Question

Four particles of masses $m,m,2m$ and $2m$ are placed at the four corners of a square of side $a$ as shown in the figure. The magnitude of the gravitational force acting on a particle of mass $m$ placed at the centre of the square is

• A. $\frac{2G{m}^2}{a^2}$
• B. $\frac{G{m}^2}{{\sqrt{2}a}^2}$
• C. $\frac{G{m}^2}{{2a}^2}$
• D. $\frac{2\sqrt{2}G{m}^2}{a^2}$

Answer 1

The correct option is D $\frac{2\sqrt{2}G{m}^2}{a^2}$

$F_1=\frac{2G{m}^2}{a^2};F_{1x}=\frac{-2G{m}^2}{a^2\sqrt{2}};F_{1y}=\frac{-2m^2}{a^2\sqrt{2}}$

$F_2=\frac{2G{m}^2}{a^2};F_{2x}=\frac{2G{m}^2}{a^2\sqrt{2}};F_{2y}=\frac{-2m^2}{a^2\sqrt{2}}$

$F_3=\frac{4G{m}^2}{a^2};F_{3x}=\frac{4G{m}^2}{a^2\sqrt{2}};F_{3y}=\frac{4G{m}^2}{a^2\sqrt{2}}$

$F_4=\frac{4G{m}^2}{a^2};F_{4x}=\frac{-4G{m}^2}{a^2\sqrt{2}};F_{4y}=\frac{4G{m}^2}{a^2\sqrt{2}}$

$\therefore F_x=F_{1x}+F_{2x}+F_{3x}+F_{4x}=0$

and $F_y=F_{1y}+F_{2y}+F_{3y}+F_{4y}=\frac{4G{m}^2}{a\sqrt{2}}$

$\therefore F=\sqrt{F_x^2+F_y^2}=\sqrt{\frac{16G^2{m}^2}{2a^2}}=\sqrt{\frac{8G^2{m}^2}{a^2}}$

$\therefore F=\frac{2\sqrt{2}Gm}{a^2}$