Orbital Period Calculator

A satellite, a moon, and a planet all obey the same rule when they orbit something bigger: how long one lap takes comes down to two things only, how far out the orbit sits and how heavy the central body is. Johannes Kepler worked this out in the early 1600s from raw planetary data, and Newton later showed why it holds using his law of gravity. Give this tool an orbital radius and a central mass and it returns the period, whether you're checking how long the ISS takes to loop the Earth, how long Mars takes to circle the Sun, or how high to park a new satellite.

What is Orbital Period?

The orbital period is simply the time for one full lap. What makes it predictable is Kepler's Third Law of Planetary Motion: square the period, cube the radius, and the two stay locked in a fixed ratio for everything orbiting the same body. Written out, the law looks like this:

P = 2\pi \sqrt{\frac{R^3}{GM}}

Where $P$ is the orbital period, $R$ is the orbital radius (distance from the center of the central body), $G$ is the gravitational constant ( $6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2}$ ), and $M$ is the mass of the central body. Notice what's missing: the mass of the orbiting object never shows up. A tiny cubesat and the much heavier ISS at the same altitude share the exact same period.

How to Use This Calculator

Fill in any two of the three values and read the third. Enter a radius and a central mass to get the period, or enter a period and a central mass to back out the radius the orbit needs. Each field takes the units you'd actually use: radius in meters, kilometers, or astronomical units; mass in kilograms, Earth masses, or solar masses; period in seconds, minutes, hours, days, or years. Change any input and the result follows.

Understanding the Formula

Take the International Space Station. It orbits about 400 km above Earth's surface, but the formula needs the distance from Earth's center, so add Earth's radius (6,371 km) to get R = 6,771,000 meters. Earth's mass is $5.972 \times 10^{24}$ kg. Plugging these in:

P = 2\pi \sqrt{\frac{(6{,}771{,}000)^3}{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}}

First, calculate $R^3 = (6.771 \times 10^6)^3 \approx 3.104 \times 10^{20}$ . Then $GM = 6.674 \times 10^{-11} \times 5.972 \times 10^{24} \approx 3.986 \times 10^{14}$ . Divide to get $\frac{R^3}{GM} \approx 778{,}725$ . Take the square root: $\sqrt{778{,}725} \approx 882.5$ seconds. Multiply by $2\pi$ : P = 5,545 seconds, or about 92.4 minutes. The real ISS comes around roughly every 92 minutes, so the math checks out.

Real-World Applications

Satellite engineers run the formula backward all the time, picking an altitude to hit a target period. A geostationary satellite has to match Earth's 24-hour spin to hover over one spot, which pins it at about 35,786 km up. GPS satellites sit lower, near 20,200 km on a 12-hour orbit that keeps enough of them in view from anywhere on the ground. Astronomers flip it the other way: time how long an exoplanet takes to cross its star, and the same relationship hands back the orbital distance. Mission planners lean on it too, for transfer orbits and for working out when two spacecraft will actually meet.

Tips for Accurate Calculations

Measure the radius from the center of the central body, not its surface. For an Earth satellite that means adding Earth's radius, 6,371 km, to the altitude. The formula assumes a circular orbit; for an elliptical one, swap in the semi-major axis (the average of the closest and farthest points) for R. It also assumes the orbiting object is light enough to ignore next to the central body, which holds for satellites and moons. For a binary star, where the two masses are comparable, replace M with their combined mass (M1 + M2).

Frequently Asked Questions

Why doesn't the orbiting object's mass matter?

Gravity accelerates all objects equally regardless of their mass. A heavier satellite needs more force to maintain its orbit, but gravity provides exactly that extra force. The mass cancels out in the derivation.

What is a geostationary orbit?

A geostationary orbit has a period of exactly 24 hours (86,400 seconds), so the satellite tracks Earth's rotation and stays over the same spot. Plug in Earth's mass and that period, and the radius works out to an altitude of about 35,786 km above the surface.

Does this work for elliptical orbits?

Yes. For elliptical orbits, replace the radius R with the semi-major axis (the average of the closest and farthest distances from the central body). Kepler's Third Law applies to all conic-section orbits.

How accurate is this formula?

For most purposes, very. Small deviations creep in from atmospheric drag on low orbits, gravitational tugs from other bodies, and relativistic effects near very massive objects. For Earth satellites, the result lands within about 1% of what's actually observed.

Can I calculate the mass of a planet from its moon's orbit?

Yes. If you know a moon's orbital period and radius, you can solve for the central mass. This is exactly how astronomers determine the mass of distant planets and stars.