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A spherically symmetric gravitational system of particles has a mass density
$$
\rho=\left\{\begin{array}{l}
\rho_0 \text { for } r \leq R \\
0 \text { for } r>R
\end{array}\right.
$$
where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by
Options:
$$
\rho=\left\{\begin{array}{l}
\rho_0 \text { for } r \leq R \\
0 \text { for } r>R
\end{array}\right.
$$
where $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $v$ as a function of distance $r(0 < r < \infty)$ from the centre of the system is represented by
- A
- B
- C
- D
Solution:
1002 Upvotes
Verified Answer
The correct answer is:
For $r \leq R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}
$$
Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$
Substituting in Eq. (i) we get, $v \propto r$
i.e., $v-r$ graph is a straight line passing through origin.
For $r>R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}
$$
or $\quad V \propto \frac{1}{\sqrt{r}}$
The corresponding $v-r$ graph will be as shown in option (c).
$\therefore$ correct option is (c).
$$
\frac{m v^2}{r}=\frac{G \cdot m m^{\prime}}{r^2}
$$
Here, $m^{\prime}=\left(\frac{4}{3} \pi r^3\right) \rho_0$
Substituting in Eq. (i) we get, $v \propto r$
i.e., $v-r$ graph is a straight line passing through origin.
For $r>R$ :
$$
\frac{m v^2}{r}=\frac{G \cdot m\left(\frac{4}{3} \pi R^3\right) \rho_0}{r^2}
$$
or $\quad V \propto \frac{1}{\sqrt{r}}$
The corresponding $v-r$ graph will be as shown in option (c).
$\therefore$ correct option is (c).
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