Two particles, each of mass m, are a distance d apart. To bring a third particle, with mass 2m, from far away to a resting point midway between the two particles the work done by an external agent does work given by:

Question

Quizplus · Accepted Answer

To bring the third particle to a resting point midway between the two particles, the work done by an external agent is equal to the gravitational potential energy of the system before and after the third particle is brought in. 
The initial gravitational potential energy is given by:
$U_1=\frac{-Gm^2}{d/2}+\frac{-Gm^2}{d/2}=-\frac{2Gm^2}{d}$
The final gravitational potential energy is given by:
$U_2=\frac{-Gm(2m)}{d/2}+\frac{-Gm(2m)}{d/2}+\frac{G(2m)^2}{d}=-\frac{4Gm^2}{d}$
Therefore, the work done by the external agent is equal to the change in gravitational potential energy:
$W=U_2-U_1=\frac{8Gm^2}{d}$
Substituting the given options, only option D equals the calculated work: 
$-8Gm(\frac{1}{d})(\frac{1}{2})(\frac{1}{2})=\frac{8Gm^2}{d}$

Two Particles, Each of Mass M, Are a Distance D