Physics - Gravitation Question with Solution | TestHub

First, calculate the mass $M(r)$ enclosed within a radius $r$ and the total mass $M_{total}$.$$M(r)={\int }_0^r\rho (x)4\pi x^{2}dx=4\pi {\rho }_0{\int }_0^r(x^2-x^3/R)dx=4\pi {\rho }_0(r^3/3-r^4/(4R))$$ Total mass $M_{total}=M(R)=4\pi {\rho }_0(R^3/3-R^3/4)=\frac{\pi {\rho }_0{R}^3}{3}$. For $r<R$, the gravitational field is $E(r)=\frac{GM(r)}{r^2}=4\pi G{\rho }_0(r/3-r^2/(4R))$. To find maximum field, $dE/dr=4\pi G{\rho }_0(1/3-2r/(4R))=0\Rightarrow r=2R/3$. So A is incorrect. For $r>R$, the gravitational field is $E(r)=\frac{G{M}_{total}}{r^2}=\frac{G(\pi {\rho }_0{R}^3/3)}{r^2}=\frac{G\pi {\rho }_0{R}^3}{3r^2}$. So C is correct. The gravitational potential at the surface is $V(R)=-\frac{G{M}_{total}}{R}=-\frac{G(\pi {\rho }_0{R}^3/3)}{R}=-\frac{G\pi {\rho }_0{R}^2}{3}$. So D is correct. For $r<R$, the potential $V(r)=V(R)-{\int }_R^{r}E(x)dx$.$$V(r)=-\frac{G\pi {\rho }_0{R}^2}{3}-4\pi G{\rho }_0{\left[\frac{x^2}{6}-\frac{x^3}{12R}\right]}_R^r$$ $$V(r)=-\frac{G\pi {\rho }_0{R}^2}{3}-4\pi G{\rho }_0\left(\frac{r^2}{6}-\frac{r^3}{12R}-\frac{R^2}{12}\right)=\frac{\pi G{\rho }_0{r}^3}{3R}-\frac{2\pi G{\rho }_0{r}^2}{3}$$ At the center ($r=0$), $V(0)=0$. So B is correct.