Consider two configurations of a system of th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For figure $(i)$,the distances are $r

Vedclass · Accepted Answer

$ - \frac{6Gm^2}{a} \left( 1 - \frac{1}{\sqrt{2}} \right) $

Vedclass · Answer

(C) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For figure $(i)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = \sqrt{a^2 + a^2} = a\sqrt{2}$.$U_1 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a\sqrt{2}} \right) = -\frac{Gm^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{Gm^2}{a} (5 + 3\sqrt{2})$.For figure $(ii)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = a$.$U_2 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a} \right) = -\frac{Gm^2}{a} (2 + 3 + 6) = -\frac{11Gm^2}{a}$.The work done by an external agent is $W = U_2 - U_1$.$W = -\frac{11Gm^2}{a} - \left( -\frac{Gm^2}{a} (5 + 3\sqrt{2}) \right) = \frac{Gm^2}{a} (-11 + 5 + 3\sqrt{2}) = \frac{Gm^2}{a} (-6 + 3\sqrt{2})$.$W = -\frac{6Gm^2}{a} \left( 1 - \frac{3\sqrt{2}}{6} \right) = -\frac{6Gm^2}{a} \left( 1 - \frac{1}{\sqrt{2}} \right)$.

Consider two configurations of a system of three particles of masses $m, 2m$ and $3m$. The work done by an external agent in changing the configuration of the system from figure $(i)$ to figure $(ii)$ is:

Similar Questions

$A$ particle is kept on the surface of a uniform sphere of mass $1000 \ kg$ and radius $1 \ m$. The work done per unit mass against the gravitational force between them is $\left[G=6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}\right]$

Two masses $m_1$ and $m_2$ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently due to gravitational attraction between them,their relative velocity of approach at a separation distance of $d$ is:

$A$ particle of mass $M$ is placed at the centre of a uniform spherical shell of mass $2M$ and radius $R$. The gravitational potential on the surface of the shell is

$A$ particle of mass $m$ is released from infinity and it moves towards a large solid sphere of mass $M$ and radius $R$ made of a jelly-like material. Neglecting the resistance offered by the material of the sphere,find the velocity of the particle when it crosses the centre of the sphere.

Give the formula of gravitational potential at distance $r$ $(r > R_E)$ from the centre of the Earth.

Vedclass Test Series

Exam Paper Generator

Online Exam Module