Three particles of mass m, 2m and 4m are placed at the corner of an equilibrium triangle of side a, Calculate :
(a) The potential energy of the system
(ii) The work done on the sysem if the side of the triangle is changed from a to 2a. Assume the potential energy to be zero when the separation is infinity.

Explanation

The potential energy of the system = sum of potential energies of all the three possible distinct pairs .
i.e., `U = U_(AB) + U_(AB) +U_(CA)`
` = (-G(m)(2m))/a +(-G(2m)(3m))/a + (-G(3m)(m))/a `
` = (-11Gm^(2))/a `
when the side a is changed to 2a , the potential energy U . = `(-11Gm^(2))/(2a)`
(ii) The work done on the system = `U.-U`
` =(11Gm^(2))/(2a)`
Note :
The work done by the system = The work done by the gravitational force
` = -(U.-U)`
` = (-11Gm^(2))/(2a)`
The work done on the system = The work done against the gravitational force
= The work done by the external force
` = U. - U `
` = (11Gm^(2))/(2a)`