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 : \(a = 4L\)