- #1
DanHelzer
- 4
- 0
Kepler's third law states T^2=(4pi^2/GM) x r^3 for CIRCULAR orbits. My question is, in the derivation for this equation ma=GMm/r^2 why can centripetal acceleration be used to replace a at m(v^2/r)=GMm/r^2 yielding v^2/r=GM/r^2 when the orbit is not circular. Planets have elliptical orbits so why can centripetal acceleration be used so that for elliptical orbits the equation for a period is (r=a) --> T^2=4pi^2/GM) x a^3.