Four particles, each of mass $m$, are placed at the corners of a square of side $a$. The minimum work needed to separate them so that they are infinitely far from each other is $(x + \sqrt2)\dfrac{Gm^2}{a}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 4
Concept. The minimum work equals minus the gravitational potential energy of the system, which is the sum over all pairs of $-\dfrac{Gm^2}{r}$.
Step 1. Count the pairs. $4$ pairs along the sides (distance $a$) and $2$ along the diagonals (distance $\sqrt2a$).
Step 2. Potential energy.
$$U = -\frac{Gm^2}{a}\left(4 + \frac{2}{\sqrt2}\right) = -(4 + \sqrt2)\frac{Gm^2}{a}$$
Step 3. Work. $W = -U = (4 + \sqrt2)\dfrac{Gm^2}{a}$, so $x = 4$.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics