Q. Two bodies of masses $m$ and $4m$ are placed at a distance $r$ . The gravitational potential at a point on the line joining them where the gravitational field is zero is,

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Solution:

$\frac{G m}{x^{2}}=\frac{G \left(\right. 4 m \left.\right)}{\left(\right. r - x \left(\left.\right)^{2}}\Rightarrow \frac{1}{x}=\frac{2}{r - x}$
$r-x=2x\Rightarrow 3x=1\Rightarrow x=\frac{1}{3}$
Solution
$V=-\frac{G m}{\left(r/3\right)}-\frac{G \left(\right. 4 m \left.\right)}{\left(\left(2r\right)/3\right)}$
$\Rightarrow V=-\frac{3 G m}{r}-\frac{6 G m}{r}=\frac{9 G m}{r}$ .