Earth has mass M_{1} and radius R_{1}. Moon h | vedclass

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Question

(B) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth's centre is given by the sum of potentials due t

Vedclass · Accepted Answer

$\left[\frac{6 G}{r}\left(M_{1}+\frac{M_{2}}{2}\right)\right]^{\frac{1}{2}}$

Vedclass · Answer

(B) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth's centre is given by the sum of potentials due to Earth and Moon:$U = -\frac{G M_{1} M}{r/3} - \frac{G M_{2} M}{2r/3} = -\frac{3 G M_{1} M}{r} - \frac{3 G M_{2} M}{2r} = -\frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.To escape to infinity,the total energy must be at least zero. Thus,the kinetic energy $K$ required is equal to the magnitude of the potential energy:$K = \frac{1}{2} M V^{2} = |U| = \frac{3 G M}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$.Solving for $V$:$V^{2} = \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right)$,$V = \left[ \frac{6 G}{r} \left( M_{1} + \frac{M_{2}}{2} \right) \right]^{\frac{1}{2}}$.Therefore,the correct option is $B$.

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