Question

Four equal masses m are kept at corners of a square of side a. If net gravitational force on a mass is given by $\bigg(\frac{2\sqrt2+1}{32}\bigg) \frac{GM^2}{L^2}$ Find the value of a in the terms of $L$.

Answer 1

$F_1=\frac{Gm^2}{a^2},F_2=\frac{Gm^2}{{2a^2}}$

$F_{net}=\frac{Gm^2}{a^2}\frac{(2\sqrt{2}+1)}{2}$

$=\left(\frac{2\sqrt{2}-1}{32}\right)\frac{Gm^2}{L^2}$
$a=4L$

The Correct Answer is : $a = 4L$