The mass density of a planet of radius $R$ 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 :  Option: 1   Option: 2 Option: 3 Option: 4

$\dpi{150} \mathrm{dm}=\rho \mathrm{d} \mathrm{v}$

$\dpi{150} \begin{array}{l} \rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) \\ \\ \therefore \quad m=\int_{0}^{r} \rho(r) \times 4 \pi r^{2} d r \\ \\ =4 \pi \rho_{0}\left(\frac{r^{3}}{3}-\frac{r^{5}}{5 R^{2}}\right) \end{array}$

$\dpi{150} \begin{array}{l} \mathrm{E}_{\mathrm{g}}=\frac{\mathrm{GM}}{\mathrm{r}^{2}} \\\\ \mathrm{E}_{\mathrm{g}}=\mathrm{G} 4 \pi \rho_{0}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^{3}}{5 \mathrm{R}^{2}}\right) \end{array}$

$\dpi{150} \begin{array}{l} \frac{d E_{g}}{d r}=\frac{1}{3}-\frac{3 r^{2}}{5 R^{2}}=0 \\ \\ \frac{1}{3}=\frac{3}{5} \frac{r^{2}}{R^{2}} \\ \\ r=\sqrt{\frac{5}{9}} R \end{array}$

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