Suppose two planets (spherical in shape) of radii $R$ and $2R$, but mass $M$ and $9M$ respectively have a centre to centre separation $8R$ as shown in the figure. A satellite of mass $m$ is projected from the surface of the planet of mass $M$ directly towards the centre of the second planet. The minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\dfrac a7\dfrac{GM}{R}}$, then the value of $a$ is ______.
[Given: The two planets are fixed in their position]
Numerical value type. Enter your answer.
Answer: 4
Neutral point at distance $x$ from the centre of $M$: $\dfrac{GM}{x^2} = \dfrac{9GM}{(8R - x)^2} \Rightarrow 8R - x = 3x \Rightarrow x = 2R$.
Beyond this point the bigger planet pulls the satellite in, so it only has to reach $x = 2R$ with (nearly) zero speed.
Potential per unit mass at the surface of $M$ ($x = R$, distance $7R$ from $9M$): $-\dfrac{GM}{R} - \dfrac{9GM}{7R} = -\dfrac{16GM}{7R}$
At $x = 2R$ (distance $6R$ from $9M$): $-\dfrac{GM}{2R} - \dfrac{9GM}{6R} = -\dfrac{2GM}{R}$
$$\frac12 v^2 = -\frac{2GM}{R} + \frac{16GM}{7R} = \frac{2GM}{7R} \Rightarrow v = \sqrt{\frac47\frac{GM}{R}}$$
So $a = 4$.
Solution by Sreeraj P, M.Sc Physics