Question

Three bodies each of mass 'm' are placed at the three corners of a square of side 'a'. The gravitational force on unit mass kept at the fourth corner is

• A. $\frac{Gm}{3a^2}$
• B. $\frac{Gm}{\sqrt{3}{a}^2}$
• C. $\frac{3Gm}{a^2}$
• D. $\frac{Gm}{a^2}(\sqrt{2}+\frac{1}{2})$

Answer 1

The correct option is D $\frac{Gm}{a^2}(\sqrt{2}+\frac{1}{2})$

the gravitational force acting on unit mass will be the vector sum of forces due to all the masses.

force due to one mass which is at distance 'a' from unit mass $\left(F\right)=\frac{Gm\left(1\right)}{a^2}$

so resultant force of the two masses at same distance $\left(F_{net1}\right)=\sqrt{F^2+F^2}=\sqrt{2}F$ $\Rightarrow F_{net1}=\frac{\sqrt{2}Gm}{(\sqrt{2}a{)}^2}=\frac{Gm}{2a^2}$

so total force on unit mass is equal to : $F_{total}=\frac{\sqrt{2}Gm}{a^2}+\frac{Gm}{2a^2}=\frac{Gm}{a^2}(\sqrt{2}+\frac{1}{2})$