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Physics Question on Gravitation

"A spherically symmetric gravitational system of particles has a mass density \rho =\Biggl \\{ \begin{array} \ \rho_0 \ \text{for} \ r \le R\\\ 0 \ \text{for} r\ > \ R \\\ \end{array} , where ρ0\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 from the centre of the system is represented by"

A

B

C

D

Answer

Explanation

Solution

"For rR r \le R mv2r=GmMr2...(i) \, \, \, \, \, \, \, \frac{mv^2}{r}= \frac{GmM}{r^2} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i) Here,M=(43πr3)ρ0Here, \, \, \, \, \, \, \, \, \, M= \bigg( \frac{4}{3} \pi r ^3 \bigg) \rho_0 Substituting in E (i), we get v \propto r For r> R, mv2r=Gm(43πR3)ρ0r2\, \, \, \, \, \, \, \, \, \, \, \, \frac{mv^2}{r}= \frac{Gm \big( \frac{4}{3} \pi R^3 \big) \rho_0 } {r^2} or v1r\, \, \, \, \, \, \, \, \, v \propto \frac{1}{ \sqrt{r} } The corresponding v-r graph will be as shown in option (c). \therefore Correct option is (c)."