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(C) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For figure $(i)$,the distances are $r

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$ - \frac{6Gm^2}{a} \left( 1 - \frac{1}{\sqrt{2}} \right) $

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(C) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For figure $(i)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = \sqrt{a^2 + a^2} = a\sqrt{2}$.$U_1 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a\sqrt{2}} \right) = -\frac{Gm^2}{a} \left( 2 + 3 + \frac{6}{\sqrt{2}} \right) = -\frac{Gm^2}{a} (5 + 3\sqrt{2})$.For figure $(ii)$,the distances are $r_{m,2m} = a$,$r_{m,3m} = a$,and $r_{2m,3m} = a$.$U_2 = -\left( \frac{G(m)(2m)}{a} + \frac{G(m)(3m)}{a} + \frac{G(2m)(3m)}{a} \right) = -\frac{Gm^2}{a} (2 + 3 + 6) = -\frac{11Gm^2}{a}$.The work done by an external agent is $W = U_2 - U_1$.$W = -\frac{11Gm^2}{a} - \left( -\frac{Gm^2}{a} (5 + 3\sqrt{2}) \right) = \frac{Gm^2}{a} (-11 + 5 + 3\sqrt{2}) = \frac{Gm^2}{a} (-6 + 3\sqrt{2})$.$W = -\frac{6Gm^2}{a} \left( 1 - \frac{3\sqrt{2}}{6} \right) = -\frac{6Gm^2}{a} \left( 1 - \frac{1}{\sqrt{2}} \right)$.

Consider two configurations of a system of three particles of masses $m, 2m$ and $3m$. The work done by an external agent in changing the configuration of the system from figure $(i)$ to figure $(ii)$ is:

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