A spherically symmetric gravitational system | vedclass

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(C) For a test mass in circular motion,the gravitational force provides the necessary centripetal force: $\frac{G M(r) m}{r^2} = \frac{m v^2}{r}$,whic

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Vedclass · Answer

(C) For a test mass in circular motion,the gravitational force provides the necessary centripetal force: $\frac{G M(r) m}{r^2} = \frac{m v^2}{r}$,which simplifies to $v = \sqrt{\frac{G M(r)}{r}}$.Case $1$: $r \leq R$The mass enclosed within radius $r$ is $M(r) = \rho_0 \cdot \frac{4}{3} \pi r^3$. Substituting this into the speed formula:$v = \sqrt{\frac{G (\rho_0 \cdot \frac{4}{3} \pi r^3)}{r}} = \sqrt{\frac{4}{3} \pi G \rho_0} \cdot r$. Thus,$v \propto r$.Case $2$: $r > R$The total mass of the system is $M = \rho_0 \cdot \frac{4}{3} \pi R^3$. The mass enclosed for $r > R$ is constant $M$. Substituting this into the speed formula:$v = \sqrt{\frac{G M}{r}}$. Thus,$v \propto \frac{1}{\sqrt{r}}$.Comparing these results,the graph shows a linear increase for $r \leq R$ and a decreasing curve proportional to $1/\sqrt{r}$ for $r > R$,which corresponds to option $C$.

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