Answer
$U = -1.4\times 10^{-6}~J$
Work Step by Step
We can find the gravitational potential energy of the three-particle system:
$U = -\frac{G~m_B~m_C}{r_{BC}}-\frac{G~m_B~m_D}{r_{BD}}-\frac{G~m_C~m_D}{r_{CD}}$
$U = -\frac{(6.67\times 10^{-11}~N~m^2/kg^2)~(35~kg)~(200~kg)}{0.80~m}-\frac{(6.67\times 10^{-11}~N~m^2/kg^2)~(35~kg)~(50~kg)}{0.40~m}-\frac{(6.67\times 10^{-11}~N~m^2/kg^2)~(200~kg)~(50~kg)}{1.20~m}$
$U = -1.4\times 10^{-6}~J$
