Question

Density of a spherical planet of radius R is given by $\rho ={\rho }_0(1+\frac{r}{R})$ where r is distance from center of planet, then

• A. Mass of planet is $\frac{7}{3}\pi {\rho }_0{R}^3$
• B. Acceleration due to gravity at surface of planet is $\frac{5}{3}G{\rho }_0\pi R$
• C. Acceleration due to gravity at center of planet is zero
• D. Potential at surface of planet is $\frac{7}{3}G\pi {\rho }_0{R}^2$

Answer 1

Answer: (A), (C)

The correct options are
A Mass of planet is $\frac{7}{3}\pi {\rho }_0{R}^3$
C Acceleration due to gravity at center of planet is zero


$dm=\rho dv={\rho }_0(1+\frac{r}{R})4\pi r^{2}dr$

$M=\int dm=4\pi {\rho }_0{\int }_0^R(r^2+\frac{r^3}{R})dr$

$M=4\pi {\rho }_0[\frac{R^3}{3}+\frac{R^4}{4R}]=4\pi {\rho }_0{R}^3\frac{7}{12}$

$g=E=\frac{GM}{R^2}=\frac{7}{3}G{\rho }_0\pi R$

at center E = 0 so g = 0

Coming to potential,
$dV=\frac{-Gdm}{R}\Rightarrow V=\frac{-GM}{R}=-\frac{7}{3}G\pi {\rho }_0{R}^2$