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(B) The gravitational potential energy $U$ of a system of particles is given by the sum of the potential energies of all possible pairs: $U = -\sum \f

Vedclass · Accepted Answer

$47$/$30$

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(B) The gravitational potential energy $U$ of a system of particles is given by the sum of the potential energies of all possible pairs: $U = -\sum \frac{Gm_i m_j}{r_{ij}}$.For a rectangle with sides $a = 3 l_o$ and $b = 4 l_o$,the diagonal $d = \sqrt{(3 l_o)^2 + (4 l_o)^2} = 5 l_o$.There are $4$ pairs with side length $3 l_o$ (two sides),$4 l_o$ (two sides),and $5 l_o$ (two diagonals).$U = -\left[ \frac{Gm^2}{3 l_o} 	imes 2 + \frac{Gm^2}{4 l_o} 	imes 2 + \frac{Gm^2}{5 l_o} 	imes 2 ight]$$U = -\frac{Gm^2}{l_o} \left[ \frac{2}{3} + \frac{2}{4} + \frac{2}{5} ight]$$U = -\frac{Gm^2}{l_o} \left[ \frac{40 + 30 + 24}{60} ight] = -\frac{94}{60} \frac{Gm^2}{l_o} = -\frac{47}{30} \frac{Gm^2}{l_o}$.The magnitude is $|U| = \frac{47}{30} \frac{Gm^2}{l_o}$.

Four particles each of mass $m$ are placed at four vertices of a rectangle having side lengths $3 l_o$ and $4 l_o$. The magnitude of the gravitational potential energy of the system in units of $\frac{Gm^2}{l_o}$ is:

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