A spherically symmetric gravitational system of particles has a mass density \rho = \ \beginmatrix \rho _0 & for & r \le R 0 & for & r \gt R \endmatrix . 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 \lt r \lt \infty) from the centre of the system is represented by

Gravitation PYQ — JEE Advanced 2008 · Shift 1 · Q52 with Solution

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Correct answer: C

Interpret the mass distribution

The density is

\rho(r)= \begin{cases} \rho_0, & r\le R \\ 0, & r>R \end{cases}

So this is a uniform solid sphere of radius $R$ and constant density $\rho_0$ .

We need the speed $V(r)$ of a test mass moving in a circular orbit under gravity.

Use gravitational force as centripetal force

For circular motion,

\frac{mV^2}{r}=F_g

where $F_g$ is the gravitational attraction on the test mass $m$ .

Because the distribution is spherically symmetric, by the shell theorem:

only the mass enclosed within radius $r$ contributes for $r<R$ ,
for $r>R$ , the sphere behaves like a point mass of total mass $M$ at the centre.

Case 1: Inside the sphere ( $r\le R$ )

Enclosed mass at radius $r$ is

M(r)=\rho_0\cdot \frac{4}{3}\pi r^3

So gravitational force on test mass $m$ is

F_g=\frac{GmM(r)}{r^2} =\frac{Gm}{r^2}\left(\frac{4}{3}\pi \rho_0 r^3\right) =\frac{4}{3}\pi G\rho_0 \, m r

Now equate with centripetal force:

\frac{mV^2}{r}=\frac{4}{3}\pi G\rho_0 \, m r

V^2=\frac{4}{3}\pi G\rho_0 r^2

Hence,

V=r\sqrt{\frac{4}{3}\pi G\rho_0}

So inside the sphere, $V\propto r$ .

Case 2: Outside the sphere ( $r>R$ )

Total mass of the sphere is

M=\rho_0\cdot \frac{4}{3}\pi R^3

Then gravitational force is

F_g=\frac{GMm}{r^2}

For circular motion,

\frac{mV^2}{r}=\frac{GMm}{r^2}

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

Substitute $M$ :

V^2=\frac{G\left(\frac{4}{3}\pi \rho_0 R^3\right)}{r}

Thus,

V=\sqrt{\frac{4}{3}\pi G\rho_0 R^3\cdot \frac{1}{r}}

So outside the sphere, $V\propto \frac{1}{\sqrt r}$ .

Check continuity at $r=R$

From inside:

V(R)=R\sqrt{\frac{4}{3}\pi G\rho_0}

From outside:

V(R)=\sqrt{\frac{4}{3}\pi G\rho_0 R^2}=R\sqrt{\frac{4}{3}\pi G\rho_0}

So the graph is continuous at $r=R$ .

Shape of the graph

Therefore:

from $r=0$ to $r=R$ , speed increases linearly with $r$ ,
at $r=R$ , speed is maximum,
for $r>R$ , speed decreases as $1/\sqrt r$ .

So the correct graph is the one that rises linearly from the origin up to $r=R$ , and then falls gradually like $1/\sqrt r$ .

This corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.

More from Gravitation

View written solutionFree

Correct answer: C

Interpret the mass distribution

The density is

\rho(r)= \begin{cases} \rho_0, & r\le R \\ 0, & r>R \end{cases}

So this is a uniform solid sphere of radius $R$ and constant density $\rho_0$ .

We need the speed $V(r)$ of a test mass moving in a circular orbit under gravity.

Use gravitational force as centripetal force

For circular motion,

\frac{mV^2}{r}=F_g

where $F_g$ is the gravitational attraction on the test mass $m$ .

Because the distribution is spherically symmetric, by the shell theorem:

only the mass enclosed within radius $r$ contributes for $r<R$ ,
for $r>R$ , the sphere behaves like a point mass of total mass $M$ at the centre.

Case 1: Inside the sphere ( $r\le R$ )

Enclosed mass at radius $r$ is

M(r)=\rho_0\cdot \frac{4}{3}\pi r^3

So gravitational force on test mass $m$ is

F_g=\frac{GmM(r)}{r^2} =\frac{Gm}{r^2}\left(\frac{4}{3}\pi \rho_0 r^3\right) =\frac{4}{3}\pi G\rho_0 \, m r

Now equate with centripetal force:

\frac{mV^2}{r}=\frac{4}{3}\pi G\rho_0 \, m r

V^2=\frac{4}{3}\pi G\rho_0 r^2

Hence,

V=r\sqrt{\frac{4}{3}\pi G\rho_0}

So inside the sphere, $V\propto r$ .

Case 2: Outside the sphere ( $r>R$ )

Total mass of the sphere is

M=\rho_0\cdot \frac{4}{3}\pi R^3

Then gravitational force is

F_g=\frac{GMm}{r^2}

For circular motion,

\frac{mV^2}{r}=\frac{GMm}{r^2}

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

Substitute $M$ :

V^2=\frac{G\left(\frac{4}{3}\pi \rho_0 R^3\right)}{r}

Thus,

V=\sqrt{\frac{4}{3}\pi G\rho_0 R^3\cdot \frac{1}{r}}

So outside the sphere, $V\propto \frac{1}{\sqrt r}$ .

Check continuity at $r=R$

From inside:

V(R)=R\sqrt{\frac{4}{3}\pi G\rho_0}

From outside:

V(R)=\sqrt{\frac{4}{3}\pi G\rho_0 R^2}=R\sqrt{\frac{4}{3}\pi G\rho_0}

So the graph is continuous at $r=R$ .

Shape of the graph

Therefore:

from $r=0$ to $r=R$ , speed increases linearly with $r$ ,
at $r=R$ , speed is maximum,
for $r>R$ , speed decreases as $1/\sqrt r$ .

So the correct graph is the one that rises linearly from the origin up to $r=R$ , and then falls gradually like $1/\sqrt r$ .

This corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.

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