Question

A spherically symmetric gravitational system of particles has a mass density $\rho =\rho _{0}$ for $r\leq R$ and $\rho =0$ for $r > R$ , where $\rho _{0}$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of the particles. Which figure represents its speed $v$ as a function of distance $r\left(0 < r < \infty\right)$ from the center of the system?

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

If $M$ is the total mass of the system of particles, then similar to a satellite revolving around the earth, the orbital velocity of the test mass is given by
$v=\sqrt{\frac{G M}{r}}$
For $r\leq R, \, \, v=\sqrt{\frac{G \cdot \frac{4}{3} \pi r^{3} \rho _{0}}{r}}$
i.e. $v \propto \sqrt{r^{2}}$ or $v \propto r$
i.e. $v$ increases linearly with $r$ up to $r = R$
Thus $\left(\right.1\left.\right)$ and $\left(\right.4\left.\right)$ are wrong but $\left(\right.2\left.\right)$ and $\left(\right.3\left.\right)$ satisfy this.
$\therefore \, \, \text{For } r > R , \, v = \sqrt{\frac{G M}{r}} \, \text{or } v \propto \frac{1}{\sqrt{r}}$
Thus is correctly shown in figure $\left(\right.2\left.\right)$ . Figure $\left(\right.3\left.\right)$ is wrong because for $r > R , \, v$ is shown as constant.