axis a must be entered in metres and the central mass M in kilograms, with G = 6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻². The period comes out in seconds in scientific notation, then in days (÷ 86,400) and in years (÷ 31,557,600, a 365.25-day year). The orbiting body's own mass is ignored, which is accurate when it is much lighter than the central body. Both inputs must be above zero.

Worked example

A body orbiting Earth (5.972e24 kg) at a distance of 384,400,000 m. Cube the axis and divide by G × M: a³ ≈ 5.680e25, G × M ≈ 3.986e14, ratio ≈ 1.425e11. Square root ≈ 377,500 s, times 2π gives 2.3719 × 10⁶ s. In days that is 27.45 days, or 0.0752 years.

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Common mistakes

  • Entering the axis in kilometres or astronomical units. It must be metres, so 384,400 km is 384400000.
  • Using the orbit's altitude above the surface. The axis is measured from the centre of the central body, so add its radius.
  • Using the orbiting object's mass instead of the central mass.
  • Typing mass without the exponent. Use scientific notation such as 5.972e24 for planets and stars.

Sources

About this calculator

Kepler's third law, in its full Newtonian form, gives the orbital period of any body orbiting a central mass as T = 2π√(a³ / GM), where a is the semi-major axis of the orbit, G is the gravitational constant, and M is the mass of the body being orbited. This works for anything from a satellite circling Earth to a planet circling the Sun to a moon circling a planet — only the central mass and orbital distance matter, not the orbiting body's own mass (as long as it's much smaller than M).

Astronomers and aerospace engineers use this formula to plan satellite orbits (a geostationary satellite is placed at the specific altitude that gives a period of one sidereal day, about 23 h 56 min, so that it keeps pace with Earth's rotation), to determine the masses of stars and planets by observing the orbital periods of their companions, and to predict how long a spacecraft transfer orbit or a newly discovered exoplanet's "year" will last.

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Comments (1)

  • Editorial team

    A quick check is a low Earth orbit. At 400 km altitude the semi-major axis is Earth's mean radius plus altitude, 6,371,000 + 400,000 = 6,771,000 m, which gives about 5,545 s, roughly 92 minutes. Putting in 42,164,000 m returns about 8.6165 × 10⁴ s, the geostationary day. If you only know an altitude, add the body's radius first. Which orbit or planet are you looking at?

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