Orbital period calculator
Pick a body and an altitude, get the orbital period and speed.
How do you calculate an orbital period?
For a circular orbit of radius r around a body of mass M: T = 2π√(r³ ÷ GM), and the orbital speed is v = √(GM ÷ r). The radius is measured from the body’s center — its radius plus the altitude (OpenStax University Physics).
Example: the International Space Station
At 400 km above Earth, the speed is about 7.67 km/s (17,200 mph) and one orbit takes about 92.6 minutes — in line with NASA’s “about every 90 minutes”, 16 orbits a day.
Does the satellite’s mass matter?
No — only the mass of the body it orbits and the distance. A small satellite and the ISS at the same height move at the same speed.
What about a geostationary orbit?
Enter an altitude of about 35,786 km above Earth: the period comes out close to one sidereal day, about 23 h 56 min, so the satellite stays over the same spot.
Which data does it use?
Masses and equatorial radii from NASA’s planetary fact sheets and G = 6.67430 × 10⁻¹¹ (CODATA). Real orbits are often elliptical; this is the circular case.
How fast does a satellite at a higher altitude move?
Slower. At 400 km the circular speed is 7.67 km/s with a 92.6-minute period; at 20,200 km (the altitude of GPS satellites) it is 3.87 km/s with a period of 11.98 hours. A higher orbit has a longer path and a weaker pull, so the period grows faster than the radius.
What is the period of a low orbit around the Moon or Mars?
117.9 minutes for a circular orbit 100 km above the Moon, and 118.4 minutes at 400 km above Mars, using the NASA body masses and radii built into the calculator.
Where the formula comes from
In a circular orbit gravity supplies exactly the centripetal force needed: GMm ÷ r² = mv² ÷ r. Cancelling the satellite’s mass m gives v = √(GM ÷ r). One lap covers 2πr, so the period is T = 2πr ÷ v = 2π√(r³ ÷ GM). Squaring shows Kepler’s third law for a given central body: T² is proportional to r³, so doubling the radius multiplies the period by 2^1.5, about 2.83.
A worked example
Take a satellite 400 km above Earth. The radius from Earth’s center is 6,378.1 + 400 = 6,778.1 km, or 6,778,137 m. With GM = G × 5.9722 × 10²⁴ kg, the speed is 7,669 m/s (7.669 km/s) and the period is 5,554 s, or 92.6 minutes. That is about 15.6 orbits per day. At 35,786 km the period becomes 23.93 hours, about one sidereal day, which is why that altitude is geostationary.
What the circular model leaves out
Real orbits are ellipses, and the semi-major axis then takes the place of r in the period formula. The calculator also ignores atmospheric drag, the slight bulge of the planet, and the pull of other bodies, all of which matter for very low or very high orbits. Treat the answer as the ideal two-body value.
Common mistakes
The most frequent error is using the altitude as the radius: r is measured from the center of the body, so Earth’s radius must be added to the altitude. A second is a unit slip, since the formula needs meters and kilograms (the calculator takes kilometers or miles for altitude and converts). Finally, remember that mass of the satellite does not appear, so a heavier satellite at the same height does not orbit any faster.
Sources: OpenStax University Physics Volume 1, chapter 13 (gravitation, including satellite orbits and Kepler’s laws); planetary masses and radii from NASA’s Planetary Fact Sheets; G from CODATA. Related: escape velocity calculator, gravitational force calculator, centripetal force calculator and weight on other planets calculator.