Gravitation Part 1

Assume that at a distance $x$ from the planet of mass $M$ , the net gravitational field becomes zero. $\therefore \;\; \frac{GM}{x^{2}} = \; \frac{G \times 9 M}{(8R-x)^{2}}$ $\Rightarrow \; \frac{1}{x^{2}} = \; \frac{9}{(8R-x)^{2}}$ $\Rightarrow \left( \; \frac{1}{x} \right) = \left( \; \frac{3}{8R-x} \right)^{2} \Rightarrow \; \frac{1}{x} = \; \frac{3}{8R-x}$ $\Rightarrow 3x = 8R - x$ $\Rightarrow 4x = 8R$ $\Rightarrow x = 2R \;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$Now, a satellite should be projected in such a way that its covers a minimum distance of $2R$ . $\therefore \; \frac{1}{2} m v^{2} \;\; - \; \frac{GMm}{R} - \; \frac{G(9M)m}{7R}$ $\;\; = \; \frac{-GMm}{2R} - \; \frac{G(9M)m}{6R}$where, $m$ is the mass of satellite. $\Rightarrow \;\; \; \frac{1}{2} v^{2} = \; \frac{2GM}{7R} \Rightarrow v = \sqrt{\; \frac{4}{7} \; \frac{GM}{R}} \;\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$According to question, the minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\; \frac{a G M}{7R}}$ . So, on comparing it with Eq. (ii), we can write $a = 4$