Orbital Period from Kepler's Third Law
Orbital Period from Kepler's Third Law
Kepler's third law connects an orbit's size directly to how long it takes to complete one revolution, and for a circular orbit this relationship reduces to a clean closed-form expression involving only the orbit radius and the gravitational parameter of the body being orbited. This period calculation underlies everything from choosing a satellite's altitude to hit a specific revisit schedule to explaining why geostationary satellites must orbit at one particular altitude (about 35,786 km) — the only radius where the orbital period exactly matches Earth's rotation.
Kepler's third law for a circular orbit gives period T = 2*pi*sqrt(a^3/mu). where a_orb is the orbit radius and mu_earth is the gravitational parameter of the central body (Earth).
Inputs
a_orb := 7000 km=7e+06 m
mu_earth := 398600 km^3/s^2=3.99e+14 m^3/s^2
The period follows purely from balancing gravitational attraction against the centripetal acceleration needed to stay in a circular path, giving a period that grows with the 3/2 power of orbit radius.
T_orb := 2·pi·sqrt(a_orb^3/mu_earth)=5828.52 s
Results
A 7,000 km orbit radius (a few hundred kilometres above Earth's surface) gives a period of roughly an hour and a half, consistent with the low Earth orbits used by the International Space Station and most Earth-observation satellites. Because period scales with the 3/2 power of radius, even modest increases in orbit altitude noticeably lengthen the period — this steep sensitivity is exactly why geostationary orbit sits at one very specific, much higher altitude.
Assumptions & limitations
- Circular orbit assumed — elliptical orbits require the semi-major axis and a more general form.
- Two-body problem — perturbations from Earth oblateness, drag, and other bodies are neglected.
- Point-mass gravitational field (spherically symmetric central body).
- mu is specific to the central body being orbited (Earth here).