Suppose two planets (spherical in shape) of r | vedclass

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(A) To reach the surface of the second planet,the satellite must cross the point where the net gravitational field is zero.Let the distance of this po

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) To reach the surface of the second planet,the satellite must cross the point where the net gravitational field is zero.Let the distance of this point from the centre of the planet of mass $M$ be $x$.$\frac{GM}{x^2} = \frac{G(9M)}{(8R-x)^2}$$\frac{1}{x} = \frac{3}{8R-x}$$8R-x = 3x \Rightarrow x = 2R$.Now,apply the law of conservation of energy between the surface of the first planet and the point $x = 2R$ (where the velocity is minimum,i.e.,$v_{min} = 0$):$E_{initial} = E_{final}$$\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(9M)m}{7R} = 0 - \frac{GMm}{2R} - \frac{G(9M)m}{6R}$$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{9GM}{7R} - \frac{GM}{2R} - \frac{9GM}{6R}$$\frac{1}{2}v^2 = \frac{GM}{R} [1 + \frac{9}{7} - \frac{1}{2} - \frac{3}{2}]$$\frac{1}{2}v^2 = \frac{GM}{R} [1 + \frac{9}{7} - 2] = \frac{GM}{R} [\frac{9}{7} - 1] = \frac{GM}{R} [\frac{2}{7}]$$v^2 = \frac{4}{7} \frac{GM}{R}$$v = \sqrt{\frac{4}{7} \frac{GM}{R}}$Comparing this with $\sqrt{\frac{a}{7} \frac{GM}{R}}$,we get $a = 4$.

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