AP® Physics C: Mechanics · Unit 2 · Topic 2.10
Kepler's Third Law for Circular Orbits
For a circular orbit, gravity supplies the centripetal force, and solving that equation shows the period squared is proportional to the orbital radius cubed: T2 = (4 π2 / GM) r3.
A planet far from the Sun moves slower and has farther to travel, so its period grows fast with distance. Both effects come from the same source: gravity gets weaker with distance, so there is less centripetal force pulling the orbiting body around, and it needs less speed to stay in orbit. Less speed over a bigger circle means a much longer trip.
The mechanism is just Newton's second law applied to circular motion. Gravity is the only force doing the pulling, so set GMm/r2 equal to the centripetal requirement mv2/r. The masses of the orbiting body cancel, leaving v2 = GM/r. Since the body covers the circle's circumference 2 π r in one period T, v = 2 π r / T. Substitute that in and solve: T2 = 4 π2 r3 / GM. That is Kepler's third law for circular orbits, and it only involves the central mass M, not the orbiting mass.
This result sits right after Newton's law of gravitation and the idea of gravitational field strength, because it is really those ideas combined with circular motion rather than a separate law. It also sets up the next step: non-circular, elliptical orbits, where the same T2 proportional to a3 relationship holds with a as the semi-major axis, and where energy and angular momentum conservation take over for the harder problems.
Key terms
- Centripetal force
- The net inward force needed to keep an object moving on a circular path at constant speed.
- GM
- The product of the gravitational constant and the central body's mass, often treated as one measured quantity for a planet.
- Orbital period T
- The time for one full trip around the orbit.
- Semi-major axis
- Half the longest diameter of an ellipse; equals the radius for a circular orbit and generalizes Kepler's third law to ellipses.
Watch it explained
- How it worksHow does setting gravity equal to mv2 over r produce a period squared proportional to radius cubed?
- Worked exampleFind the orbital period of a satellite circling 7000 km from the center of the Earth.
- Worked exampleFind the orbital radius a satellite needs so that its period is exactly 24 hours.
- Going furtherWhat happens to the orbital period at fixed radius when the central body's mass doubles?
Check yourself
End of topic.