Question

Four masses (each of m) are placed at the vertices of a regular pyramid (triangular base) of side 'a'. Find the work done by the system while taking them apart so that they form the pyramid of side '2a'.

• A. $\frac{3G{m}^2}{a}$
• B. $-\frac{3G{m}^2}{a}$
• C. $-\frac{2G{m}^2}{a}$
• D. $\frac{2G{m}^2}{a}$

Answer 1

Answer: (B)

$-\frac{3G{m}^2}{a}$

The initial gravitational potential is given as

$-6\frac{G{m}^2}{a}$

Final gravitational potential is

$-6\frac{G{m}^2}{2a}$

Change in potential is

$-6\frac{G{m}^2}{2a}-(-6\frac{G{m}^2}{a})$

or

$6\frac{G{m}^2}{a}-6\frac{G{m}^2}{2a}=6\frac{G{m}^2}{2a}=3\frac{G{m}^2}{a}$

Thus the external work done would be

$-3\frac{G{m}^2}{a}$