$$\require{cancel}$$
If $$F = - \frac{GMn}{r^2}$$ then $$V = \frac{GMn}{r}$$, and hence
$V' = - \dfrac{GMn}{r} + \dfrac{L^2}{2mr^2}. \tag{21.3.1}\label{eq:21.3.1}$
I sketch this in Figure XXI.1. The total energy (potential + kinetic) is constant (independent of $$r$$) and is greater than (or equal to) the potential energy. If the total energy is less than zero, you can see from the graph that $$r$$ has a lower (perihelion) and upper (aphelion) limit; this corresponds to an elliptic orbit. But if the total energy is positive, $$r$$ has a lower limit, but no upper limit; this corresponds to a hyperbolic orbit. If the total energy is equal to the minimum of $$V'$$, only one value of $$r$$ is possible, and the orbit is a circle.