Question

The gravitational potential energy of an isolated system of three particles, each of mass $m$, at the three corners of an equilateral triangle of side $l$ is

• A. $-\frac{G{m}^2}{l}$
• B. $-\frac{G{m}^2}{2l}$
• C. $-\frac{2G{m}^2}{l}$
• D. $-\frac{3G{m}^2}{l}$

Answer 1

Answer: (D)

$-\frac{3G{m}^2}{l}$

$\text{Gravitational potential energy of a pair of particles}=-\frac{G{m}^2}{l}$
$\text{Since,we have three pairs,therefore the total gravitational potential energy is}=-\frac{3G{m}^2}{l}$