Orbital Period Calculator

For a small body orbiting a much larger mass, the period T satisfies T = 2 x pi x sqrt(a^3 / (G x M)), where a is the semi-major axis, M is the central mass and G is the gravitational constant. The period grows with the three-halves power of the orbit size.

It is half the longest diameter of the elliptical orbit and sets the orbit's overall size. For a circular orbit it equals the orbit radius, measured centre to centre between the two bodies.

In this simplified form it does not. The equation assumes the central mass dominates, as with a satellite around Earth or a planet around the Sun, so the small body's own mass drops out of the result.

Meters for the semi-major axis and kilograms for the central mass, the SI units that pair with the built-in gravitational constant. The period comes back in seconds and is also converted to days and Julian years.

Yes. The fields accept forms like 1.496e11 and 5.972e24, which is the practical way to type astronomical distances and masses. Very large or small results are shown in scientific notation too.

Using Earth's average orbital radius of 1.496e11 m and the Sun's mass of 1.989e30 kg, the formula returns about 3.156e7 seconds, which is roughly 365 days, essentially one year as expected.

The gravitational constant is G = 6.674e-11 N m^2/kg^2. Years are Julian years of exactly 365.25 days, so the days and years cards stay consistent with each other.

Yes, as long as you supply the semi-major axis rather than the closest or farthest distance. Kepler's third law depends only on the semi-major axis, so eccentricity does not change the period.

No. The whole calculation runs locally in your browser, so the orbit values you enter are never uploaded.