Two planets have masses M and 16M and their radii are $$a$$ and $$2a$$, respectively. The separation between the centres of the planets is $$10a$$. A body of mass $$m$$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is:

Let $$x$$ be the distance of the neutral point from the center of the smaller planet ($$M$$).

$$\frac{GMm}{x^2} = \frac{G(16M)m}{(10a - x)^2} \implies \frac{1}{x} = \frac{4}{10a - x} \implies 10a - x = 4x \implies x = 2a$$

Distance of the neutral point from the center of the larger planet: $$10a - x = 8a$$

Using conservation of energy from the surface of the larger planet to the neutral point: $$E_i = E_f$$

$$\frac{1}{2}mv^2 - \frac{GMm}{10a - 2a} - \frac{G(16M)m}{2a} = -\frac{GMm}{2a} - \frac{G(16M)m}{8a}$$

$$\frac{1}{2}v^2 - \frac{GM}{8a} - \frac{16GM}{2a} = -\frac{GM}{2a} - \frac{16GM}{8a}$$

$$\frac{1}{2}v^2 - \frac{GM}{a}\left(\frac{1}{8} + 8\right) = -\frac{GM}{a}\left(\frac{1}{2} + 2\right)$$

$$\frac{1}{2}v^2 - \frac{65GM}{8a} = -\frac{5GM}{2a}$$

$$\frac{1}{2}v^2 = \frac{65GM}{8a} - \frac{20GM}{8a} = \frac{45GM}{8a} \implies v^2 = \frac{45GM}{4a} \implies v = \frac{3}{2}\sqrt{\frac{5GM}{a}}$$