The mass density of a planet of radius R vari | vedclass

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(B) The gravitational field $E$ at a distance $r$ inside the planet is given by Gauss's Law for gravity: $E(4\pi r^{2}) = 4\pi G M(r)$,where $M(r)$ is

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$r = \sqrt{\frac{5}{9}} R$

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(B) The gravitational field $E$ at a distance $r$ inside the planet is given by Gauss's Law for gravity: $E(4\pi r^{2}) = 4\pi G M(r)$,where $M(r)$ is the mass enclosed within radius $r$.$M(r) = \int_{0}^{r} \rho(r) 4\pi r^{2} dr = 4\pi \rho_{0} \int_{0}^{r} \left(1 - \frac{r^{2}}{R^{2}}\right) r^{2} dr$.$M(r) = 4\pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^{5}}{5R^{2}} \right]$.Thus,$E = \frac{G M(r)}{r^{2}} = 4\pi G \rho_{0} \left( \frac{r}{3} - \frac{r^{3}}{5R^{2}} \right)$.To find the maximum field,set $\frac{dE}{dr} = 0$:$\frac{dE}{dr} = 4\pi G \rho_{0} \left( \frac{1}{3} - \frac{3r^{2}}{5R^{2}} \right) = 0$.$\frac{1}{3} = \frac{3r^{2}}{5R^{2}} \Rightarrow r^{2} = \frac{5R^{2}}{9} \Rightarrow r = \sqrt{\frac{5}{9}} R$.

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:

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