Derive Keplers 3rd law

Equate gravitational force (GMm/r^2) to centripetal force (mv^2/r). Rearrange to get v^2=GM/r. Due to the approximated circular orbit v=2pir/T so v^2=4pi^2r^2/T^2. Therefore GM/r=4pi^2r^2/T^2. Rearranging gives GM=4pi^2r^3/T^2. Rearranging futher gets T^2=(4pi^2/GM)r^3. Hence T^2 is proportional to r^3.

MS
Answered by Michael S. Physics tutor

4917 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

A mass, m, is resting on a slope being slowly tilted upwards from horizontal. The static friction co-efficient is 0.3 and the dynamic friction co-efficient is 0.2: at what angle will the mass begin to slip?


On the line of centres between the Earth and the Moon, there is a point where the net gravitational force is zero. Given that the distance between the two is 385,000 km, and that the Earth has a mass 81x that of the Moon, how far is this point from Earth?


Why the Newton's second law of motion important?


In terms of particles, explain how resistance arises in metal conductors and why does this resistance increases with temperature.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning