Q 11-07-003NEETNEET 2026Top questionEasy
In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius $R$ is proportional to :
Answer: (B) $R^{3/2}$
Gravity provides the centripetal force:
$$\frac{GMm}{R^2} = m\omega^2 R = m\frac{4\pi^2}{T^2}R \;\Rightarrow\; T^2 = \frac{4\pi^2}{GM}R^3$$
So $T \propto R^{3/2}$ (Kepler's third law).
Solution by Sreeraj P, M.Sc Physics