Two planets have masses M and 16, M and their | vedclass

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(B) To reach the smaller planet,the body must cross the point of zero gravitational field (neutral point) between the two planets.Let the distance of

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$\frac{3}{2} \sqrt{\frac{5GM}{a}}$

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(B) To reach the smaller planet,the body must cross the point of zero gravitational field (neutral point) between the two planets.Let the distance of the neutral point from the center of the smaller planet be $x$.Equating the gravitational fields from both planets at this point:$\frac{GM}{x^2} = \frac{G(16M)}{(10a - x)^2}$Taking the square root on both sides:$\frac{1}{x} = \frac{4}{10a - x} \implies 10a - x = 4x \implies x = 2a$.Now,apply the Law of Conservation of Mechanical Energy between the surface of the larger planet and the neutral point.The potential at the surface of the larger planet (radius $2a$,mass $16M$) is $V_L = -\frac{G(16M)}{2a} - \frac{GM}{8a} = -\frac{8GM}{a} - \frac{GM}{8a} = -\frac{65GM}{8a}$.The potential at the neutral point (distance $8a$ from larger,$2a$ from smaller) is $V_P = -\frac{G(16M)}{8a} - \frac{GM}{2a} = -\frac{2GM}{a} - \frac{GM}{2a} = -\frac{5GM}{2a}$.Using $KE_i + PE_i = KE_f + PE_f$,where $KE_f = 0$ at the neutral point:$\frac{1}{2}mv^2 + m V_L = 0 + m V_P$$\frac{1}{2}v^2 = V_P - V_L = -\frac{5GM}{2a} - (-\frac{65GM}{8a}) = \frac{-20GM + 65GM}{8a} = \frac{45GM}{8a}$.$v^2 = \frac{45GM}{4a} \implies v = \sqrt{\frac{45GM}{4a}} = \frac{3}{2} \sqrt{\frac{5GM}{a}}$.

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