Expression for Gravitational Potential Energy

To find an explicit formula for gravitational potential energy, consider the work done AGAINST the Earth's gravitational force $\vec{F}_g$ while displacing an object through a small displacement $d\vec{r}$; this work appears as an increase $dU=-\vec{F}_g\cdot d\vec{r}$ in the potential energy of the Earth-object system (the negative sign appears because dU is the work done by US, the external agent, working AGAINST the gravitational force). For a displacement from an initial position $r_i$ to a final position $r_f$, the total change in potential energy is obtained by integrating: $\Delta U=\int_{r_i}^{r_f}dU=-\int_{r_i}^{r_f}\vec{F}_g\cdot d\vec{r}.$

The Earth's gravitational force on an object of mass m at distance r is $\vec{F}_g=-\dfrac{GMm}{r^2}\hat{r}$ (the negative sign here shows the force points OPPOSITE to $\hat{r}$, i.e. towards the Earth's centre, while $\hat{r}$ itself points outward from the centre to the object). Since the displacement $d\vec{r}$ is along $\hat{r}$ for this radial problem, $\vec{F}_g\cdot d\vec{r}=-\dfrac{GMm}{r^2}dr$, and substituting: $\Delta U=-\int_{r_i}^{r_f}\left(-\dfrac{GMm}{r^2}\right)dr=GMm\int_{r_i}^{r_f}\dfrac{dr}{r^2}=GMm\left[-\dfrac{1}{r}\right]_{r_i}^{r_f}=GMm\left(\dfrac{1}{r_i}-\dfrac{1}{r_f}\right).$ …