Question

Three particles of identical masses $m$ are kept at the vertices of an equilateral triangle of each side length $a$. The gravitational force of attraction on any one of the particles is:

• A. $\sqrt{2}\frac{G{m}^2}{a^2}$
• B. $\sqrt{3}\frac{G{m}^2}{a^2}$
• C. $\frac{3G{m}^2}{a^2}$
• D. $\frac{2G{m}^2}{a^2}$

Answer 1

Answer: (B)

$\sqrt{3}\frac{G{m}^2}{a^2}$

$F_{net}=\sqrt{(F_1^2+F_2^2+2F_1{F}_{2}cos\theta )}$

Here, $F_1=F_2=\frac{Gmm}{r^2}$ and $\theta ={60}^0$
Therefore, on solving we get, $F_{net}=\sqrt{3}\frac{G{m}^2}{a^2}$