Gravitation

We have ρ = $\rho_0$ for r ≤ R = 0 for r > R For r ≤ R ; We have $E_g$ = $-\frac{GMr}{R^3}$ Therefore, the gravitational force experienced by a mass m is $F_g$ = $mE_g$ = $-\frac{GMmr}{R^3}$ The negative sign signifies an attractive force. Therefore, for circular motion, we have $\frac{GMmr}{R^3}$ = $\frac{mv^2}{r}$ where M = $\frac{4}{3}\pi R^3\rho_0$ ⇒ $\frac{4}{3}\pi Gmr\rho_0$ = $\frac{mv^2}{r}$ ⇒ V = $\sqrt{\left(\frac{4}{3}\right)\pi G\rho_0 r^2}$ (1) ⇒ V α r For r > R : We have $E_g$ = $-\frac{GM}{r^2}$ Therefore, $F_g$ = $mE_g$ = $\frac{GMm}{r^2}$ = $\frac{mv^2}{r}$ $\frac{GM}{r^2}\left(\frac{4}{3}\pi R^3\rho_0\right)$ = $\frac{mv^2}{r}$ From Eq. (1), we get V = $\sqrt{\frac{\frac{4}{3}\pi G\rho_0 R^3}{r}}$ ⇒ V α $\frac{1}{\sqrt{r}}$ Therefore, the correct plot is as shown in option (C):