Context: I've been playing Kerbal Space Program again, and am trying to parameterize the space of all possible orbits in a nice way. Or at least, figure out what it's homeomorphic to.
TLDR: I can show that for a fixed orbital energy $E$, the space of possible 2D orbits is:
- $S^2$ if $E < 0$ (i.e. a closed orbit)
- $S^1 \times \Bbb R$ if $E \ge 0$ (i.e. an escape trajectory / flyby).
But I can't unify these two regimes into a single space.
I'm starting with 2D orbits, which have 3 degrees of freedom. Some possible choices of parameter are:
- $a$ -- the semimajor axis (positive for ellipses, negative for hyperbolas)
- $e$ -- the eccentricity
- $\omega$ -- the argument of periapsis, i.e., what angle is your "closest approach point" at
- $E$ -- the energy of the orbit, equal to $-\mu/2a$
- $h$ -- the angular momentum of the orbit, equal to $\sqrt{\mu a (1-e^2)}$
($\mu$ is a constant physical parameter of the body you're orbiting)
I can show that the collection of elliptical orbits with a fixed $a > 0$ (equivalently, a fixed $E < 0$) is homeomorphic to $S^2$, as follows:
Let $\vec{e}$ be a vector with length $e$, pointing in the direction of periapsis. This is unambiguous for circles, despite their lack of unique periapsis, because then $\vec{e} = 0$. Since $h^2 = \mu a (1 - e^2)$, we can rearrange that into $e_x^2 + e_y^2 + \frac{h^2}{\mu a} = 1$. Clearly this is an ellipsoid, and since $\vec{e}$ and $h$ together determine the orbit (make sure to check special cases like circular and radial trajectories), we get that the space of orbits is homeomorphic to $S^2$.
Likewise, this shows that hyperbolic orbits form a hyperboloid of one sheet, which is homeomorphic to $S^1 \times \Bbb R$. (Note that $a < 0$, so that $h^2$ term is negative.)
Parabolas are a slightly different animal. Since $a = \infty$, our constraint simplifies to $e_x^2 + e_y^2 + 0h^2 = 1$. So the length of $\vec{e}$ is required to be $1$, and we have no constraints on $h$ at all.
This is unlike our earlier cases, where we had two degrees of freedom for eccentricity, and a choice of $\vec{e}$ would determine our $h$ up to sign.
This means that I'm having a lot of trouble incorporating $a$ as a parameter here. Well, since $a$ behaves poorly at the elliptic -> parabolic -> hyperbolic transition (looks like $1/x$, in that it grows rapidly, passes through infinity, and then becomes negative), I've been trying to use $E = -\mu/2a$ as a parameter instead; it's probably nicer. So my constraint is then $e_x^2 + e_y^2 - \frac{2E}{\mu^2} h^2 = 1$.
As I vary $E$ across all of $\Bbb R$, I should be able to glue the level surfaces into a $3$-manifold -- the shape of the whole space.
As $E$ slides up through $(-\infty, 0)$, things are easy -- each level surface of $E$ is an ellipsoid, and so our manifold so far is a thickened sphere, $S^2 \times (-\infty, 0)$.
But when $E = 0$, we suddenly get $S^1 \times \Bbb R$, and I don't see how to glue this to the outside of the thickened sphere in a way that preserves continuity of the parameters.
If I can figure that out, the hyperbolic ($E > 0$) cases will probably be fine, since they are also shaped like $S^1 \times \Bbb R$. So I should just be able to glue that layer down onto the parabolic level surface.
At the end, I suspect I'll have $S^2 \times \Bbb R$, but without an explicit homeomorphism I'm having trouble convincing myself.
A few notes:
- we need to use $h$ in order to distinguish parabolic orbits, since they all have $a = \infty$, $E = 0$, and $e = 1$.
- we need to use $a$ or $E$ to distinguish radial orbits, since they all have $e = 1$ and $h = 0$.
- I thought about using the periapsis $q = a(1-e)$ as a parameter, but it behaves poorly when the eccentricity passes through $0$.