Define gravitational potential energy.

Explanation

Consider two masses `m_(1)andm_(2)` are initially separated by a distance r'. `m_(1)` is assumed to be fixed in its position.
(a)
To move the mass `m_(2)` through an infinitestimal displacement `dvec(r)` from `vec(r)` to `vec(r)+dvec(r)` (shown in the Figure above), work has to be done externally. This infinitesimal work is given by
`dW=vec(F)_(ext.)dvec(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)dvec(r)" "...(3)`
And `dvec(r)=drhat(r)" "...(4)`
`rArr" "dW=(Gm_(1)m_(2))/(r^2)hat(r).(drhatr)" "...(5)`
`hat(r).hat(r)=1" "`(since both are unit vectors)
`therefore" "dW=(Gm_(1)m_(2))/(r^2)dr" "...(6)`
Hence the total work done for displacing the particle from r' to r is
`W=overset(r)underset(r')(int)dW=overset(r)underset(r')(int)(Gm_(1)m_(2))/(r^2)dr" "...(7)`
`W=-((Gm_(1)m_(2))/(r))_(r')^(r)`
`W=-(Gm_(1)m_(2))/(r)+(Gm_(1)m_(2))/(r')" "...(8)`
`W=U(r)-U(r')`
where, `U(r)=(-Gm_(1)m_(2))/(r)`