A spherically symmetric gravitational system | vedclass

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(C) For a spherically symmetric system,the gravitational force on a test mass $m$ at a distance $r$ is provided by the mass $M(r)$ contained within a

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(C) For a spherically symmetric system,the gravitational force on a test mass $m$ at a distance $r$ is provided by the mass $M(r)$ contained within a sphere of radius $r$.Case $1$: For $r \leq R$,the mass $M(r) = \rho_0 \cdot \frac{4}{3} \pi r^3$.The gravitational force is $F = \frac{G M(r) m}{r^2} = \frac{G m}{r^2} \cdot \rho_0 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi G \rho_0 m r$.For circular motion,the centripetal force is $\frac{m V^2}{r} = F = \frac{4}{3} \pi G \rho_0 m r$.Thus,$V^2 = \frac{4}{3} \pi G \rho_0 r^2$,which implies $V \propto r$.Case $2$: For $r > R$,the total mass $M = \rho_0 \cdot \frac{4}{3} \pi R^3$ acts as a point mass at the center.The gravitational force is $F = \frac{G M m}{r^2}$.For circular motion,$\frac{m V^2}{r} = \frac{G M m}{r^2}$,which gives $V^2 = \frac{G M}{r}$.Thus,$V \propto \frac{1}{\sqrt{r}}$.Therefore,the speed $V$ increases linearly with $r$ for $r \leq R$ and decreases as $1/\sqrt{r}$ for $r > R$. This corresponds to the graph in option $C$.

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