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(D) The gravitational potential at the mid-point $P$ due to two fixed particles of mass $m$ is given by:$V = V_1 + V_2 = -\frac{Gm}{l/2} - \frac{Gm}{l

Vedclass · Accepted Answer

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

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(D) The gravitational potential at the mid-point $P$ due to two fixed particles of mass $m$ is given by:$V = V_1 + V_2 = -\frac{Gm}{l/2} - \frac{Gm}{l/2} = -\frac{4Gm}{l}$The gravitational potential energy of the particle of mass $m_0$ at point $P$ is:$U = m_0 V = -\frac{4Gmm_0}{l}$For the particle to just escape to infinity,its total energy at the point of projection must be equal to its total energy at infinity,which is zero.By the law of conservation of energy:$K_i + U_i = K_f + U_f$$\frac{1}{2} m_0 v^2 + \left( -\frac{4Gmm_0}{l} \right) = 0 + 0$$\frac{1}{2} m_0 v^2 = \frac{4Gmm_0}{l}$$v^2 = \frac{8Gm}{l}$$v = \sqrt{\frac{8Gm}{l}} = 2 \sqrt{\frac{2Gm}{l}}$Thus,the minimum velocity is $2 \sqrt{\frac{2Gm}{l}}.$

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

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