Question 131 of 223

Four spheres each of mass $m$ form $a$ square of side $d$ (as shown in figure). A fifth sphere of mass $M$ is situated at the centre of square. The total gravitational potential energy of the system is

A$-\dfrac{Gm}{d}\left[\left(4+\sqrt{2}\right)m+4\sqrt{2}M\right]$B$-\dfrac{Gm}{d}\left[\left(4+\sqrt{2}\right)M+4\sqrt{2}m\right]$C$-\dfrac{Gm}{d}\left[3{m}^{2}+4\sqrt{2}M\right]$D$-\dfrac{Gm}{d}\left[6{m}^{2}+4\sqrt{2}M\right]$

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Answer: (A) $-\dfrac{Gm}{d}\left[\left(4+\sqrt{2}\right)m+4\sqrt{2}M\right]$

The total gravitational potential energy of the system is
${U}_{system}=\left({U}_{12}+{U}_{23}+{U}_{34}+{U}_{41}\right)+\left({U}_{13}+{U}_{24}\right)+\left({U}_{15}+{U}_{25}+{U}_{35}+{U}_{45}\right)$
$=-\dfrac{4Gmm}{d}-\dfrac{2Gmm}{d\sqrt{2}}-\dfrac{4\sqrt{2}GMm}{d}$
$=-\dfrac{\left(4+\sqrt{2}\right)G{m}^{2}}{d}-\dfrac{4\sqrt{2}GMm}{d}$
$=-\dfrac{Gm}{d}\left[\left(4+\sqrt{2}\right)m+4\sqrt{2}M\right]$