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.
• B.
• C.
• D.

Answer 1

The correct option is A

The force at P due to A is

$F_A$ = $\frac{G{m}^2}{{\left(AP\right)}^2}$ = $\frac{G{m}^2}{{2d}^2}$

along PA . The force at P due to C is

$F_C$ = $\frac{G{m}^2}{{\left(CP\right)}^2}$ = $\frac{G{m}^2}{{\left(2d\right)}^2}$

Along PC. The force at P due to B is

$F_B$ = $\frac{G{m}^2}{{\left(d\right)}^2}$ along PB.

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

Component at $F_A$ along PB = $F_A$ cos ${45}^{\circ }$ = $\frac{G{m}^2}{{2\sqrt{2}{d}^2}^{}}$

Component at $F_C$ along PB = $F_C$ cos ${45}^{\circ }$ = $\frac{G{m}^2}{{2\sqrt{2}{d}^2}^{}}$

Hence , the resultant of the three forces is $\frac{G{m}^2}{d^2}$($\frac{1}{\sqrt{2}}$ + $\frac{1}{\sqrt{2}}$ + 1 ) = $\frac{G{m}^2}{d^2}$(1 + $\frac{1}{\sqrt{2}}$) Along PB