Kepler's Third Law Calculator

The law, checked against the real solar system

Run each planet's measured orbital period backward through the law, a = T to the power 2/3, and out comes the distance its orbit must have. The periods below are the same sidereal values used across this site, and the last column is the mean distance printed on each planet's fact page.

Earth's row is exact by definition, since both the year and the astronomical unit are calibrated on its orbit. Eight of the nine land on their fact-page figure at its stated rounding. Pluto misses by a hair, 39.466 implied against 39.48 measured, about three hundredths of a percent, and the miss is genuine rather than a rounding accident. The law is exact only for two bodies, and the solar system is a crowd; the accuracy note below has the details.

Worked examples

Each follows the exact arithmetic the calculator runs, with results shown to 4 significant figures.

Mars around the Sun

Mars orbits at a semi-major axis of 1.524 AU. Cubing 1.524 gives 3.540, and the square root of that is 1.881, so Mars takes 1.881 years, about 687 days. Its fact page lists 687.0 days.

Jupiter, solved in reverse

Jupiter's period is 11.862 years. Squaring gives 140.71, and the cube root of that is 5.201, so the law puts Jupiter at 5.201 AU. Its fact page lists 5.20 AU. The third decimal hides a real effect, covered in the accuracy note above, because Jupiter's own mass is large enough to shift the answer by a few hundredths of a percent.

A planet around a red dwarf

Take a planet at 0.05 AU, closer in than Mercury, around a red dwarf of half the Sun's mass. With the mass set to 0.5, the law gives the square root of 0.05 cubed over 0.5, which is 0.01581 years, or about 5.8 days. Compact orbits around small stars run fast, which is why so many known exoplanets have periods of days rather than years.

The International Space Station

The station flies about 420 km up. Adding Earth's 6,371 km mean radius gives a semi-major axis of 6,791 km, and the full formula returns 5,569 seconds, about 93 minutes per orbit. The crew circles Earth more than fifteen times a day.

A geostationary satellite

A satellite that hovers over one spot must match Earth's rotation against the stars. Enter a semi-major axis of 42,164 km and the answer is 86,164 seconds, which is 23 hours 56 minutes, the sidereal day rather than the 24 hour solar day. The figure is often quoted as an altitude of 35,786 km; that is height above the equator, where Earth's radius is 6,378 km, and the two add to the same 42,164 km from the center.

The Moon, and why it misses

At a mean distance of 384,400 km, the formula returns 27.45 days. The true sidereal month is 27.32 days, and the gap is the law being honest about its assumptions. The plain formula treats the satellite as weightless next to its planet, but the Moon carries about 1.2 percent of Earth's mass, and the proper two-body form divides by the sum of both masses. Folding that in brings the answer to about 27.29 days; the Sun's steady pull on the Moon's orbit stretches the real month the rest of the way to 27.32.

Frequently asked questions

What is Kepler's third law?

The square of a body's orbital period is proportional to the cube of its orbit's semi-major axis, the long radius of its ellipse. Larger orbits are slower twice over, since the path is longer and the body also moves more slowly along it. For planets around the Sun, with periods in years and distances in astronomical units, the rule becomes period squared equals distance cubed, with no constant to remember.

What units make T^2 = a^3 work?

Years and astronomical units, for anything orbiting the Sun. Both units are defined by Earth's own orbit, so Earth calibrates the constant to exactly 1. In any other units, or around any other central body, the full form applies, with the period equal to 2 pi times the square root of the semi-major axis cubed divided by GM, the central body's gravitational parameter.

How do I use it for satellites?

Switch the calculator to Around Earth and enter either the altitude above the surface or the semi-major axis measured from Earth's center. An altitude has Earth's mean radius of 6,371 km added to it, and the period follows from the full formula with Earth's GM of 398,600 cubic kilometers per second squared. The International Space Station, at about 420 km up, comes out near 93 minutes.

Why does a geostationary orbit take 23 h 56 m and not 24 h?

A geostationary satellite has to match Earth's rotation with respect to the stars, and Earth turns once in 23 hours 56 minutes 4 seconds, the sidereal day. The familiar 24 hour day is slightly longer because Earth also advances along its orbit each day and needs about four extra minutes of turning to face the Sun again. Enter a semi-major axis of 42,164 km and the calculator lands on the sidereal day.

Does the law work for exoplanets?

Yes, once the central star's mass is accounted for, since the constant in the law encodes that mass. Set the mass in solar masses and the calculator solves period equals the square root of distance cubed over mass. This is how orbit sizes are found in practice, because a transit or wobble measurement gives the period directly, the star's mass comes from its spectrum, and the law hands back the distance.

Is the law exact?

Only for two idealized bodies, one orbiting the other with nothing else around. Real planets tug on one another, and a planet's own mass slightly strengthens the attraction the law attributes to the Sun alone, so measured orbits miss the pure rule by small fractions of a percent. In the table above the largest gap belongs to Pluto, whose implied 39.466 AU sits about three hundredths of a percent from the measured 39.48.

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