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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

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

List-$I$	List-$II$
$(A)$ Kinetic energy of planet	$(1)$ $-\frac{GMm}{a}$
$(B)$ Gravitational potential energy of Sun-planet system	$(2)$ $\frac{GMm}{2a}$
$(C)$ Total mechanical energy of planet	$(3)$ $\frac{GM}{r}$
$(D)$ Escape energy at the surface of planet for unit mass object	$(4)$ $-\frac{GMm}{2a}$

$A$ 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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