Why do the orbit equations have to be symmetric about two axes even the orbit is not bounded?

Question

In the book of Classical Mechanics by Goldstein, at page 88, it is given that:

$$ \frac{d^{2} u}{d t^{2}}+u=-\frac{m}{l^{2}} \frac{d}{d u} V\left(\frac{1}{u}\right) . $$ The preceding equation is such that the resulting orbit is symmetric about two adjacent turning points. To prove this statement, note that i= the orbit is symmetrical.

However, the orbit might not be bounded, so there might not be two turning point; just one. In such a case, how can we argue that the orbit equation always has two turning points and it is symmetric about both axes?

Answer 1

The author isn't using the elliptic orbit of the 1/r potential (which would have changed the statement of symmetry to be "around the semiminor and semimajor axes the paths are symmetric"), because he is still talking about a generalized central force, hence turning points are not necessarily axes. Fig 3.12 shows two examples combined: once for the periapsis (which would look more like a semimajor axis, if it were corresponding to a 1/r potential) and again on the adjacent turning point (would be semiminor axis), so that $u$ is the same in going to the two turning points on either side of each of the $\theta=0$ points. The adjacent turning points are on either side of one turning point, an unbounded path is still symmetric as Goldstein later analyzes scattering and states between eq. 3.93 and 3.94: "[T]he orbit must be symmetric about the direction of the periapsis," even though he is talking about scattering the symmetry is the same. The turning points in fig 3.12 can be considered each as $\theta=0$, by coordinate freedom, and so each path has a symmetry under $\theta=-\theta$ swap.