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(A) The gravitational potential energy $U$ of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For the given square configuratio

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$2$

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(A) The gravitational potential energy $U$ of a system of particles is given by $U = -\sum \frac{G m_i m_j}{r_{ij}}$.For the given square configuration with side $d$,there are four sides of length $d$ and two diagonals of length $\sqrt{2}d$.The masses at the corners are $m_1 = m$,$m_2 = M-m$,$m_3 = m$,and $m_4 = M-m$.The potential energy is:$U = -\frac{G}{d} [m(M-m) + (M-m)m + m(M-m) + (M-m)m] - \frac{G}{\sqrt{2}d} [m^2 + (M-m)^2]$$U = -\frac{G}{d} [4m(M-m)] - \frac{G}{\sqrt{2}d} [m^2 + M^2 - 2Mm + m^2]$$U = -\frac{G}{d} [4Mm - 4m^2 + \frac{1}{\sqrt{2}}(M^2 - 2Mm + 2m^2)]$To maximize $U$ (or minimize $|U|$),we differentiate $U$ with respect to $m$ and set it to zero:$\frac{dU}{dm} = -\frac{G}{d} [4M - 8m + \frac{1}{\sqrt{2}}(-2M + 4m)] = 0$$4M - 8m - \sqrt{2}M + 2\sqrt{2}m = 0$$M(4 - \sqrt{2}) = m(8 - 2\sqrt{2})$$M(4 - \sqrt{2}) = 2m(4 - \sqrt{2})$$\frac{M}{m} = 2$Thus,$x = 2$.

$A$ body of mass $2M$ is split into four masses $\{m, M-m, m, M-m\}$,which are placed at the corners of a square of side $d$ as shown in the figure. The ratio $\frac{M}{m}$ for which the gravitational potential energy of the system is maximum is $x:1$. The value of $x$ is ..... .

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