Three masses of $1kg, 2kg$ and $3kg$ are placed at the vertices of an equilateral triangle of side $1m$. Find the gravitational potential energy of this system.

Take $G=6.67\times {10}^{-11}N·m^2/k{g}^2$

$U=-G\left(\frac{m_3{m}_2}{n_2}+\frac{m_2{m}_1}{r_{31}}+\frac{m_2{m}_1}{r_{21}}\right)$

Here, $$\begin{gathered} r_{32}=r_{31}=r_{21}=1.0m m_1=1kg \\ m_2=2kg \end{gathered}$$

and            $m_3=3kg$.

Substituting the values, we get

    $U=-(6.67\times {10}^{-11})\left(\frac{3\times 2}{1}+\frac{3\times 1}{1}+\frac{2\times 1}{1}\right)$

or $U=-7.337\times {10}^{-10}J$