The masses and radii of the Earth and Moon ar | vedclass

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Question

(A) To escape to infinity, the total mechanical energy of the body must be at least zero.Let the body be projected with speed $v$ from a distance $r_1

Vedclass · Accepted Answer

$\sqrt{\frac{6G}{d} (M_1 + 2M_2)}$

Vedclass · Answer

(A) To escape to infinity, the total mechanical energy of the body must be at least zero.Let the body be projected with speed $v$ from a distance $r_1 = 2d/3$ from $M_1$. The distance from $M_2$ is $r_2 = d - 2d/3 = d/3$.The potential energy $U$ at this point is $U = -\frac{GM_1m}{r_1} - \frac{GM_2m}{r_2} = -\frac{GM_1m}{2d/3} - \frac{GM_2m}{d/3} = -\frac{3GM_1m}{2d} - \frac{3GM_2m}{d} = -\frac{3Gm}{2d} (M_1 + 2M_2)$.The kinetic energy is $K = \frac{1}{2}mv^2$.For the body to escape to infinity, $K + U = 0$.$\frac{1}{2}mv^2 - \frac{3Gm}{2d} (M_1 + 2M_2) = 0$.$\frac{1}{2}v^2 = \frac{3G}{2d} (M_1 + 2M_2)$.$v^2 = \frac{6G}{d} (M_1 + 2M_2)$.$v = \sqrt{\frac{6G}{d} (M_1 + 2M_2)}$.

The masses and radii of the Earth and 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 body of mass $m$ should be projected from a distance $2d/3$ from the centre of $M_1$ so as to escape to infinity is:

Similar Questions

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