Orbital Period Calculator
Compute orbital period using validated scientific equations. See step-by-step derivations, unit analysis, and reference values.
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer
Orbital Period Calculator
Calculator
Adjust values & calculateEnter your values below. Every result is computed in your browser โ no data is sent to any server.
Formula: T = 2 pi sqrt(a^3 / (G M))
Worked example โ Orbital period: 365.25 days (1 year) confirming Kepler's Third Law
Formula
T = 2 pi sqrt(a^3 / (G M))
Where T is the orbital period in seconds, a is the semi-major axis in meters, G is the gravitational constant (6.674 x 10^-11 m^3 kg^-1 s^-2), and M is the mass of the central body in kilograms. This is derived from Kepler's Third Law combined with Newton's law of gravitation.
Worked Examples
Example 1: Earth's Orbital Period
Problem:Calculate the orbital period of Earth orbiting the Sun at 1 AU (1.496 x 10^11 m) with the Sun's mass of 1.989 x 10^30 kg.
Solution:T = 2 x pi x sqrt(a^3 / (G x M)) T = 2 x pi x sqrt((1.496e11)^3 / (6.674e-11 x 1.989e30)) T = 2 x pi x sqrt(3.348e33 / 1.327e20) T = 2 x pi x sqrt(2.524e13) T = 2 x pi x 5.024e6 T = 31,557,600 seconds = 365.25 days
Result:Orbital period: 365.25 days (1 year) confirming Kepler's Third Law
Example 2: ISS Orbital Period
Problem:The ISS orbits at 420 km altitude. With Earth's radius of 6,371 km and mass of 5.972 x 10^24 kg, calculate its orbital period.
Solution:Semi-major axis = 6,371 + 420 = 6,791 km = 6,791,000 m T = 2 x pi x sqrt((6.791e6)^3 / (6.674e-11 x 5.972e24)) T = 2 x pi x sqrt(3.134e20 / 3.986e14) T = 2 x pi x sqrt(7.862e5) T = 2 x pi x 886.7 T = 5,571 seconds = 92.8 minutes
Result:ISS orbital period: 92.8 minutes | Velocity: 7.66 km/s | ~15.5 orbits per day
Frequently Asked Questions
What is Kepler's Third Law and how does it calculate orbital period?
Kepler's Third Law of Planetary Motion states that the square of a planet's orbital period is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T squared equals 4 pi squared times a cubed divided by G times M, where T is the orbital period, a is the semi-major axis (the average distance from the central body), G is the gravitational constant, and M is the mass of the central body. This elegant relationship means that if you know the distance of an orbiting body from its parent and the mass of the parent, you can calculate exactly how long one complete orbit takes. Johannes Kepler discovered this empirical relationship in 1619, and Isaac Newton later provided the theoretical foundation by deriving it from his law of universal gravitation, showing that it applies to any two gravitationally bound objects in the universe.
What is the semi-major axis and how does eccentricity affect orbits?
The semi-major axis is half the longest diameter of an elliptical orbit, representing the average distance between the orbiting body and the central body over one complete orbit. For a perfectly circular orbit, the semi-major axis equals the orbital radius. Eccentricity measures how elongated an orbit is, ranging from 0 for a perfect circle to values approaching 1 for extremely elongated ellipses. Earth has an eccentricity of 0.0167, making its orbit nearly circular, while Pluto has an eccentricity of 0.248, causing significant variation in its distance from the Sun. Eccentricity does not affect the orbital period directly, as Kepler's Third Law depends only on the semi-major axis and central mass. However, eccentricity determines the variation in orbital velocity, with objects moving faster at periapsis (closest approach) and slower at apoapsis (farthest point) to conserve angular momentum.
How are orbital periods used to discover exoplanets?
Orbital periods are fundamental to exoplanet detection through the transit method, where astronomers measure periodic dips in a star's brightness as a planet crosses in front of it. The time between transits directly gives the orbital period, and using Kepler's Third Law with the known stellar mass, scientists can calculate the planet's distance from its star. The Kepler Space Telescope discovered over 2,600 confirmed exoplanets using this technique. The radial velocity method also relies on orbital periods by detecting the periodic wobble of a star caused by an orbiting planet's gravitational pull. Shorter orbital periods are easier to detect because multiple transits can be observed in less time. This observational bias means that close-in hot Jupiters with periods of a few days were among the first exoplanets discovered, while detecting Earth-like planets with year-long periods requires years of continuous observation.
What determines the orbital period of satellites around Earth?
Satellite orbital periods around Earth depend entirely on their altitude above the surface, which determines the semi-major axis. Low Earth orbit satellites at 400 km altitude, like the International Space Station, complete one orbit in approximately 93 minutes traveling at 7.66 km/s. Medium Earth orbit GPS satellites at 20,200 km altitude have periods of about 12 hours. At exactly 35,786 km altitude, a satellite achieves geostationary orbit with a period matching Earth's 24-hour rotation, appearing stationary above a fixed point on the equator. This principle is used for communications and weather satellites. The Moon orbits at 384,400 km with a period of 27.3 days. For any circular orbit, doubling the altitude more than doubles the period because of the cube-root relationship in Kepler's law. No satellite can orbit below approximately 160 km altitude because atmospheric drag would quickly deorbit it.
How do orbital velocities change at different points in an elliptical orbit?
In an elliptical orbit, velocity varies continuously according to the vis-viva equation: v equals the square root of G times M times the quantity 2 over r minus 1 over a, where r is the current distance and a is the semi-major axis. At periapsis, the closest point, the orbiting body moves fastest because gravitational potential energy has converted to kinetic energy. At apoapsis, the farthest point, the body moves slowest as kinetic energy has converted back to potential energy. For Earth orbiting the Sun, the velocity varies from 30.29 km/s at perihelion to 29.29 km/s at aphelion, a difference of about 3.4 percent due to Earth's low eccentricity. Comets with highly eccentric orbits show extreme velocity variations, with some reaching over 100 km/s at perihelion while crawling at just a few km/s at aphelion billions of kilometers away.
How do orbital velocities relate to altitude?
Orbital velocity decreases with altitude: v = sqrt(GM/r), where G is the gravitational constant, M is Earth's mass, and r is orbital radius. Low Earth orbit (400 km) requires about 7.67 km/s. Geostationary orbit (35,786 km) requires only 3.07 km/s. Escape velocity from Earth's surface is 11.2 km/s.
References
Background & Theory
History
Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer ยท Editorial policy
Related Calculators
๐งฎSatellite Speed Calculator
Calculate the orbital speed of a satellite at any altitude above Earth.
๐งฎSpace Mission Cost Calculator
Estimate space mission costs from payload mass, orbit, and launch vehicle selection.
๐งฎSpace Suit Air Supply Calculator
Calculate breathable air supply duration for EVA from tank pressure, volume, and consumption rate.
๐งฎBlackbody Peak Wavelength Calculator
Calculate blackbody peak wavelength with inputs, formulas, and instant results.
๐งฎParallax Distance Calculator
Calculate parallax distance with inputs, formulas, and instant results.
๐งฎRedshift Calculator
Calculate redshift with inputs, formulas, and instant results.
๐งฎField of View Calculator
Calculate field of view with inputs, formulas, and instant results.