Q. A spherically symmetric gravitational system of particles has a mass density $\rho = \left\{ \begin{array}{r} \rho_{0}\ for\ r \leq R \\ 0\ 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) [image:8_image31.emf]
B) [image:8_image32.emf]
C) [image:8_image33.emf]
D) [image:8_image34.emf]
Explanation
(c) $\mathbf{For\ }\mathbf{r \leq R:}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{mv^{2}}{r} = \frac{Gmm^{'}}{r^{2}}$
$Here,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m^{'} = \left( \frac{4}{3}{\pi r}^{3} \right)\rho_{0}$
Substituting in Eq. (i) we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v \propto r$
$ie,\ v–r$ graph is a straight line passing through orgine.
$\mathbf{For\ }\mathbf{r} > R:$
$\frac{{mv}^{\mathbf{2}}}{r} = \frac{G\ m\left( \frac{4}{3}\pi R^{3} \right)\rho_{0}}{r^{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v \propto \frac{1}{\sqrt{r}}$
The corresponding $v–r$ graph will be as shown in option (c).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.