Three particles of mass $m$ are placed at the vertices of an equilateral triangle of side length $a$. A fourth particle of mass $m$ is placed at the centroid of the triangle. What is the gravitational potential energy of the system?

Solution

1. The potential energy of the three particles at the vertices of an equilateral triangle with side length $a$ is $U_{vertices}=-3\frac{G{m}^2}{a}$.
2. The distance from the centroid to each vertex is $r=\frac{a}{\sqrt{3}}$, so the potential energy between the centroid particle and the three vertex particles is $$\begin{array}{rl}{U}_{centroid}& =-3\frac{G{m}^2}{a/\sqrt{3}}\\ & =-3\sqrt{3}\frac{G{m}^2}{a}\end{array}$$.
3. Summing these components gives the total potential energy $$\begin{array}{rl}U& =-\frac{3G{m}^2}{a}-\frac{3\sqrt{3}G{m}^2}{a}\\ & =-\frac{3G{m}^2(1+\sqrt{3})}{a}\end{array}$$.
4. Hence the answer is (C).