Three identical spheres of mass $m$ are placed at the vertices of an equilateral triangle of side $a$. When released, they interact only through gravitational force and collide after a time $T = 4$ seconds. If the sides of the triangle are increased to length $2a$ and the masses of the spheres are made $2m$, then they will collide after ______ seconds.
Numerical value type. Enter your answer.
Answer: 8
The collision time can depend only on $G$, $m$ and $a$. Let $T \propto G^xm^ya^z$:
$$[T] = (M^{-1}L^3T^{-2})^x M^y L^z \Rightarrow -2x = 1,\ -x + y = 0,\ 3x + z = 0$$
$$x = -\tfrac12,\quad y = -\tfrac12,\quad z = \tfrac32 \Rightarrow T \propto \sqrt{\frac{a^3}{Gm}}$$
With $a \to 2a$ and $m \to 2m$: $T' = 4\times\sqrt{\dfrac{8}{2}} = 4\times2 = 8\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics