How the orbital period calculator works
Every orbit, from a cubesat skimming the atmosphere to the Moon making its stately monthly lap, obeys one rule that Johannes Kepler worked out in 1619 and Newton later explained: the square of the orbital period is proportional to the cube of the orbit's size. Feed this page an altitude and a central body and it solves that law for you, showing every step with your own numbers. The only two things the answer depends on are the size of the orbit (the semi-major axis, a) and how strongly the central body pulls (its standard gravitational parameter, mu).
Notice what is missing from that list: the satellite. Its mass appears nowhere in the math. A one kilogram cubesat and the 420 tonne International Space Station at the same altitude take exactly the same time to go around, for the same reason the hammer and the falcon feather hit the lunar dust together when Apollo 15's David Scott dropped them in 1971. Gravity accelerates everything equally, so the thing doing the falling never gets a vote. An orbit is just falling that keeps missing the ground, and it is the most counterintuitive true thing on this page.
The formula
Term by term: T is the orbital period in seconds; a is the semi-major axis, the orbit's size measured from the central body's center (for a circular orbit it is simply the orbit radius; for an ellipse it is the average of the closest and farthest distances); and μ (mu) is the standard gravitational parameter, GM, the gravitational constant times the body's mass, in km³/s². The second equation is vis-viva: the speed at any point of the orbit, where r is the current distance from the center. On a circular orbit r = a and it collapses to v = √(μ⁄a).
These are the standard values this calculator uses, from the NASA and JPL fact sheets. Radii are equatorial. Jupiter gets one honest footnote: a gas giant has no surface, so its "radius" is the altitude of the 1 bar pressure level, the cloud tops.
| Body | mu = GM (km³/s²) | Equatorial radius (km) |
|---|---|---|
| Earth | 398,600.4418 | 6,378.137 |
| The Moon | 4,902.800 | 1,737.4 |
| Mars | 42,828.37 | 3,396.2 |
| Jupiter | 126,686,534 | 71,492 (cloud tops) |
| The Sun | 132,712,440,018 | 695,700 |
Why mu instead of separate G and M? Because mu is what we can actually measure. Tracking a spacecraft's path pins down GM to a dozen digits, while G by itself is one of the least precisely known constants in physics. Orbital mechanics quietly sidesteps the problem by never separating the two.
Worked example
The ISS flies at roughly 420 km altitude. How long is one lap of Earth?
Distance from Earth's center: r = 420 + 6,378.137 = 6,798.137 km. (The 420 km altitude is about 261 miles up.)
Circular orbit, so the semi-major axis is the same: a = 6,798.137 km.
Kepler's third law: T = 2π√(6,798.137³ ⁄ 398,600.4418) = 2π × 887.8 = 5,578.2 seconds.
Convert: 5,578.2 seconds is 92.97 minutes, moving at v = √(398,600.4418 ⁄ 6,798.137) = 7.66 km/s, about 17,129 mph.
That is just under 16 laps a day, which is why ISS astronauts get 16 sunrises every 24 hours.
The altitude vs radius trap
The single most common mistake with Kepler's law: plugging in the altitude where the math wants the distance from the center. Satellite altitudes are quoted from the surface because that is how humans think, but gravity pulls from the center of the planet, so the law needs altitude plus the body's radius. Skip the addition and the numbers go absurd: treat the ISS's 420 km altitude as its orbit radius and Kepler cheerfully reports a period of about 86 seconds, for an "orbit" buried more than 5,900 km inside the planet. This calculator does the addition for you and shows it as the first step, so the trap is disarmed before it springs. If your figure is already measured from the center (semi-major axes usually are; the GPS constellation's 26,560 km is one), subtract the radius before entering it, or pick the custom body option and leave the radius blank.
Why geostationary satellites sit at exactly 35,786 km
A satellite that hangs motionless over one spot must complete an orbit in the time Earth takes to rotate once. Here is the subtle part: that is not 24 hours. The 24 hour solar day is one rotation relative to the Sun, but Earth also travels along its orbit each day, so it has to turn about one extra degree to face the Sun again. One rotation relative to the stars, which is the frame orbits actually live in, takes the sidereal day: 23 h 56 min 4.09 s, or 86,164.09 seconds.
Solve Kepler's law backwards for that period and out comes a = ∛(μ(T⁄2π)²) = 42,164 km from Earth's center. Subtract Earth's 6,378 km radius and you land on 35,786 km of altitude: the geostationary belt, sometimes called the Clarke orbit after Arthur C. Clarke, who worked out in 1945 that relay stations parked there could cover the globe. Enter 35,786 km above and this page will recognize the orbit and tell you so. Use 24 hours instead of the sidereal day and you miss the belt by about 77 km of altitude, which in this business is a lot: a "stationary" satellite built on the solar day would drift around the planet once a year.
Why everything in low Earth orbit takes about 90 minutes
The ISS, the Hubble telescope, Starlink satellites, and every crewed capsule since Gagarin all lap the planet in roughly an hour and a half, and Kepler's law explains why with one glance. The period depends on the distance from Earth's center, and down in low orbit the altitude is a rounding error on top of Earth's 6,378 km radius. Run the numbers: 200 km altitude gives 88.49 minutes, 1,000 km gives 105.12 minutes. The entire band that nearly all satellites inhabit spans barely a quarter of an hour of period. Even the theoretical limit, a drag-free orbit skimming the surface at zero altitude, takes 84.49 minutes. There is simply no room in the geometry for a 20 minute orbit or a 3 hour one down there, and since mass never enters the law, this is true for every object in LEO regardless of what it weighs.
Elliptical orbits: sprint at the bottom, coast at the top
Most real orbits are not perfect circles. An ellipse has a low point (periapsis; perigee around Earth) and a high point (apoapsis; apogee), and Kepler's period law handles it with one graceful move: use the average of the two distances as the semi-major axis, a = (rp + ra) ⁄ 2. Two wildly different-looking orbits with the same average size share exactly the same period.
Speed is where ellipses get dramatic. The vis-viva equation says speed depends on where you are right now, and the difference can be huge: a geostationary transfer orbit dipping to 200 km and climbing to 35,786 km crosses its low point at 10.24 km/s and drifts over the top at 1.6 km/s, better than six times slower. Fastest at periapsis, always; this is Kepler's second law (equal areas in equal times) wearing its everyday clothes. Comets are the extreme case: Halley's comet whips through its months near the Sun and then spends the balance of its 76 year period crawling out beyond Neptune, moving at a walking-pace crawl by orbital standards.
Where this math bends (honestly)
This page runs pure two-body Newtonian gravity: one satellite of negligible mass around one perfectly spherical body. Reality perturbs it. Earth's equatorial bulge (the J2 effect) slowly twists orbit planes, whispers of atmosphere drag on low satellites, and the Sun and Moon tug on everything; mission planners layer all of that on top of Kepler. You can even see the "negligible mass" assumption strain in our own reference table: it gives the Moon a 27.45 day period around Earth, while the true sidereal month is 27.32 days. The gap exists because the Moon is 1.2 percent of Earth's mass, not zero, and the two really orbit their common center. For everything else on this page's scale, satellites, stations, and probes, the error is tiny and the answers here are the same ones the textbooks give. Getting to orbit in the first place is a different problem entirely, priced in the delta-v calculator with the rocket equation, and leaving Earth behind for the planets is the space travel time calculator's department.