Physics - Gravitation Question with Solution | TestHub

$\rho =\frac{{\rho }_0}{R^2}{r}^2$. Mass within radius r: $M(r)={\int }_0^r\frac{{\rho }_0}{R^2}{r}^{\prime 2}4\pi r^{\prime 2}d{r}^{\prime }=\frac{4\pi {\rho }_0}{5R^2}{r}^5$.

Inside the planet $g=\frac{GM(r)}{r^2}=\frac{4\pi G{\rho }_0}{5R^2}{r}^3$, i.e. g varies as the cube of the distance - (B) is correct.

Total mass $M=\frac{4\pi {\rho }_0{R}^3}{5}$, so $v_{esc}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{8\pi G{\rho }_0{R}^2}{5}}$. Option (A) has the wrong factor (4 instead of 8) - (A) is wrong.

Since $v_{esc}=\sqrt{2GM/R}$ depends only on total mass and radius, it equals that of a uniform planet of the same M and R - (C) is correct.

Energy required $=\frac{1}{2}m{v}_{esc}^2\propto m$, so the ratio for masses m and 2m is 1:2 - (D) is correct.