91Ó°ÊÓ

Mass of particle A$=m_A$

Mass of particle B,$m_B=3m_A$

Mass of particle C,$m_c=75m_A$

Distance between particle A and particle B$=d\left(in+xdiection\right)$

This problem is based on Newton’s law of gravitation. Net force is zero when both forces have the same magnitude and are in the opposite direction.

Formula:

Gravitational force of attraction,$F=\frac{GMm}{r^2}$ (i)

Using equation (i),Force between A and C is given by:

$F_{AC}=\frac{G{m}_A{m}_C}{x^2}$

Using equation (i), Force between A and B is given by:

$F_{AB}=\frac{G{m}_B{m}_A}{d^2}$

We have,

$$\begin{gathered} F_{AC}=F_{AB} \\ \frac{G{m}_A{m}_C}{x^2}=\frac{G{m}_{B}mA}{d^2} \end{gathered}$$

Hence,

$$\begin{gathered} \frac{x^2}{d^2}=\frac{m_C}{m_B} \\ \frac{x^2}{d^2}=\frac{75}{3} \\ x=Â\pm 5d \end{gathered}$$

Force Between A&B is directed alongpositive xaxis, hence force between B and C should be along negativex axis so that the net force is zero.

Therefore, the separation between particles A & C is -5d