Question

Two bodies 'A' and 'B' having masses M and 3M respectively are kept at a distance 'y' apart. A small particle is to be placed so that the net gravitational force on it due to A and B is zero. Its distance from mass A will be

• A. $x=\frac{y}{1+\sqrt{3}}$
• B. $x=\frac{y}{1+\sqrt{6}}$
• C. $x=\frac{y}{1+\sqrt{2}}$
• D. $x=\frac{y}{1+\sqrt{4}}$

Answer 1

The correct option is A $x=\frac{y}{1+\sqrt{3}}$

Let m be the mass of the particle which is placed at a distance x from A.
$F_1=\frac{GMm}{x^2}$towards A and
$F_2=\frac{G\left(3M\right)m}{(y-x{)}^2}$ towards B
The net force = 0, if $F_1=F_2$
$\Rightarrow \frac{GMm}{x^2}=\frac{G\left(3M\right)m}{(y-x{)}^2}$
$\Rightarrow \sqrt{3}x=y-x$
$\Rightarrow x=\frac{y}{1+\sqrt{3}}$