Kepler's third law states that the square of an orbital period is proportional to the cube of the orbit's semi-major axis. Johannes Kepler found the relationship empirically in 1619 from Tycho Brahe's observations, without knowing why it held. Newton later showed it falls directly out of universal gravitation, and in doing so turned it from a pattern in a table of planets into a general rule that works for moons, satellites, exoplanets and binary stars alike. This Kepler's third law calculator solves the relationship for whichever of the three quantities you are missing, and tells you which form of the law it used.
Arb Digital builds free calculators that expose their method rather than hiding it. That matters here because the law has two common expressions that look completely different. The solar-system convenience form, T² = a³ ÷ M with years, astronomical units and solar masses, is compact and hides all the constants. The general form, T = 2π√(a³ ÷ GM), works for any central body but requires SI units throughout. Both are implemented below, both give the same answer, and the result panel names the one that produced your number.
What This Kepler's Third Law Calculator Does
Choose a unit system, choose which of the three quantities to solve for, fill in the other two, and read the answer. In solar-system mode the inputs are years, astronomical units and solar masses, which is how astronomy problems are usually posed. In general mode the inputs are seconds, metres and kilograms, which is what you need for a satellite around Earth or a moon around Jupiter.
Internally the tool always works in SI and always uses the general form, converting your solar-system inputs on the way in and back out again. That is a deliberate design choice: the convenience form is really the general form with the constants pre-divided out, so implementing one equation and converting units is both simpler and less error-prone than maintaining two separate code paths that could drift apart.
The results grid gives the period in days regardless of your input units, because days are the practical unit for most orbits; the mean orbital speed, computed as the circumference of a circular orbit of radius a divided by the period; the semi-major axis in kilometres; and the central mass in kilograms. Those last two make cross-checking against published figures straightforward without any manual conversion.
How to Use It
- Pick the unit system that matches your problem. Planets around the Sun are easiest in solar-system units. Anything else — a satellite, a moon, a star's companion — belongs in the general form.
- Choose what you are solving for. Period from orbit and mass, orbit from period and mass, or mass from period and orbit. The third case is how astronomers actually weigh distant objects.
- Enter the semi-major axis, not the altitude. For a satellite this is the planet's radius plus the mean altitude. Entering altitude alone is the single most common error on this page's subject.
- Use the sidereal period. That is one full orbit measured against the fixed stars. A synodic period, measured against a moving observer, is a different number.
- Check the results grid against a published value. If your answer for a known object is off by a large factor, the cause is almost always a units mistake or an altitude-for-radius substitution.
The Formula: How Kepler's Third Law Is Calculated
The general form is T = 2π√(a³ ÷ GM), where T is the orbital period in seconds, a is the semi-major axis in metres, M is the mass of the central body in kilograms and G is the gravitational constant. This tool uses G = 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻², the 2022 CODATA recommended value published by NIST. Rearranged, a = ∛(GM·T² ÷ 4π²) and M = 4π²·a³ ÷ (G·T²).
Newton's derivation is short enough to follow. For a circular orbit the gravitational force supplies the centripetal force: GMm/r² = mv²/r. The orbiting mass m cancels immediately, which is why it does not appear in the law. Substituting v = 2πr/T and rearranging gives T² = 4π²r³/(GM), which is Kepler's third law with the constant of proportionality made explicit. The result generalises from circles to ellipses with the semi-major axis replacing the radius, as NASA sets out in its overview of orbits and Kepler's laws.
The solar-system form drops out when you measure T in years, a in astronomical units and M in solar masses. All the constants collapse to one and the law reads simply T² = a³ ÷ M. For a planet orbiting the Sun, M = 1, so T² = a³ — Kepler's original statement. Mars sits at 1.524 AU, so its period is √(1.524³) = √3.539 = 1.881 years, which is 687 days. That matches the observed value, and you can reproduce it above with the Mars preset.
Work the general form too. A satellite in low Earth orbit at 420 km altitude has a semi-major axis of about 6,791 km, or 6.791 × 10⁶ m. Earth's mass is about 5.972 × 10²⁴ kg, so GM = 3.986 × 10¹⁴. Cube the axis: 3.132 × 10²⁰. Divide by GM: 7.857 × 10⁵. Take the square root: 886.4. Multiply by 2π: 5,569 seconds, or about 92.8 minutes. That is the familiar hour-and-a-half orbit of the International Space Station, derived from nothing but a mass and a distance.
Semi-Major Axis Is Not Radius, Altitude or Distance
Most errors with this law are geometric rather than arithmetic. The semi-major axis is half the longest diameter of the orbital ellipse, and it equals the average of the periapsis distance — closest approach — and the apoapsis distance at the far point. It is not the current distance between the two bodies, which changes continuously through the orbit, and it is not the closest approach.
For satellites, altitude is the number that appears in press releases and the number that must not go into this equation. Altitude is measured from the surface; the semi-major axis is measured from the centre. A satellite at 420 km altitude has a semi-major axis of 6,791 km, sixteen times larger. Using 420 km directly gives a period of about 5.7 minutes, an answer so wrong it is easy to spot — which is fortunate, because subtler versions of the same error are not.
One more nuance: a highly elliptical orbit has the same period as a circular orbit with the same semi-major axis, even though its shape is completely different. Eccentricity does not appear in Kepler's third law at all. It affects where the body is at any moment, which is Kepler's second law, but not how long the circuit takes.
Weighing the Universe With a Period and a Distance
Solving for M is the most scientifically interesting use of this law, because it is the primary way masses are measured across astronomy. You cannot put a star on a scale. You can, however, watch something orbit it, measure the period and the separation, and read the mass straight out of M = 4π²a³ ÷ (GT²).
This is how the mass of the Sun was first established, how the masses of planets with moons are determined, and how the masses of binary stars and exoplanet host stars are found today. It is also how the supermassive black hole at the centre of our galaxy was measured — by tracking individual stars through their orbits around it over decades and applying exactly this equation.
The one caveat is the two-body assumption. Strictly the law involves the sum of both masses, M₁ + M₂, and both bodies orbit their common centre of mass. When the orbiting object is far lighter than the central one, as with a planet around a star, the difference is negligible. For a binary star with two comparable masses, or for the Earth–Moon system where the Moon is a bit over one percent of Earth's mass, using only the larger mass introduces a real error, and the full form is required.
Why the Orbiting Body's Mass Cancels Out
This surprises people, and it is the same fact Galileo demonstrated with falling objects. Gravitational force is proportional to the orbiting mass, and the acceleration produced by a force is inversely proportional to that same mass, so the two dependencies cancel exactly. A one-kilogram object and a ten-tonne object at the same altitude follow identical orbits.
The practical consequences are large. Satellite orbits can be planned without knowing the spacecraft's mass. A tool bag dropped by an astronaut stays in a nearly identical orbit rather than falling away. And the crew of a space station float not because gravity is absent but because they and the station accelerate identically — a point covered further in the gravitational force calculator.
The cancellation fails when a non-gravitational force is significant. Atmospheric drag in low orbit depends on cross-sectional area relative to mass, so a light, bulky object decays faster than a dense, compact one at the same altitude. Radiation pressure has a similar mass-dependent effect on very light objects. Kepler's third law describes pure two-body gravitation and knows nothing about either.
Reading the Orbital Speed Figure
The grid reports mean orbital speed as 2πa ÷ T, which is exact for a circular orbit and a good average for a mildly elliptical one. For an eccentric orbit the instantaneous speed varies considerably — fastest at periapsis, slowest at apoapsis — because equal areas are swept in equal times.
Two relationships are worth carrying away from that number. Orbital speed falls as the square root of distance, so a satellite twice as far out moves about 41 percent slower and takes far longer per orbit for both reasons. And escape velocity at any radius is exactly √2 times the circular orbital speed there, which the escape velocity calculator works directly. For the circular-motion mechanics behind these figures, see the centripetal force calculator and the angular velocity calculator. If you need to move a period between hours, days and years, the time converter handles that, and the scientific notation converter is useful for the very large numbers this subject generates. Everything else is on the free tools hub.
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Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Entering altitude instead of semi-major axis — add the central body's radius. For low Earth orbit the difference is a factor of sixteen.
- Mixing unit systems — years with metres, or kilograms with astronomical units, produces a confidently wrong answer. Set the unit system first, then fill the fields.
- Using T² = a³ for a non-solar system — that simplified form assumes a central mass of one solar mass. Around any other body the general GM form is required.
- Using a synodic period — the law needs the sidereal period, measured against the fixed stars rather than against a moving observer such as Earth.
- Ignoring the second mass in a binary — the law strictly uses the sum of both masses, which matters whenever the two are comparable.
Related Free Tools From Arb Digital
Kepler's third law comes straight out of Newtonian gravitation, which the gravitational force calculator handles directly. For the speed needed to leave an orbit entirely, use the escape velocity calculator; for the force holding a body in a circular path, the centripetal force calculator; and for rotation rates, the angular velocity calculator. Practical conversions live in the time converter and the scientific notation converter. The complete set is on the free online tools hub.
Frequently Asked Questions
It always computes in SI using the general form, T equals two pi times the square root of a cubed divided by GM. Solar-system inputs are converted in and out around that core, and the result panel names which unit system produced your answer.
Because gravitational force is proportional to that mass and the acceleration it produces is inversely proportional to it, so the two cancel exactly. A light satellite and a heavy one at the same semi-major axis have the same period.
It is half the longest diameter of the orbital ellipse, equal to the average of the closest and farthest distances from the central body. For a circular orbit it is simply the radius. It is never the altitude above a surface.
No. Kepler's third law contains no eccentricity term, so a circular orbit and a highly elliptical one with the same semi-major axis have identical periods. Eccentricity changes where the body is at a given moment, not how long the circuit takes.
Yes, and that is how such masses are actually measured. Solve for the central mass from an observed period and semi-major axis. The result is strictly the sum of both masses, which is a good approximation whenever the orbiting body is much lighter.
Because that simplified statement assumes years, astronomical units and a central mass of exactly one solar mass. Change the central body and the constant of proportionality changes with it, so the general GM form is needed.
No. Kepler's third law describes pure two-body gravitation. Drag in low orbit depends on a spacecraft's area-to-mass ratio and steadily reduces the semi-major axis, which the law itself cannot represent.
This tool is provided for educational and study use. It applies an idealised two-body Newtonian model and is not intended for mission planning, navigation or any operational orbital determination.
