Gravitational Potential Energy

The gravitational force is conservative. We can see that it only depends on the end points by considering the work it takes to go between two points

$$W ~=~ \int_A^B {\bf F}({\bf r}) \cdot d{\bf l}$$ (1.16)

${\bf F}$ points in the radial direction, that is in the negative ${\bf r}$ direction. The dot product ${\bf F}(r) \cdot d{\bf l}~=~ \vert F(r)\vert dl \cos\theta$. But ${\hat r}\cdot d{\bf l}~=~ dl \cos\theta$ is the component of $d{\bf l}$ in the radial direction. Call it $dr$. So

$$W ~=~ \int_{r_A}^{r_B} F(r) dr$$ (1.17)

In other words, the integral only depends on the radial coordinate $r$, and is therefore independent of any meanderings at different angles, that is it's independent of the path.

From the definition of potential energy, we know that the potential energy is defined in relation to some reference point, say at radius $r_i$. Let's set the potential $U$ at radius $r_i$ equal to zero so

$$U(r) -U(r_i) ~=~ U(r) ~=~ - \int_{r_i}^{r} F(r) dr$$ (1.18)

But $F(r)$ point towards the center so it is

$$F(r) ~=~ -{GMm\over r^2}$$ (1.19)

Plugging this in we have

$$U(r) ~=~ GMm \int_{r_i}^{r} {dr\over r^2} ~=~ GMm ({1\over r_i}-{1\over r})$$ (1.20)

To make life simple lets get rid of the first term by starting at infinity, so that $1/r_i ~=~ 0$. Then

$$U(r) ~=~ -{GMm \over r}$$ (1.21)

So after all this, the final form for the potential energy is pretty simple. The potential energy decreases as the two objects get closer together. It is inversely proportional to $r$, unlike the force which is inversely propotional to $r^2$. It is shown in figure 1.5.