Question

Two stationary particles of masses $M_1$ and $M_2$ are at distance d apart. A third particle, lying on the line joining the particles, experiences no resultant gravitational force. The distance of this particle from $M_1$ is

• A. $d\left(\frac{\sqrt{M_2}}{\sqrt{M_1}-\sqrt{M_2}}\right)$
• B. $d\left(\frac{\sqrt{M_1}}{\sqrt{M_1}+\sqrt{M_2}}\right)$
• C. $d\left(\frac{\sqrt{M_1}}{\sqrt{M_1}-\sqrt{M_2}}\right)$
• D. $d\left(\frac{M_1}{M_1+M_2}\right)$

Answer 1

The correct option is B $d\left(\frac{\sqrt{M_1}}{\sqrt{M_1}+\sqrt{M_2}}\right)$

Force on $m$ towards $M_1$,
$F_1=\frac{G{M}_{1}m}{r^2}$
Force on $m$ towards $M_2$,
$F_2=\frac{G{M}_{2}m}{(d-r{)}^2}$
As there is no resultant force
$\left|\overrightarrow{F_1}\right|=\left|\overrightarrow{F_2}\right|$
$\Rightarrow \frac{G{M}_{1}m}{r^2}=\frac{G{M}_{2}m}{(d-r{)}^2}$
$\Rightarrow {\left(\frac{d-r}{r}\right)}^2=\frac{M_2}{M_1}$
$\Rightarrow \frac{d}{r}-1=\frac{\sqrt{M_2}}{\sqrt{M_1}}$
$\frac{d}{r}=\frac{\sqrt{M_2}+\sqrt{M_1}}{\sqrt{M_1}}$
$\Rightarrow r=d\left[\frac{\sqrt{M_1}}{\sqrt{M_1}+\sqrt{M_2}}\right]$