Question

Three particles A, B and C each of mass m, are placed in a line with AB = BC = d. Find the gravitational force on a fourth particle P of same mass, placed at a distance d from the particle B on the perpendicular bisector of the line AC.

• A. $⟮\frac{\sqrt{2}+1}{\sqrt{2}}⟯\frac{m}{d^2}$ (along PB)
• B. $⟮\frac{\sqrt{2}-1}{\sqrt{2}}⟯\frac{G{m}^2}{d^2}$ (along PB)
• C. $⟮\frac{\sqrt{2}-1}{\sqrt{2}}⟯\frac{m^2}{d}$ (along PB)
• D. $⟮\frac{1+1}{\sqrt{2}}⟯\frac{G{m}^2}{d^2}$ (along PB)

Answer 1

The correct option is D $⟮\frac{1+1}{\sqrt{2}}⟯\frac{G{m}^2}{d^2}$ (along PB)

$Fa=\frac{G{m}^2}{(AP{)}^2}$

$=\frac{G{m}^2}{2d^2}\text{along PA}$

The force at $P$ due to $C$ is

$Fc=\frac{G{m}^2}{(CP{)}^2}$

$=\frac{G{m}^2}{2d^2}\text{along PC}$.

The force at $P$ due to $B$ is

$Fb=\frac{G{m}^2}{d^2}\text{along PB}$

Now resultant of $Fa,Fb$ and $Fc$ will be along with $PB$

So $\angle APB=\angle BPC={45}^{\circ }$

So $Fa$ along $PB=Facos{45}^{\circ }=\frac{G{m}^2}{2\surd 2d^2}$

So $Fc$ along $PB=Fccos{45}^{\circ }=\frac{G{m}^2}{2\surd 2d^2}$

So Fb along $PB=\frac{G{m}^2}{d^2}$

The resultant of the three forces will be

$\frac{G{m}^2}{d^2}(\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}+1})$

$\frac{G{m}^2}{d^2}\left(\frac{1+1}{\sqrt{2}}\right)$ net force along $PB.$