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

The Correct Option is B

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:

\[ a = 4L \]