Question

One dimension.In the figure, two point particles are fixed on anxaxis separated by distanced. ParticleAhas mass${\mathit{m}}_{\mathbf{A}}$M and particle Bhas mass$\mathbf{3}\mathbf{.}\mathbf{00}{\mathit{m}}_{\mathbf{A}}$. A third particle C, of mass$\mathbf{750}{\mathit{m}}_{{\mathbf{A}}_{\mathbf{}}}$, is to be placed on the xaxis and near particles Aand B. In terms of distance d, at what xcoordinate should Cbe placed so that the net gravitational force on particle Afrom particles Band Cis zero?

Short answer

The x-coordinate of particle C is -5dat which the net gravitational force on particle A from particles B and C is zero.

Step-by-step solution

The given data

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$=d\left(in+xdiection\right)$

Understanding the concept of Newton’s law of gravitation

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)

Calculating the value of x-coordinate of particle C

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,

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

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