The masses m in figures (A) and (B) are ident | vedclass

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(B) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.For case $(A)$,we have three pairs of

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$U_A < U_B$

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(B) The gravitational potential energy of a system of particles is given by $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.For case $(A)$,we have three pairs of masses with distances $d, d,$ and $\sqrt{d^2+d^2} = d\sqrt{2}$.$U_A = -\frac{Gm^2}{d} - \frac{Gm^2}{d} - \frac{Gm^2}{d\sqrt{2}} = -\frac{2Gm^2}{d} - \frac{Gm^2}{d\sqrt{2}}$.For case $(B)$,we have three pairs of masses with distances $d, d,$ and $2d$.$U_B = -\frac{Gm^2}{d} - \frac{Gm^2}{d} - \frac{Gm^2}{2d} = -\frac{2Gm^2}{d} - \frac{Gm^2}{2d}$.Comparing the two,since $\frac{1}{\sqrt{2}} \approx 0.707$ and $\frac{1}{2} = 0.5$,we have $\frac{1}{\sqrt{2}} > \frac{1}{2}$.Therefore,$-\frac{Gm^2}{d\sqrt{2}} This implies $U_A < U_B$.

The masses $m$ in figures $(A)$ and $(B)$ are identical. The gravitational potential energy in the two cases are $U_A$ and $U_B$. Then

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