Orbital Period Calculator

Calculate the orbital period of a body from its semi-major axis using Kepler's third law, for any central mass in solar masses.

Orbital period1 years
In days
365.3

Kepler's third law: T² ∝ a³ / M (T in years, a in AU, M in solar masses).

What the Orbital Period Calculator Does

This orbital period calculator estimates how long an object takes to complete one full orbit around a central body, using Kepler's third law. You enter the orbit's average distance (semi-major axis) and the mass of the body being orbited, and it returns the period.

It is built for astronomy students, amateur stargazers, physics teachers, and anyone curious about how planets, moons, asteroids, or exoplanets move. Because it works in solar-system units, it is especially handy for quick checks on planets orbiting the Sun or stars of known mass.

How It Works: Kepler's Third Law Formula

Kepler's third law states that the square of an orbital period is proportional to the cube of the orbit's semi-major axis. When you use convenient astronomical units, the proportionality constant becomes 1, which simplifies the math considerably.

The formula this calculator uses is:

T = sqrt(a^3 / M)

where T is the orbital period in years, a is the semi-major axis (average orbital distance) in astronomical units (AU), and M is the mass of the central body in solar masses. One AU equals the average Earth-Sun distance (about 149.6 million km), and one solar mass equals the mass of the Sun.

Why Earth Equals One Year

The units are chosen so that Earth's orbit acts as a built-in reference point. Earth orbits at a = 1 AU around the Sun, which has a mass of M = 1 solar mass.

Plugging those in: T = sqrt(1^3 / 1) = sqrt(1) = 1 year. This confirms the formula and gives you an intuitive baseline. Any orbit farther out than Earth's takes longer than a year; any orbit closer in takes less.

Worked Example: Mars and Jupiter

Consider Mars, which orbits the Sun at an average distance of about 1.524 AU. With M = 1 solar mass:

T = sqrt(1.524^3 / 1) = sqrt(3.54) = 1.88 years. That matches Mars's real orbital period of roughly 687 days.

Now try Jupiter at a = 5.20 AU: T = sqrt(5.20^3 / 1) = sqrt(140.6) = 11.86 years. This is very close to Jupiter's measured period of about 11.86 years, showing how reliable the law is across very different orbits.

Tips, Common Mistakes, and What Affects the Result

The formula assumes the orbiting object's mass is tiny compared with the central body, so it works best for planets, moons, asteroids, and small probes. Keep these points in mind:

  • Use the semi-major axis, not the perihelion or aphelion distance. For an elliptical orbit, a is the average of the closest and farthest points.
  • Match your units. This version needs AU and solar masses; mixing in kilometers or kilograms without converting will give wrong answers.
  • Set M correctly. For anything orbiting the Sun, M = 1. For a moon orbiting a planet, or a planet orbiting another star, use that body's mass in solar masses instead.
  • Eccentricity does not change the period. A circular and a stretched ellipse with the same semi-major axis share the same orbital period.
  • Results are idealized. They ignore the small pull of other bodies (perturbations) and relativistic effects, so real periods can differ slightly.