Q 11-07-185JEE MainJEE Main 2019 (9 Apr, Shift 2)Medium
A test particle is moving in a circular orbit in the gravitational field produced by a mass density $\rho(r) = \dfrac{K}{r^2}$. Identify the correct relation between the radius $R$ of the particle's orbit and its period $T$:
Answer: (C) $T/R$ is a constant
Mass inside radius $R$:
$$M(R) = \int_0^R\frac{K}{r^2}4\pi r^2\,dr = 4\pi KR$$
For the circular orbit,
$$\frac{mv^2}{R} = \frac{GM(R)m}{R^2} = \frac{4\pi GKm}{R} \Rightarrow v^2 = 4\pi GK$$
The speed is independent of $R$, so $T = \dfrac{2\pi R}{v} \propto R$: $T/R$ is constant.
Solution by Sreeraj P, M.Sc Physics