Masses and radii of Earth and Moon are M_1, M | vedclass

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(A) The gravitational potential energy $U$ of a mass $m$ placed at the midpoint between $M_1$ and $M_2$ is the sum of the potentials due to both masse

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$\sqrt{\frac{4G(M_1 + M_2)}{d}}$

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(A) The gravitational potential energy $U$ of a mass $m$ placed at the midpoint between $M_1$ and $M_2$ is the sum of the potentials due to both masses:$U = -\frac{G M_1 m}{d/2} - \frac{G M_2 m}{d/2} = -\frac{2 G m}{d} (M_1 + M_2)$To escape the system,the total energy must be at least zero. By the law of conservation of energy,the kinetic energy provided must equal the magnitude of the potential energy:$\frac{1}{2} m v_e^2 = |U| = \frac{2 G m}{d} (M_1 + M_2)$Solving for the escape velocity $v_e$:$v_e^2 = \frac{4 G (M_1 + M_2)}{d}$$v_e = \sqrt{\frac{4 G (M_1 + M_2)}{d}}$

Masses and radii of Earth and Moon are $M_1, M_2$ and $R_1, R_2$ respectively. The distance between their centers is $d$. The minimum velocity given to a mass $m$ from the midpoint of the line joining their centers so that it escapes the gravitational field of both is:

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