The change in potential energy of a system corresponding to a conservative force is defined as

`U_f-U_i=int_(U_i)^(U_f) dU=- intvec F * vec dr=-W`

i.e., The change in potential energy is equal to negative of work done

Let a particle of mass `m_1` be kept fixed at point `A` and another particle of mass ` m_2` is taken from a point `B` to `C` . Initially, the distance between the particle is `r_1` & finally it becomes `AC = r_2` . We have to calculate the change in gravitational potential energy of the system of two particles.

Consider a small displacement when the distance between the particles changes from `r` to `r + dr` In the figure this corresponds to the second particle going from `D` to `L`

Force acting on second particle is

`vec F=(Gm_1m_2)/r^2` along `vec(DA) :. dW=vec F * vec(dr)=-(Gm_1m_2)/r^2 dr`

`dU=- dW = int (Gm_1m_2 dr)/r^2`

`int_(r_1)^(r_2)dU=Gm_1m_2 int _(r_1)^(r_2)(dr)/r^2=Gm_1m_2(-1/r)_(r_1)^(r_2)`

`U(r_2)-U(r_1)=Gm_1m_2(1/r_1-1/r_2)`...............(i)

We choose the potential energy of the two particles system to be zero when the distance between them is infinity. This means `U(oo)=0`

Now in equation (i)

Take `r_1=r` and `r_2=oo`

`U(oo) - U(r) = Gm_1m_2(1/r-1/oo)` `:. U(r)=(-Gm_1m_2)/r`

Hence when two masses `m_1` and `m_2` separated by a distance `r` their gravitational potential energy is

`U(r)=(-Gm_1m_2)/r`