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A spherically symmetric gravitational system of particles has a mass density $\rho=\left\{\begin{array}{ll}\rho_{0} & \text { for } \mathrm{r} \leq \mathrm{R} \\ 0 & \text { for } \mathrm{r}>\mathrm{R}\end{array}\right.$
where $\mathrm{r}_{0}$ is a constant. A test mass can undergo
circular motion under the influence of the gravitational field of particles. Its speed $\mathrm{V}$ as a function of distance $r(0 Options:
where $\mathrm{r}_{0}$ is a constant. A test mass can undergo
circular motion under the influence of the gravitational field of particles. Its speed $\mathrm{V}$ as a function of distance $r(0
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1141 Upvotes
Verified Answer
The correct answer is:
Given mass density \(P=\left\{P_{0}\right.\) for \(r \leq R O\) for \(\left.r>R\right\}\)
Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).
\(\begin{aligned}
&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\
&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\
&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\
&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }
\end{aligned}\)
Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.
Since gravitational depends on mass, which is zero for \(r>R\), hence it only acts for \(r \leq R\).
\(\begin{aligned}
&\overrightarrow{\mathrm{F}}_{\text {graw }}=\overrightarrow{\mathrm{F}}_{\text {circular }} \\
&\Rightarrow \frac{\mathrm{GM}_{\text {sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}} \\
&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text {sphere }}}{\mathrm{R}} \text { since } \mathrm{r} \leq \mathrm{R} \\
&\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text { straight line }+\mathrm{ve} \text { slope }
\end{aligned}\)
Beyond \(\mathrm{r}=\mathrm{R}\), mass density \(=0 \Rightarrow\) acceleration due to gravity \(=0 \Rightarrow\) velocity stays constant.
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