Question 133 of 223

A body of mass $\left(2M\right)$ splits into four masses ${m,M-m,m,M-m},$ which are rearranged to form a square as shown in the figure. The ratio of $\dfrac{M}{m}$ for which, the gravitational potential energy of the system becomes maximum is $x:1.$ The value of $x$ is ________.

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Answer: 2

$U=\dfrac{G{m}^{2}}{a}+\dfrac{G(M-m{)}^{2}}{a}+\dfrac{2Gm(M-m)}{a}+\dfrac{2Gm(M-m)}{\sqrt{2}a}$

For maxima or minima of potential energy $\dfrac{dU}{dm}=0$
$\dfrac{dU}{dm}=\dfrac{G}{a}(2m-2(M-m)+(M-2m)2+\sqrt{2}(M-2m)$
$(4m-4m-2\sqrt{2}m)+(-2M+2M+\sqrt{2}M)=0;\sqrt{2}M=2\sqrt{2}m;\dfrac{M}{m}=2$