Kepler’s Third Law Calculator (period ⇄ semi-major axis)

Understanding the Kepler's third law calculator — concept and principle

Kepler's third law links the size of an orbit to the time it takes to go around once. Given the semi-major axis of an orbit in astronomical units (AU) and the mass of the central body in solar masses, this calculator returns the orbital period in years and days. It is handy for planets, asteroids, comets, moons around a planet expressed in solar units, and exoplanets around other stars.

An orbit 1.524 times wider takes 1.88 times as long — Mars’s 687-day year.

How it is calculated

T² = a³ ÷ M, so T = √(a³ ÷ M).

T is the period in years.
a is the semi-major axis (half the longest diameter of the ellipse) in AU; 1 AU is the mean Earth–Sun distance, defined as exactly 149,597,870,700 m.
M is the central mass in solar masses.

The period in days is T × 365.25, using the Julian year.

The principle

Johannes Kepler found from Tycho Brahe's observations, and published in 1619, that the square of each planet's period is proportional to the cube of its mean distance from the Sun. Newton later derived it from his law of gravitation: for a circular orbit, gravity provides the centripetal force, GM ÷ r² = (2π ÷ T)² r, which gives T² = 4π² r³ ÷ GM. The same result holds for ellipses when r is replaced by the semi-major axis a.

Choosing years, AU and solar masses makes the constant 4π² ÷ G equal to 1 for the Earth around the Sun, leaving the tidy form T² = a³ ÷ M. Strictly, M is the sum of both masses, but for planets around a star the smaller mass is negligible.

Worked example

Inputs: a = 1.524 AU (about Mars's orbit), M = 1 solar mass.
a³ = 1.524³ = 3.539606.
T = √(3.539606 ÷ 1) ≈ 1.881384018 years.
In days: 1.881384018 × 365.25 ≈ 687.1755127 days.

Each point is the calculator re-run with only “Semi-major axis (AU)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Semi-major axis (AU)	Period (years)
0.381	0.2351730023
0.762	0.6651696986
1.14	1.221994766
1.52	1.881384018
1.91	2.629314098
2.29	3.456323141
3.05	5.321357589

How to read the result

The result is close to Mars's measured sidereal period of about 687 days. Because T grows as a^1.5, an orbit twice as large takes 2.83 times as long: Jupiter at 5.2 AU needs about 11.86 years, while Mercury at 0.387 AU needs about 88 days. A heavier central star shortens the period for the same orbit size; around a star of 4 solar masses, a 1 AU orbit takes only half a year.

Common mistakes

Entering the average distance of a strongly elliptical orbit instead of the semi-major axis.
Entering kilometres instead of AU, or kilograms instead of solar masses.
Expecting the synodic period (time between oppositions as seen from Earth) instead of the sidereal period.

Keep in mind: This is the two-body law. Perturbations from other planets, a companion mass that is not negligible, or relativistic effects change real periods slightly, so precise ephemerides use detailed numerical models.

Frequently asked questions

Can I use this for moons around Jupiter?

Yes, if you express Jupiter's mass in solar masses (about 0.000955) and the orbit in AU; the formula is the same.

How do I find the semi-major axis from a period?

Rearrange to a = (M × T²)^(1/3). An object with a 2-year period around the Sun has a ≈ 1.587 AU.

Does the shape of the orbit change the period?

No. Orbits with the same semi-major axis have the same period, whether circular or very elongated.

Sources: Johannes Kepler, Harmonices Mundi (1619) · IAU 2012 Resolution B2 (definition of the astronomical unit) · Halliday, Resnick and Walker, Fundamentals of Physics, chapter on gravitation

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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