Calculateus

Kepler's Third Law (Orbital Period) Calculator

Calculate the orbital period of a body from its semi-major axis and the mass of the object it orbits.

Result

Orbital Period
0.9999 years
In Days
365.2 days

About the Kepler's Third Law Calculator

The Kepler's Third Law Calculator finds how long it takes an object to complete one orbit around a central body, based on the size of its orbit and the central body's mass. It applies the full, mass-dependent version of Kepler's third law derived from Newtonian gravity, not just the simplified ratio form used for comparing planets within the same solar system. Enter the mass of the object being orbited and the orbit's semi-major axis in astronomical units to get the orbital period in years and days.

How It Works

You provide two inputs: the mass of the central body in kilograms (with reference values given for the Sun and Earth) and the semi-major axis of the orbit in astronomical units (AU), where 1 AU is the Earth-Sun distance. The calculator converts the semi-major axis to meters, then applies the formula derived from Newton's law of gravitation to compute the orbital period in seconds, before converting that into both years and days for easier reading.

Period = 2 pi x sqrt(semi-major axis in meters cubed / (G x central mass in kg)), where G = 6.6743 x 10^-11 N m^2/kg^2. Semi-major axis is converted from AU using 1 AU = 1.495978707 x 10^11 m.

Formula & Methodology

To reproduce this by hand, first convert the semi-major axis from AU to meters by multiplying by 1.495978707 x 10^11. Cube that distance, divide by the product of the gravitational constant G (6.6743 x 10^-11) and the central mass in kilograms, and take the square root of the result. Multiply that by 2 pi to get the orbital period in seconds. From there, dividing by 86,400 gives the period in days, and dividing by 365.25 x 86,400 gives the period in years.

Examples

Earth's orbit around the Sun

Using the Sun's mass of 1.989 x 10^30 kg and a semi-major axis of 1 AU, the calculator returns an orbital period of almost exactly 1.0000 years (about 365.3 days), matching Earth's known orbital period since these are Earth's own real orbital parameters.

A hypothetical moon at 1 AU from Earth

Swapping the central mass to Earth's roughly 5.972 x 10^24 kg while keeping the same 1 AU semi-major axis gives a vastly longer period, since a much smaller central mass provides much weaker gravitational pull to hold an object in such a wide orbit, illustrating how central mass and period trade off in the formula.

Advantages

  • Uses the full physical form of Kepler's third law with an explicit gravitational constant and central mass, rather than a simplified ratio that only works when comparing bodies orbiting the same star.
  • Handles any central mass, from planets to stars, making it usable for solar system objects, exoplanet orbits, or hypothetical scenarios alike.
  • Reports the result in both years and days, matching how astronomical periods are typically discussed and compared.

Common Mistakes

  • Entering the central mass in the wrong units or forgetting the exponent, since these masses are typically expressed in scientific notation like 1.989 x 10^30 kg and a missing exponent changes the answer by many orders of magnitude.
  • Confusing the semi-major axis with the average or minimum orbital distance; for an elliptical orbit, the semi-major axis is specifically half the longest diameter of the ellipse, not the perigee or apogee distance.
  • Applying this formula to a binary system of two similar-mass bodies without accounting for both masses, since the derivation assumes the central body's mass dominates.

Edge Cases to Watch For

  • Both the central mass and the semi-major axis must be greater than zero; the calculator returns an error for zero or negative values in either field, since neither has physical meaning at or below zero for this formula.
  • This calculator assumes the orbiting body's own mass is negligible compared to the central body's mass, which holds well for planets orbiting stars or moons orbiting planets, but breaks down for binary systems of comparable mass, where both masses need to be combined.
  • The formula assumes a two-body system with no other significant gravitational influences, ignoring perturbations from other planets or bodies that would alter a real orbit's period over time.

Common Use Cases

  • Astronomy students and educators working through orbital mechanics problems involving Kepler's laws.
  • Space enthusiasts estimating the orbital period of a known exoplanet, moon, or hypothetical satellite from its orbital distance.
  • Anyone comparing how orbital period changes with distance from a star or with the mass of the central body.
Written & fact-checked by the Calculateus TeamLast updated August 5, 2026How we verify our formulas

Frequently asked questions

What does Kepler's Third Law describe?

It relates a body's orbital period to the size of its orbit: the square of the period is proportional to the cube of the semi-major axis (T² ∝ a³), with the constant of proportionality depending on the mass being orbited. It's how astronomers calculate orbital periods for planets, moons, and exoplanets from their orbital distance alone.

Related calculators

Ohm's Law CalculatorCalculate voltage, current, resistance, or power using Ohm's Law - solve for whichever value is missing.Projectile Motion CalculatorCalculate the range, maximum height, and flight time of a projectile launched at an angle.Free Fall & Kinematics CalculatorCalculate the distance fallen and final velocity of an object in free fall after a given time.Force Calculator (Newton's Second Law)Calculate force, mass, or acceleration using Newton's second law of motion (F = ma).

Conclusion

The Kepler's Third Law Calculator turns Newton's derivation of Kepler's law into a direct period calculation for any central mass and orbital distance. It is a practical way to see how gravity, mass, and distance combine to set the rhythm of an orbit.