Question

Find the gravitational potential energy of a system of four particles, each having mass m, placed at the vertices of a square of side l at the centre of the square.

• A.
• B.
• C.
• D. zero

Answer 1

The correct option is

A

The system has four pairs with distance I and two diagonal pairs with distance $\sqrt{2}l$.

$\therefore$ U = $-4\frac{G{m}^2}{l}-2\frac{G{m}^2}{\sqrt{2}l}=-\frac{2G{m}^2}{l}(2+\frac{1}{\sqrt{2}})=-5.41\frac{G{m}^2}{l}$

The gravitational potential at the center of the square is (r = $\frac{\sqrt{2}}{2}l$)

V = Algebraic sum of potential due to each particle

$\Rightarrow V=-\frac{4\sqrt{2}Gm}{l}$.