Two stationary particles of masses $M_{1}$ and $M_{2}$ are $' d'$ distance apart. A third particle lying on the line joining the particles, experiences no resultant gravitational force. What is distance of this particle from $M_{1}$ ?

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$d [\frac{\sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}]$

$d [\frac{2 \sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}]$

$d [\frac{4 \sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}]$

$d [\frac{\sqrt[3]{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}]$

answer is A.

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Detailed Solution

Let $m$ be the mass of the third particle
Force on $m$ towards $M_{1}$ is $F_{1}$ $= \frac{G M_{1} m}{r^{2}}$

Force on $m$ towards $M_{2}$ is $F_{2} = \frac{G M_{2} m}{{(d - r)}^{2}}$
Since net force on $m$ is zero $∴ F_{1} = F_{2}$
$\Rightarrow \frac{G M_{1} m}{r^{2}} = \frac{G M_{2} m}{{(d - r)}^{2}} \Rightarrow {(\frac{d - r}{r})}^{2} = \frac{M_{2}}{M_{1}} \Rightarrow \frac{d}{r} - 1 = \frac{\sqrt{M_{2}}}{\sqrt{M_{1}}} \Rightarrow r = d [\frac{\sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}]$