# Is a complex Keplerian eccentricity $e=\sqrt{1+\frac{2\varepsilon h^2}{\mu^2}}$ possible?

Wikipedia provides the following formula for the Keplerian orbital eccentricity:

$$e=\sqrt{1+\frac{2\varepsilon h^2}{\mu^2}}$$

Considering that a bound object means negative energy, if the magnitude of the ratio in the square root is bigger than 1, then the eccentricity turns out to be complex.

So, what am I missing?

At any given point during the orbit, the velocity can be decomposed into a radial component $v_r$ and an azimuthal component $v_\phi$, and it is only the latter that contributes to the angular momentum, so that $$h = L/m = rp_\phi/m = r v_\phi$$ at all times, but both components contribute to the kinetic energy, $$\frac12 v^2 = \frac12 v_r^2 + \frac12 v_\phi^2.$$ Here the second term can be re-cast in terms of the angular momentum and the distance, giving rise to the so-called centrifugal barrier for the dynamics, in which the specific energy reads \begin{align} \varepsilon & = \frac12 v_r^2 + \frac12 v_\phi^2 - \frac{\mu}{r} \\ & = \frac12 v_r^2 + \frac12 \frac{h^2}{r^2} - \frac{\mu}{r} \\ & \geq \frac12 \frac{h^2}{r^2} - \frac{\mu}{r}, \end{align} because the radial kinetic energy term $\frac12 v_r^2$ is nonnegative.
This means that the specific energy must be no smaller than $f(r) = \frac12 \frac{h^2}{r^2} - \frac{\mu}{r}$, i.e. that it must be no more negative than $-\left| \frac12 \frac{h^2}{r^2} - \frac{\mu}{r} \right|$. As it turns out, this is a simple fraction of $r$ which (so long as $h\neq 0$) always has a minimum at some radius $r=r_*$. To find this minimum, you simply set the derivative of $f(r)$ to zero, \begin{align} 0 & = f'(r_*) \\ & = -2 \frac12 \frac{h^2}{r_*^3} + \frac{\mu}{r_*^2} \end{align} and you solve, giving trivially $r_* = \frac{h^2}{\mu}$. If you put this back into the above inequality, you get \begin{align} \varepsilon & \geq f(r) \geq f(r_*) = \frac12 \frac{h^2\mu^2}{h^4} - \frac{\mu^2}{h^2} = -\frac{\mu^2}{2h^2}, \end{align} and if you re-phrase this a bit, you get that $$\frac{2\varepsilon h^2}{\mu^2} + 1 \geq 0$$ for all possible orbits, which rules out any complex-valued eccentricities.