Orbital period calculator for planets, moons and satellites

The orbital period is the time an object takes to go once round its orbit. When an object circles a much heavier mass M at a distance r from its centre, Kepler's third law gives T = 2π√(r³ ÷ GM), where G = 6.674 × 10⁻¹¹ N m²/kg². A satellite 420 km above the Earth, 6,791 km from its centre, takes about 92.8 minutes. This orbital period calculator works out T from the radius and the central mass, or the radius from the period, with every step.

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Kepler's Third Law Calculator

Apply Kepler's third law of planetary motion

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How to use the orbital period calculator

1. Open the Kepler's Third Law Calculator and choose Solve for Time (the period) or Solve for Radius.
2. Type the mass of the central body, the one being orbited, as a number times a power of ten: 5.9722 and 24 for the Earth. Type 0 as the power if there isn't one. Mass can be in kg, g, tonnes, lb or solar masses (M☉).
3. For the period, type the orbital radius, measured from the centre of the central body, in m, km, miles or astronomical units (AU). For the radius, type the period in seconds, hours, days or years.
4. Read the answer in seconds or metres, with Also equal to listing it in hours, days and years, or km, miles and AU. The working shows r³ and T², and for the period it converts to years.
5. The graph plots period against orbital radius for your central mass, with your orbit marked. For practice, press Random problem.

The orbital period formula

It comes from setting the gravitational force equal to the centripetal force needed for a circular orbit, and it is Kepler's third law: T² is proportional to r³. The orbiting object's own mass cancels, so a tiny satellite and the Moon at the same distance would take the same time, as long as the orbiting object is much lighter than the central one. For an elliptical orbit, use the semi-major axis, half the longest width of the ellipse, as r. For the derivation and the ratio form, see the Kepler's third law guide.

Here x is the orbital radius in thousands of kilometres and y is the period in minutes, around the Earth. Orbits further out are slower in two ways: the path is longer and the speed is lower, so the period grows faster than the radius.

Worked examples you can check in the calculator

Example 1: the Earth around the Sun

The calculator opens on the Sun, 1.989 × 10³⁰ kg, and the Earth's distance, 1.496 × 10¹¹ m. It gives T = 3.155 × 10⁷ s, which it converts to 0.9999 years, or 365.2 days. Change the mass unit to M☉ and the distance unit to AU and type 1 and 1 (with 0 as each power of ten) to get the same answer.

Example 2: a satellite in low Earth orbit

A satellite orbits 420 km above the surface. Add the Earth's mean radius, 6,371 km, to get r = 6,791 km = 6.791 × 10⁶ m. With M = 5.9722 × 10²⁴ kg:

That is 92.8 minutes, so it goes round about 15 times a day.

Example 3: the radius for a one-day orbit

A geostationary satellite must go round once per sidereal day, the time the Earth takes to turn once: 23.9345 hours. Choose Solve for Radius, set the time unit to hours, and type 23.9345 with the Earth's mass. The calculator gives r = 42,164,279 m, about 42,164 km from the Earth's centre. Take away the Earth's equatorial radius, 6,378 km, and the satellite is 35,786 km above the equator.

Example 4: the Moon, and when the formula is off

The Moon is 3.844 × 10⁸ m from the Earth. With the Earth's mass the calculator gives 2.372 × 10⁶ s, or 27.45 days. NASA gives 27.3 days. The gap is because the Moon is not tiny compared with the Earth: about 1.2% of its mass. For two bodies the formula really uses their combined mass, M + m, and with 5.9722 + 0.073 = 6.045 × 10²⁴ kg it gives 27.29 days. For a satellite or a planet around the Sun, the difference is too small to matter.

Example 5: Mars, in AU and solar masses

Mars is 228.0 million km from the Sun, which is 1.524 AU. With 1 M☉ and 1.524 AU, the calculator gives 1.88 years (687.1 days); NASA lists 687.0 days. In years and AU around the Sun, the formula becomes T² = r³: 1.524³ = 3.54, and √3.54 = 1.88.

What the calculator doesn't do

• Orbital speed. Work it out from the answer: v = 2πr ÷ T. For Example 2, v = 2π × 6.791 × 10⁶ ÷ 5569 = 7,661 m/s.
• Altitude. It wants the radius from the centre. Add the planet's radius to a height above the surface first, and subtract it from the answer.
• Two comparable masses. It uses the central mass only. For a binary star or the Earth and Moon, add the two masses yourself and type the total.
• The mass from an orbit. Rearrange by hand: M = 4π²r³ ÷ (GT²).

Common mistakes

• Using the altitude as r. r is from the centre of the planet, not from its surface.
• Using the satellite's mass. M is the mass being orbited. The orbiting mass cancels out.
• Days or years in the formula. In SI the period must be in seconds. The calculator converts for you if you set the unit.
• Forgetting the square root. 4π²r³ ÷ GM is T², not T.
• Kilometres for metres. G is in SI units, so r must be in metres unless you set the unit.

Read next

Physics · Fields and gravityKepler's third law: formula, derivation and examplesKepler's third law links orbital period and radius. Get the formula, the A-level derivation, four worked examples and a free calculator.