Question

Four particles each of mass m are placed at the vertices of a square of side l. The potential energy of the system is

• A. $-\frac{\sqrt{2}G{m}^2}{l}(2-\frac{1}{\sqrt{2}})$
• B. $-\frac{2G{m}^2}{l}(2+\frac{1}{\sqrt{2}})$
• C. $-\frac{\sqrt{2}G{m}^2}{l}(\sqrt{2}+\frac{1}{\sqrt{2}})$
• D. $-\frac{2G{m}^2}{l}(\sqrt{2}-\frac{1}{\sqrt{2}})$

Answer 1

The correct option is B $-\frac{2G{m}^2}{l}(2+\frac{1}{\sqrt{2}})$

From figure $AB=BC+CD+AD=l$
$\therefore AC=BD=\sqrt{l^2+l^2}=l\sqrt{2}$
The potential engery of the system of four particles each of mass m placed at the vertices A, B, C and D of square is
$U=(-\frac{G\times m\times m}{AB})+(-\frac{G\times m\times m}{AC})+(-\frac{G\times m\times m}{AD})+(-\frac{G\times m\times m}{BC})+(-\frac{G\times m\times m}{BD})+(-\frac{G\times m\times m}{CD})$
$=(-\frac{G{m}^2}{l})+(-\frac{G{m}^2}{l\sqrt{2}})+(-\frac{G{m}^2}{l})+(-\frac{G{m}^2}{l})+(-\frac{G{m}^2}{l\sqrt{2}})+(-\frac{G{m}^2}{l})$
$=-\frac{4G{m}^2}{l}-\frac{2G{m}^2}{l\sqrt{2}}=-\frac{2G{m}^2}{l}(2+\frac{1}{\sqrt{2}})$