Question 86 of 223

The mass density of a planet of radius R varies with the distance r from its centre as $\rho (r)={\rho }_{0}\left(1-\dfrac{{r}^{2}}{{R}^{2}}\right)$ Then the gravitational field is maximum at:

A$r=\sqrt{\dfrac{3}{4}}R$B$r=R$C$r=\dfrac{1}{\sqrt{3}}R$D$r=\sqrt{\dfrac{5}{9}}R$

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Answer: (D) $r=\sqrt{\dfrac{5}{9}}R$

$dm=\rho \times 4{\pi x}^{2}dx={\rho }_{0}\left(1-\dfrac{{x}^{2}}{{r}^{2}}\right)\times 4{\pi x}^{2}dx$
$m=4\pi \rho {\int }_{0}^{r}\left({x}^{2}-\dfrac{{x}^{4}}{{R}^{2}}\right)dxs$
$m=4{\pi \rho }_{0}\left|\dfrac{{r}^{3}}{3}-\dfrac{{r}^{5}}{5{R}^{2}}\right|$
$E=\dfrac{Gm}{{r}^{2}}=\dfrac{G}{{r}^{2}}\times 4{\pi \rho }_{0}\left(\dfrac{{r}^{3}}{3}-\dfrac{{r}^{5}}{5{R}^{2}}\right)$
$E=4{\pi G\rho }_{0}\left(\dfrac{r}{3}-\dfrac{{r}^{3}}{5{R}^{2}}\right)$
$E$ is maximum when $\dfrac{dE}{dt}=0\Rightarrow \dfrac{dE}{dr}=4{\pi G\rho }_{0}\left(\dfrac{1}{3}-\dfrac{3{r}^{2}}{5{R}^{2}}\right)=0$
$\Rightarrow r=\dfrac{\sqrt{5}}{3}R$
${E}_{\max }=4{\pi G\rho }_{0}\times \dfrac{\sqrt{5R}}{3}\left[\dfrac{1}{3}-\dfrac{1}{5}\times \dfrac{5}{9}\right]$
${E}_{\max }=\dfrac{8\sqrt{5}}{27}{\pi G\rho }_{0}R$