Satellite orbital period and altitude calculator
Enter a satellite's altitude to get its orbital period and speed, or enter a period to find the altitude, for a circular orbit around the Earth.
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Orbital period
Circular, two-body orbits around a spherical Earth. Real orbits are slightly elliptical and perturbed by the Earth's shape, the Moon, the Sun and air drag, so periods differ by seconds. Not for orbit determination or navigation.
How to calculate an orbital period
- Choose what you know: the altitude above the Earth's surface or the orbital period.
- Type the value and its unit, or pick a ready-made orbit such as the ISS, GPS or a geostationary satellite.
- Read the period, the speed, the orbit radius and the number of orbits per day.
The formulas
For a circular orbit with radius a (the distance from the centre of the Earth), Kepler's third law gives the period T = 2π √(a³ / μ), and the orbital speed is v = √(μ / a). Here μ = GM = 398,600.4418 km³/s² is the Earth's geocentric gravitational constant (WGS 84 / EGM96) and a = R + h, where R = 6,378.137 km is the WGS 84 equatorial radius and h the altitude. To find the altitude from a period, the tool inverts the law: a = (μ T² / 4π²)^(1/3).
Orbits per day = 86,400 s / T. A geostationary orbit has T equal to one sidereal day, 86,164.0905 s, which gives a = 42,164 km and h = 35,786 km. These equations come from Kepler (1619) and Newton and are given, for instance, in D. A. Vallado, Fundamentals of Astrodynamics and Applications, and in NASA's Basics of Space Flight.
The model has no air drag, no Earth oblateness (J2), no lunar or solar gravity and no eccentricity, so a real satellite's period differs a little; the ISS, for example, is listed at 92.8 to 93.0 minutes. Not for orbit determination or navigation.
Questions
What is Kepler's third law?
The square of an orbit's period is proportional to the cube of its semi-major axis. For a circular orbit around the Earth, T = 2π √(a³ / μ), where a is the distance from the Earth's centre and μ is the Earth's gravitational parameter, 398,600.4418 km³/s².
Why is the radius used, not the altitude?
The law uses the distance from the centre of the Earth. The tool adds the Earth's equatorial radius of 6,378.137 km to the altitude you enter. Using a different radius, such as the mean one, changes the answer a little.
How long is the period of the ISS?
At an altitude of about 420 km the period is about 93 minutes, so the station circles the Earth about 15.5 times a day at 7.66 km/s (27,600 km/h).
What is a geostationary orbit?
It is the circular orbit over the equator at 35,786 km altitude, where the period equals one sidereal day (23 h 56 min 4 s), so a satellite stays above the same point on the ground. Enter 86,164 s to find it.
Does a higher orbit go faster?
No, slower. Orbital speed v = √(μ / a) falls as the orbit gets bigger, while the period increases. A low satellite circles the Earth in about 90 minutes, whereas the Moon takes 27 days.
Why is there a warning below 160 km?
Air drag at such altitudes is high enough that a satellite re-enters within days or hours. The formula still gives a period but a stable orbit is not realistic.