The masses and radii of the Earth and the Moo | vedclass

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(B) The potential energy $(PE)$ of a particle of mass $m$ placed at a point midway between the Earth and the Moon is given by:$U = -\left(\frac{GM_1 m

Vedclass · Accepted Answer

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

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(B) The potential energy $(PE)$ of a particle of mass $m$ placed at a point midway between the Earth and the Moon is given by:$U = -\left(\frac{GM_1 m}{d/2} + \frac{GM_2 m}{d/2}\right) = -\frac{2Gm(M_1 + M_2)}{d}$The particle will escape to infinity if its total energy becomes at least zero.By the law of conservation of energy,$U + K = 0$,where $K$ is the kinetic energy $\frac{1}{2}mv^2$.$-\frac{2Gm(M_1 + M_2)}{d} + \frac{1}{2}mv^2 = 0$$\frac{1}{2}mv^2 = \frac{2Gm(M_1 + M_2)}{d}$$v^2 = \frac{4G(M_1 + M_2)}{d}$$v = \sqrt{\frac{4G(M_1 + M_2)}{d}}$

The masses and radii of the Earth and the Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is

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