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

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

Vedclass · Answer

(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:$U = -\left(\frac{GM_1 m}{d/2} + \frac{GM_2 m}{d/2}\right) = -\frac{2Gm(M_1 + M_2)}{d}$To escape to infinity,the total energy of the particle must be at least zero.$U + K = 0$$-\frac{2Gm(M_1 + M_2)}{d} + \frac{1}{2}mv^2 = 0$$\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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