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(D) Let the mass of the projected particle be $m_0$. The distance of the particle from each fixed mass $m$ is $l/2$.The gravitational potential energy

Vedclass · Accepted Answer

$2\sqrt{\frac{2Gm}{l}}$

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(D) Let the mass of the projected particle be $m_0$. The distance of the particle from each fixed mass $m$ is $l/2$.The gravitational potential energy $U$ of the particle $m_0$ at the midpoint is given by:$U = -\frac{Gmm_0}{l/2} - \frac{Gmm_0}{l/2} = -\frac{2Gmm_0}{l} - \frac{2Gmm_0}{l} = -\frac{4Gmm_0}{l}$.For the particle to escape to infinity,its total mechanical energy must be at least zero.$E_{total} = KE + PE = 0$$\frac{1}{2}m_0v^2 + U = 0$$\frac{1}{2}m_0v^2 - \frac{4Gmm_0}{l} = 0$$\frac{1}{2}m_0v^2 = \frac{4Gmm_0}{l}$$v^2 = \frac{8Gm}{l}$$v = \sqrt{\frac{8Gm}{l}} = 2\sqrt{\frac{2Gm}{l}}$.

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to:

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