Derive the expression for gravitational potential energy.

Explanation

The gravitational force is a conservation force and hence we can define a gravitational potential energy associated with this conservation 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.

Two distant masses changing the linear distance
To move the mass through an infinitesimal displacement from 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") = (Gm_(1)m_(2))/(r^(2)) vec(F)_(G)" "` ...(2)
Substituting equation (2) in (1), we get
`dW = (Gm_(1)m_(2))/(r^(2)) hat(r ) . dvec(r )`
` d vec(r ) = dr hat(r) rArr dW = (Gm_(1)m_(2))/(r^(2)) hat(r ) . (dr hat(r ))`
`hat(r ) . hat(r ) = 1` (Since both are unit vectors )
`dW = (Gm_(1)m_(2))/(r^(2)) dr`
Thus the total work done for displacing the particle from r. to r is
`W = int_(r)^(r) (Gm_(1)m_(2))/rdr `
`W = -((Gm_(1)m_(2))/r)_(r )^(r ) `
`W = - (Gm_(1)m_(2))/r + (Gm_(1)m_(2))/(r.) `
`W = U(r ) - U (r. ) `
where `U(r ) = (-Gm_(1)r_(2))/r`
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.

Case for calculation of workdone by gravity
Case 1: If `r < r. : `Since gravitational force is attractive, is attracted by Then can move from r. to r without any external work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.
Case 2 : If` r > r.` : Work has to be done against gravity to move the object from r. to r. Therefore work is done on the body by external force and hence work done is positive.