Orbital Period Calculator
Work out how long one orbit takes at a given altitude, or the altitude needed for a chosen period, such as the 35,786 km of a geostationary orbit, using Kepler’s third law.
Kepler’s third law
Johannes Kepler found in 1619 that the square of a planet’s period is proportional to the cube of its distance from the Sun. Newton showed why, and the exact form for any body is T = 2π√(a³/μ), where a is the semi-major axis and μ is the gravitational parameter of the central body. The period depends only on the semi-major axis: a circular orbit and an elongated ellipse with the same a take exactly the same time. The calculator works in either direction. Given an altitude it returns the period; given a period it solves a = ∛(μ(T/2π)²) and subtracts the planet’s radius to give the altitude above the surface.
The fastest possible orbit of Earth is one that just grazes the surface at the equatorial radius of 6,378.137 km. It would take 84.49 minutes, and no circular orbit can be quicker, because any lower path runs into the ground. A 400 km orbit lasts 92.56 minutes, which works out to 15.56 orbits per day.
Finding geostationary altitude
A geostationary satellite stays above one point on the equator, so it must complete one orbit in exactly the time the Earth takes to turn once relative to the stars. That is the sidereal day of 23.9344696 hours, or 86,164.09 seconds, not the 24-hour solar day. Putting that period into the inverse formula gives a semi-major axis of 42,164.2 km. Subtracting the equatorial radius leaves an altitude of 35,786 km, the figure quoted in every communications satellite brochure. The orbital speed there is 3.075 km/s.
If you use 24 hours instead, the answer comes out 35,863 km, about 77 km too high. A satellite parked there would complete one orbit slightly slower than the Earth turns and would drift westward by nearly one degree a day. Station-keeping engineers care about far smaller errors than that, so the sidereal day matters.
Periods in practice
Orbit designers pick periods for a purpose. Sun-synchronous Earth observation satellites around 700 to 800 km have periods near 100 minutes and pass over each place at the same local solar time. Navigation constellations such as GPS use a period of half a sidereal day, so the pattern of satellites over a ground station repeats every day four minutes earlier by the clock. Molniya orbits are highly elliptical with the same 12-hour period, spending most of each orbit slowly over high northern latitudes. Each of these starts from Kepler’s third law and a chosen period. A common misconception is that an elliptical orbit with the same apoapsis as a circle has the same period. It does not: the ellipse is shorter on average, so its semi-major axis and its period are smaller.
How to use it
- Choose the central body.
- Choose what to solve for. The default solves for altitude from a period, so you can check the geostationary example immediately.
- For altitude from period, enter the period in hours. The default 23.9344696 h is one sidereal day and gives 35,786 km.
- For period from altitude, enter the circular orbit altitude in kilometres and read the period in hours and minutes, the number of orbits per day and the circular speed.
- If the result shows a negative altitude, the period you asked for is shorter than a surface-grazing orbit, and no such orbit exists around that body.
Questions from the flight desk
How do you calculate the orbital period of a satellite?
Use Kepler’s third law, T = 2π√(a³/μ). For Earth μ is 398,600.4418 km³/s² and a is the orbit radius, the planet radius plus the altitude. At 400 km, a is 6,778.137 km and the period comes out at 92.56 minutes. For an elliptical orbit use the semi-major axis, the average of the periapsis and apoapsis radii.
Why is geostationary orbit at 35,786 km?
Because that is the only circular equatorial orbit whose period equals one sidereal day, 23.9344696 hours. Solving Kepler’s third law for that period gives an orbit radius of 42,164 km, and subtracting the Earth’s equatorial radius of 6,378 km leaves 35,786 km. Any lower orbit runs ahead of the turning Earth and any higher one falls behind.
What is the difference between a sidereal day and a solar day?
A solar day of 24 hours is the time from noon to noon. Because the Earth also moves along its orbit, it must turn a little more than once for the Sun to return to the same place. One rotation relative to the distant stars, the sidereal day, takes 23 hours 56 minutes 4 seconds. Orbits are timed against the stars.
Does an elliptical orbit have a different period from a circular one?
Only if its semi-major axis is different. Period depends on a alone, so a circle of radius 10,000 km and an ellipse with periapsis 8,000 km and apoapsis 12,000 km from the centre have the same period. The spacecraft on the ellipse simply moves faster near the bottom and slower near the top.
Disclaimer: Orbital mechanics results are ideal two-body figures with instantaneous burns, computed from the values you enter and the constants stated on each page. They are not mission analysis, flight dynamics or launch safety advice. Real orbits are perturbed by Earth’s oblateness, the Moon, the Sun and atmospheric drag, and real burns carry gravity and steering losses. Never use these numbers to operate a spacecraft or to fly a rocket; follow the safety code of your rocketry association and the advice of qualified engineers.