IIT-JEE 2008 · Paper 1 · Physics Q29

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PhysicsSingle correct+3 / -1Question 29 of 69

A spherically symmetric gravitational system of particles has a mass density
$\rho=\begin{cases}\rho_0&\text{for }r\le R\\0&\text{for }r>R\end{cases}$
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

V against r: rises linearly from the origin to r = R, then stays constant.

V against r: very large near r = 0, falling steeply and levelling off to a constant beyond R.

V against r: rises linearly from the origin to a peak at r = R, then decreases (as 1/√r) for r > R.

V against r: constant from 0 to R, then decreasing for r > R.

Answer and worked solution

Official answer: C

Orbital speed $V=\sqrt{\dfrac{GM(r)}r}$ .

Inside ( $r\le R$ ): $M(r)=\frac43\pi\rho_0r^3$ , so $V\propto r$ , rising linearly from zero.
Outside: $M$ is constant, so $V\propto\dfrac1{\sqrt r}$ , decreasing.

The speed peaks at $r=R$ : graph (C).

(C)

Topics:Gravitational field and potential Satellites and orbital motion

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IIT-JEE 2008 · Paper 1, Physics question 29