Derive the expression for gravitational potential energy.

Explanation

(i) Two masses `m_(1)` and `m_(2)` are initially separated by a distance r. Assume `m_(1)` to be fixed in its position, work must be done on `m_(2)` to move the distance from 'r' to r' to r as shown in figure (a).

(iii) To move the mass `m_(2)` through an infinitesimal displacement `d vec(r)` from r to `vec(r) + d vec(r)` (shown in the Figure (b)). work has to be done externally. This infinitesimal work is given by
`dW = vec(F)_(ext) . d vec(r)" ".......(1)`
The work is done against the gravitational force, therefore,
`|vec(F)_(ext)| = |vec(F)_(G)| = (Gm_(1)m_(2))/(r^2)" ".......(2)`
Substituting Equation (2) in (1), we get
`dW = (Gm_(1)m_(2))/(r^2)hat(r). d vec(r)`
`dvec(r) = dr hat(r) , (Gm_(1)m_2)/(r^2) hat(r) . d vec(r) = dW`
`hat(r) . hat(r)` = 1 (since both are unit vectors)
`:. dW = (Gm_(1)m_(2))/(r^2) dr`
(iii) Thus the total work done for displaceing the particle form r' to r is
`W = int_(r')^(r) dW = int_(r')^(r) (Gm_(1)m_2)/(r^2) dr`
`W = -((Gm_(1)m_2)/(r^2))_(r')^(r)`
`W = -(Gm_(1)m_2)/(r)+(Gm_(1)m_2)/(r')`
`W = U(r) - U(r')`, Where `U(r) = (-Gm_(1)m_2)/(r)`
(iv) This work done W gives the gravitational potential energy difference of the system of masses `m_(1)` and `m_(2)` when the separation between them are r and r' respectively.