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(A) Let the point where the gravitational field is zero be at a distance $x$ from the mass $4 \,m$. The distance from the mass $9 \,m$ will be $(r - x

Vedclass · Accepted Answer

$\frac{-25 G m}{r}$

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(A) Let the point where the gravitational field is zero be at a distance $x$ from the mass $4 \,m$. The distance from the mass $9 \,m$ will be $(r - x)$.At this point,the magnitudes of the gravitational fields due to both masses are equal:$\frac{G(4 \,m)}{x^2} = \frac{G(9 \,m)}{(r - x)^2}$$\frac{4}{9} = \left(\frac{x}{r - x}\right)^2$Taking the square root on both sides:$\frac{2}{3} = \frac{x}{r - x}$$2(r - x) = 3x \Rightarrow 2r - 2x = 3x \Rightarrow 5x = 2r \Rightarrow x = \frac{2r}{5}$So,the distance from $4 \,m$ is $\frac{2r}{5}$ and from $9 \,m$ is $r - \frac{2r}{5} = \frac{3r}{5}$.The gravitational potential $V$ at this point is the sum of the potentials due to both masses:$V = -\frac{G(4 \,m)}{x} - \frac{G(9 \,m)}{r - x}$$V = -\frac{G(4 \,m)}{\frac{2r}{5}} - \frac{G(9 \,m)}{\frac{3r}{5}}$$V = -\frac{20Gm}{2r} - \frac{45Gm}{3r} = -\frac{10Gm}{r} - \frac{15Gm}{r} = -\frac{25Gm}{r}$

Two bodies of masses $4 \,m$ and $9 \,m$ are separated by a distance $r$. The gravitational potential at a point on the line joining them where the gravitational field becomes zero is:

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