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(B) To escape the gravitational influence of both masses,the particle must reach a point where the total gravitational potential energy is zero,which

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$V = \sqrt{\frac{4G(M_1 + M_2)}{r}}$

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(B) To escape the gravitational influence of both masses,the particle must reach a point where the total gravitational potential energy is zero,which is at infinity. By the law of conservation of energy,the total energy at the midpoint must be equal to the total energy at infinity (which is $0$).At the midpoint,the distance from each mass is $r/2$.The total energy at the midpoint is the sum of kinetic energy and gravitational potential energy:$E_i = \frac{1}{2}mV^2 - \frac{GM_1m}{r/2} - \frac{GM_2m}{r/2}$Setting the total energy to zero:$\frac{1}{2}mV^2 - \frac{2GM_1m}{r} - \frac{2GM_2m}{r} = 0$$\frac{1}{2}mV^2 = \frac{2Gm}{r}(M_1 + M_2)$$V^2 = \frac{4G(M_1 + M_2)}{r}$$V = \sqrt{\frac{4G(M_1 + M_2)}{r}}$

The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

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