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(D) Let the point $P$ be the neutral point where the gravitational field is zero. Let $r_1$ be the distance from mass $m$ and $r_2$ be the distance fr

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$-\frac{16 G m}{R}$

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(D) Let the point $P$ be the neutral point where the gravitational field is zero. Let $r_1$ be the distance from mass $m$ and $r_2$ be the distance from mass $9m$.The condition for zero gravitational field is:$\frac{G m}{r_1^2} = \frac{G (9m)}{r_2^2}$$\frac{1}{r_1^2} = \frac{9}{r_2^2} \implies \frac{r_2}{r_1} = 3 \implies r_2 = 3 r_1$Since $r_1 + r_2 = R$,we have $r_1 + 3 r_1 = R \implies 4 r_1 = R \implies r_1 = \frac{R}{4}$ and $r_2 = \frac{3R}{4}$.The gravitational potential $V$ at point $P$ is the sum of potentials due to both masses:$V = -\frac{G m}{r_1} - \frac{G (9m)}{r_2}$$V = -\frac{G m}{R/4} - \frac{9 G m}{3R/4}$$V = -\frac{4 G m}{R} - \frac{36 G m}{3R} = -\frac{4 G m}{R} - \frac{12 G m}{R}$$V = -\frac{16 G m}{R}$

Two bodies of mass $m$ and $9 m$ are placed at a distance $R$. The gravitational potential on the line joining the bodies where the gravitational field equals zero,will be ($G=$ gravitational constant):

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