Question

Four identical particles of mass $m$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $\left( \frac{2 \sqrt{2} + 1}{32} \right) \frac{G m^2}{L^2},$ the length of the sides of the square is:

• A. $\frac{L}{2}$
• B. 4L
• C. 3L
• D. 2L

Answer 1

Answer: (B)

4L

Answer 2

Given four masses $m$ are placed at the corners of a square with side $a$. The net gravitational force on one mass due to the other three masses is given by:

$F_{\text{net}} = \sqrt{2}F + F'$

Where:

$F = \frac{Gm^2}{a^2}, \quad F' = \frac{Gm^2}{(\sqrt{2}a)^2}$

Substituting values:

$F_{\text{net}} = \sqrt{2}\frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2}$

Equating:

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

Solving gives:

$a = 4L$