Q 11-07-169JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
The mass density of a spherical galaxy varies as $\dfrac{K}{r}$ over a large distance $r$ from its center. In that region, a small star is in a circular orbit of radius $R$. Then the period of revolution, $T$ depends on $R$ as:
Answer: (A) $T^{2} \propto R$
Mass inside radius $R$: $M(R) = \displaystyle\int_0^R \frac{K}{r}\,4\pi r^{2}\,dr = 2\pi K R^{2}$.
For the circular orbit, $\dfrac{v^{2}}{R} = \dfrac{GM(R)}{R^{2}} = 2\pi GK$, so $v^{2} = 2\pi GKR$.
$$T^{2} = \frac{4\pi^{2}R^{2}}{v^{2}} = \frac{2\pi R}{GK} \;\Rightarrow\; T^{2}\propto R$$
Solution by Sreeraj P, M.Sc Physics