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 the surface of the smaller planet, the minimum firing speed needed is:
Answer: (D) $\dfrac32\sqrt{\dfrac{5GM}{a}}$
The body must just get past the neutral point, where the two pulls balance. At distance $x$ from the smaller planet:
$$\frac{GM}{x^2} = \frac{16GM}{(10a - x)^2} \Rightarrow 10a - x = 4x \Rightarrow x = 2a$$
so the neutral point is $8a$ from the larger planet. Beyond it the smaller planet pulls the body in.
Potential energy at the surface of the larger planet:
$$U_1 = -\frac{16GMm}{2a} - \frac{GMm}{8a} = -\frac{65GMm}{8a}$$
At the neutral point:
$$U_2 = -\frac{16GMm}{8a} - \frac{GMm}{2a} = -\frac{20GMm}{8a}$$
$$\frac12mv^2 = U_2 - U_1 = \frac{45GMm}{8a} \Rightarrow v = \sqrt{\frac{45GM}{4a}} = \frac32\sqrt{\frac{5GM}{a}}$$
Solution by Sreeraj P, M.Sc Physics