A spherically symmetric gravitational system of particles has a mass density where is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed as a function of distance from the centre

Question

A spherically symmetric gravitational system of particles has a mass density $ρ = {\begin{matrix} ρ_{0} & for & r \leq R \\ 0 & for & r > R \end{matrix}$
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 – [2008]

Smart Achievers · Accepted Answer

(3)

$For r \geq R$

$\frac{m v^{2}}{r} = m \times [\frac{G M}{r^{2}}] where M is the total mass of the spherical system.$

$∴$ $v \propto \frac{1}{\sqrt{r}}$ $(∵ M = (\frac{4}{3} π R^{3}) ρ_{0})$

$For r < R$

$∴$ $\frac{m v^{2}}{r} = m [\frac{G M r}{R^{3}}] \Rightarrow v \propto r$

i.e., $v – r$ graph is a straight line passing through origin.

Solution

1

2

3

4

Want Us to Email you About Special Offers & Updates?

Solution

1

2

3

4

Ans. (3) For r≥R mv2r=m×[GMr2] where M is the total mass of the spherical system. ∴ v∝1r (∵M=(43πR3)ρ0) For r<R ∴ mv2r=m[GMrR3] ⇒ v∝r i.e., v–r graph is a straight line passing through origin.

Want Us to Email you About Special Offers & Updates?