In 1619 Johannes Kepler noticed that the planets farther from the Sun take disproportionately longer to go around it: square the period and you get a number proportional to the cube of the orbit’s size. Newton later explained why, and his version of the law turns it into a practical tool. With this calculator you can find how long an orbit takes, how big an orbit must be for a chosen period, or how massive a planet or star is from the motion of something circling it.
How to use the orbital period calculator
- Choose to Solve for the period, the orbit size or the central mass.
- Pick what the object is Orbiting around: the Sun, Earth, the Moon, Mars, Jupiter or Saturn, or enter a custom central mass.
- For a listed body, choose whether to give the Orbit size as a semi-major axis or as the altitude of a circular orbit above the surface.
- Enter the known values. The period accepts days, hours, minutes, seconds or years; sizes accept au, km, m, mi or Earth radii.
- Optionally add the Eccentricity and the Orbiting body’s mass, then pick the result unit. The best-fit setting switches between minutes, hours, days and years automatically.
Kepler’s third law formula
T is the period, a the semi-major axis (the radius, for a circle), M the central mass, m the orbiting mass and G = 6.67430 × 10⁻¹¹ N·m²/kg². For an elliptical orbit, the speed at any distance r comes from the vis-viva equation, v = √(GM(2/r − 1/a)).
Worked example
Earth's year
The defaults are an orbit around the Sun with a = 1 au = 149,597,870,700 m.
T = 2π √((1.495979 × 10¹¹)³ ÷ (6.6743 × 10⁻¹¹ × 1.98841 × 10³⁰)) = 31,558,200 s = 365.257 days
Adding Earth's own mass (1 Earth mass) gives 365.256 days, the sidereal year. Enter Earth's eccentricity, 0.0167, and the calculator puts perihelion at 147.1 million km, where Earth moves at 30.29 km/s, and aphelion at 152.1 million km, at 29.29 km/s.
Weighing Jupiter. Its moon Io orbits at 421,700 km every 1.769138 days. Solve for the central mass and the calculator returns 1.8985 × 10²⁷ kg, within 0.02% of Jupiter’s accepted mass. Astronomers measure the masses of planets, stars and even black holes this way.
Common orbits around Earth
| Orbit | Altitude | Period |
|---|---|---|
| Low Earth orbit (ISS) | 420 km | about 93 minutes |
| GPS satellites | 20,200 km | about 11 h 58 min |
| Geostationary | 35,786 km | 23.934 hours |
| The Moon | 384,400 km mean distance | about 27.3 days |
Altitudes are measured from the equator; the GPS period is half a sidereal day by design. For the Moon, the two-body formula with the Moon’s mass included gives 27.28 days against a measured 27.32; much of the gap comes from the Sun’s pull, which this model ignores.
Why bigger orbits are so much slower
Orbital speed falls as 1/√r while the distance around the orbit grows in proportion to r, so the period grows as r3/2. Neptune is about 30 times farther from the Sun than Earth, and its year is about 301.5 ≈ 165 Earth years. The table under the result shows this for every planet.
What the model leaves out
The calculation treats the two bodies as point masses with nothing else around. Atmospheric drag slowly lowers satellites in low orbit; Earth’s equatorial bulge makes orbits precess; and other planets tug on each other. For the speed needed to leave orbit entirely, see the escape velocity calculator, and for the force holding an orbit together, the gravitational force calculator and the centripetal force calculator.
Frequently asked questions
What is Kepler's third law?
The square of an orbit's period is proportional to the cube of its semi-major axis. Newton showed the constant of proportionality is 4π² ÷ G(M + m), which is why the law can also be used to weigh planets and stars.
How long does the International Space Station take to orbit Earth?
At about 420 km altitude the calculator gives 92.97 minutes, so the crew sees about 15.5 sunrises a day. Real ISS altitude drifts between roughly 400 and 430 km, changing the period by a minute or so.
What altitude is a geostationary orbit?
About 35,786 km above the equator, a radius of 42,164 km. The period there is 23.934 hours, one sidereal day, so a satellite keeps pace with Earth's spin and appears fixed in the sky.
Does the orbiting object's mass matter?
Only when it is not tiny compared with the central body. A satellite's mass is irrelevant, but the Moon is 1.2% of Earth's mass and Jupiter is 0.1% of the Sun's. Enter the optional mass for those cases; the calculator then uses M + m.
What does eccentricity change?
Not the period: any ellipse with the same semi-major axis takes the same time. Eccentricity sets how stretched the orbit is, so the calculator uses it to report the closest and farthest distances and the speeds there.