JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Gravitation

A spherically symmetric gravitational system of particles has a mass density

ρ = {ρ_{0} 0 for r \leq R for r > R

where

ρ_{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

Answer: (C)

Solution

For

r \leq R

, the gravitational force on a test mass at distance

r

is due to the mass enclosed within radius

r

M (r) = \frac{4}{3} π r^{3} ρ_{0}

. For circular motion,

\frac{GM ( r )}{r ^{2}} = \frac{v ^{2}}{r}

, giving

v^{2} = \frac{4 π G ρ _{0} r ^{2}}{3}

, so

v \propto r

. For

r > R

, the total mass is constant:

M = \frac{4}{3} π R^{3} ρ_{0}

. Thus

\frac{GM}{r ^{2}} = \frac{v ^{2}}{r}

, giving

v^{2} = \frac{GM}{r}

, so

v \propto \frac{1}{r}

. Graph (C) shows this behavior.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option