The mass and radius of the Earth and Moon are | vedclass

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(A) Let the mass $m$ be at a distance $r_1 = \frac{2d}{3}$ from the centre of $M_1$. Then,its distance from the centre of $M_2$ is $r_2 = d - \frac{2d

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$\left[\frac{3 G(M_1+2 M_2)}{d}\right]^{\frac{1}{2}}$

Vedclass · Answer

(A) Let the mass $m$ be at a distance $r_1 = \frac{2d}{3}$ from the centre of $M_1$. Then,its distance from the centre of $M_2$ is $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.To escape the gravitational influence of the system,the total energy of the body must be at least zero at infinity.By the law of conservation of energy,the kinetic energy provided at the initial position must equal the magnitude of the gravitational potential energy at that position:$\frac{1}{2} m v_e^2 = \frac{G M_1 m}{r_1} + \frac{G M_2 m}{r_2}$Substituting the values of $r_1$ and $r_2$:$\frac{1}{2} m v_e^2 = \frac{G M_1 m}{(2d/3)} + \frac{G M_2 m}{(d/3)}$$\frac{1}{2} m v_e^2 = \frac{3 G M_1 m}{2d} + \frac{3 G M_2 m}{d}$$\frac{1}{2} m v_e^2 = \frac{3 G m}{2d} (M_1 + 2 M_2)$$v_e^2 = \frac{3 G (M_1 + 2 M_2)}{d}$$v_e = \left[\frac{3 G (M_1 + 2 M_2)}{d}\right]^{\frac{1}{2}}$

The mass and radius 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 $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is:

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