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Two bodies of masses $4 \mathrm{~m}$ and $9 \mathrm{~m}$ are separated by a distance ' $r$ '. The gravitational potential at a point on this line joining them where the gravitational field becomes zero is
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The correct answer is:
$\frac{-25 G m}{r}$
At the position when the gravitational field is zero. (Distance $x$ is measured from $4 \mathrm{~m}$ mass)
$$
\begin{aligned}
& \frac{G(4 \mathrm{~m})}{x^2}=\frac{G(9 \mathrm{~m})}{(r-x)^2} \\
& \frac{4}{9}=\left(\frac{x}{r-x}\right)^2 \Rightarrow \frac{2}{3}=\frac{x}{r-x} \\
& 2 r-2 x=3 x \Rightarrow 2 r=5 x \Rightarrow x=\frac{2 r}{5}
\end{aligned}
$$
The point $P$ is at a distance $\frac{2 r}{5}$ from mass $4 \mathrm{~m}$ and $\left(r-\frac{2}{5} r\right)=\frac{3 r}{5}$ from mass $m$.
$$
\begin{aligned}
& v=-\frac{G(4 \mathrm{~m})}{\left(\frac{2}{5} r\right)}-\frac{G(9 \mathrm{~m})}{\frac{3 r}{5}}=\frac{-5 G m}{r}\left[\frac{4}{2}+\frac{9}{3}\right] \\
& =\frac{-5 G m}{r}[2+3] \quad=-25 \frac{G m}{r}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{G(4 \mathrm{~m})}{x^2}=\frac{G(9 \mathrm{~m})}{(r-x)^2} \\
& \frac{4}{9}=\left(\frac{x}{r-x}\right)^2 \Rightarrow \frac{2}{3}=\frac{x}{r-x} \\
& 2 r-2 x=3 x \Rightarrow 2 r=5 x \Rightarrow x=\frac{2 r}{5}
\end{aligned}
$$
The point $P$ is at a distance $\frac{2 r}{5}$ from mass $4 \mathrm{~m}$ and $\left(r-\frac{2}{5} r\right)=\frac{3 r}{5}$ from mass $m$.
$$
\begin{aligned}
& v=-\frac{G(4 \mathrm{~m})}{\left(\frac{2}{5} r\right)}-\frac{G(9 \mathrm{~m})}{\frac{3 r}{5}}=\frac{-5 G m}{r}\left[\frac{4}{2}+\frac{9}{3}\right] \\
& =\frac{-5 G m}{r}[2+3] \quad=-25 \frac{G m}{r}
\end{aligned}
$$
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