Derive the expression for gravitational potential energy.

Explanation

The gravitational force is a conservative force and hence we can detine a gravitational potential energy associated with this conservative force field.
Two masses `m_(1)` and `m_(2)` are initially separated by a distance `r^(.)` . Assuming `m_(1)` to be fixed in its position, Work must be done on `m_(2)` to move the distance from `r^(.)` to r.

To move the mass `m_(2)` through an intinitesimal to be done displacement `dvecr` from `vecr` to `vecr + dvecr`, work has enternally. This infinitesimal work is given by
`dW=vecF_(ext).dvecr` ...(1)
The work is done against the gravitational force, therefore,
`vecF_(ext)=(Gm_(1)m_(2))/(r^(2))hatr` ...(2)
Substituting equation (2) in (1), we get
`dW=(Gm_(1)m_(2))/(r^(2))hatr.dvecr` ...(3)
`dvecr=drhatr implies dW=(Gm_(1)m_(2))/(r^(2))hatr.(drhatr)`
`hatr.hatr=1` (Since both are unit vectors)
`:. dW=(Gm_(1)m_(2))/(r^(2))dr` ....(4)
Thus the total work done for displacing the particle from `r^(.)` to r is
`W=int_(r^(.))^(r)(Gm_(1)m_(2))/(r^(2))dr` ...(5)
`W=-((Gm_(1)m_(2))/(r))_(r^(.))^(r)`
`W=(Gm_(1)m_(2))/(r)+(Gm_(1)m_(2))/(r^(.))` ...(6)
`W=U(r)-U(r^(.))`
where `U(r)=(-Gm_(1)m_(2))/(r)` ...(7)
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.