Question:
The mass density of a planet of radius $\mathrm{R}$ varies with the distance $r$ from its centre as
$\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) .$ Then the gravitational field
is maximum at:
Correct Option: , 2
Solution:
E $4 \pi r^{2}=\int \rho_{0} 4 \pi r^{2} d r$
$\Rightarrow \mathrm{Er}^{2}=4 \pi \mathrm{G} \int_{0}^{r} \rho_{0}\left(1-\frac{\mathrm{r}^{2}}{\mathrm{R}^{2}}\right) \mathrm{r}^{2} \mathrm{dr}$
$\Rightarrow \mathrm{E}=4 \pi \mathrm{G} \rho_{0}\left(\frac{\mathrm{r}^{3}}{3}-\frac{\mathrm{r}^{5}}{5 \mathrm{R}^{2}}\right)$
$\frac{\mathrm{dE}}{\mathrm{dr}}=0 \quad \therefore \mathrm{r}=\sqrt{\frac{5}{9}} \mathrm{R}$