Earth has mass M_1 and radius R_1,and the moo | vedclass

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(D) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth and $2r/3$ from the moon is given by:$U = -\frac

Vedclass · Accepted Answer

$\left[\frac{6 G}{r}\left(M_1+\frac{M_2}{2}\right)\right]^{1 / 2}$

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(D) The gravitational potential energy $U$ of the body of mass $M$ at a distance $r/3$ from the Earth and $2r/3$ from the moon is given by:$U = -\frac{G M_1 M}{r/3} - \frac{G M_2 M}{2r/3} = -\frac{3 G M_1 M}{r} - \frac{3 G M_2 M}{2r} = -\frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.To escape to infinity,the total energy must be at least zero. If $v$ is the required escape speed,the kinetic energy is $\frac{1}{2} M v^2$.By conservation of energy: $\frac{1}{2} M v^2 + U = 0$.$\frac{1}{2} M v^2 = \frac{3 G M}{r} \left( M_1 + \frac{M_2}{2} \right)$.Solving for $v$:$v^2 = \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right)$.$v = \left[ \frac{6 G}{r} \left( M_1 + \frac{M_2}{2} \right) \right]^{1/2}$.

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