Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length 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 also the masses of the spheres are made 2m, then they will collide after _________ seconds.

The only quantities that decide the collapse-time for three identical masses released from the corners of an equilateral triangle are

• the initial side length $$a$$,
• the mass of each sphere $$m$$,
• the universal constant of gravitation $$G$$.

Let the required time be $$T$$.  Assume a dependence of the form

$$T \propto G^{\,p}\,m^{\,q}\,a^{\,r}$$

Write dimensions (M ≡ mass, L ≡ length, T ≡ time):

$$[T] = T^{1},\quad [G]=L^{3}M^{-1}T^{-2},\quad [m]=M^{1},\quad [a]=L^{1}$$

Equating powers of the fundamental dimensions,

$$L: \; 0 = 3p + r$$ $$M: \; 0 = -p + q$$ $$T: \; 1 = -2p$$

Solving,

$$p = -\frac12,\quad q = -\frac12,\quad r = \frac32$$

Hence

$$T \propto G^{-1/2}\,m^{-1/2}\,a^{3/2} \;=\;k\,\sqrt{\dfrac{a^{3}}{G\,m}}$$

for some dimensionless constant $$k$$ that is the same for both experiments because the geometry is identical.

Initial situation: side $$=a$$, mass $$=m$$, time $$=T = 4\,$$s.

New situation: side $$=2a$$, mass $$=2m$$. Using the proportionality,

$$\dfrac{T'}{T}=\dfrac{\sqrt{(2a)^{3}/(G\,2m)}}{\sqrt{a^{3}/(G\,m)}}$$

$$\qquad=\,\dfrac{2^{3/2}}{2^{1/2}} = 2$$

Therefore $$T' = 2T = 2 \times 4\text{ s} = 8\text{ s}$$.

So the three spheres will collide after 8 seconds.