Question

Two bodies A and B of masses $m$ and $2m$ respectively are kept a distance $d$ apart. Where should a small particle be placed, so that the net gravitational force on it due to the bodies A and B is zero?

• A. $\frac{d}{(1+\sqrt{2})}$
• B. $\frac{d}{(1-\sqrt{2})}$
• C. $\frac{d}{(1+\sqrt{3})}$
• D. $\frac{2d}{(1+\sqrt{3})}$

Answer 1

The correct option is A $\frac{d}{(1+\sqrt{2})}$

It is clear that the particle must be placed on the line $AB$ between the bodies A and B. Suppose it is at a distance $x$ from A.
Let its mass be $m$

$F_2=\frac{Gm{m}^{\prime }}{x^2}$
and that due to B is
$F_2=\frac{G\left(2m\right)m^{\prime }}{(d-x{)}^2}$ towards B.
The net force will be zero if $F_1=F_2$
Thus, $\frac{Gm{m}^{\prime }}{x^2}=\frac{G\left(2m\right)m^{\prime }}{(d-x{)}^2}$
or $(d-x{)}^2=2x^2$
or $d-x=\pm \sqrt{2}x$
or $d=(1\pm \sqrt{2})x.$
Thus, $x=\frac{d}{1+\sqrt{2}}$ or $\frac{d}{1-\sqrt{2}}$
As $x$ cannot be negative,
$x=\frac{d}{1+\sqrt{2}}$