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(D) The gravitational potential $V$ at a point due to a mass $m$ at distance $r$ is given by $V = -\frac{Gm}{r}$.Here,$m = 1 \ kg$. The objects are pl

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$4G$

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(D) The gravitational potential $V$ at a point due to a mass $m$ at distance $r$ is given by $V = -\frac{Gm}{r}$.Here,$m = 1 \ kg$. The objects are placed at $x = \pm 1, \pm 2, \pm 4, \pm 8, \ldots \ m$.Since there are two objects at each distance $r$ (one at $+r$ and one at $-r$),the total potential $V_{total}$ at $x=0$ is:$V_{total} = \sum -\frac{Gm}{r_i} = -G(1) \left[ \frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \ldots ight]$$V_{total} = -G \left[ 2 \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots ight) ight]$The term in the bracket is a geometric progression with first term $a=1$ and common ratio $r=1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.Therefore,$V_{total} = -G 	imes 2 	imes 2 = -4G$.The magnitude of the potential is $|V_{total}| = 4G$.

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