1
$\begingroup$

A recent question includes the passage:

From a topological point of view, it seems likely that any rigid body would have to admit at least one stable low orbit, regardless of mass concentrations. I haven't gone over the math, though. Has it already been done?

What is does the phrase "topological point of view" mean when applied to two-body orbits in this context, lunar or otherwise?

I know that topological concepts such as bifurcation are important in three or more body orbits (e.g. this stuff or the image below from here), but in this context it's just a (presumably) massless body in orbit around another rigid body.

CR3BP and bifucation

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ The idea I think is to consider the space of possible orbits andshow that because there are orbits that are perturbed (by the non-spherical components of the gravity of the central mass) in one way and other orbits perturned the other way, then somewhere in between there must be one that isn't perturbed. Given that there are several possible independent forms of perturbation, this might well not be true, but the topology of the space of all possible orbits will tell you. $\endgroup$
    – Steve Linton
    Commented Apr 24, 2018 at 15:10
  • $\begingroup$ @SteveLinton Can you give an example of what a simple example of a "space of possible orbits" might be? For a simple central force, I suppose it might have five dimensions ignoring time. But for a more complicated gravity field, would the number of dimensions in this "space" even be finite? Or is the space actually always just a six dimensional state vector at t=0 (starting point)? $\endgroup$
    – uhoh
    Commented Apr 24, 2018 at 22:47

1 Answer 1

Reset to default
2
$\begingroup$

Start by considering a large mass, spherical of radius $r$ or point, at the origin. Note that at perihelion or aphelion the relative motion of an orbiting object is tangent to the center of mass.

Any orbit with distance $A > r$ from the origin to aphelion and distance $P$ to perihelion is characterized entirely by a point on in the unit sphere (corresponding, say, to the point of perihelion) along with a point on the unit circle (giving the direction of tangent motion at perihelion).

We can therefore think about the set of $A,P$ orbits in terms of unit vector fields on the unit 2-sphere. More generally, the set of orbits with energy $E$ and perihelion distance $P$ will have the same property.

For a spherical center mass the two are completely equivalent. If the central mass is not spherical, there is still an equal-energy surface $S_E$ of perihelia. $S_E$ will no longer be spherical but will (usually?) be topologically the same as a 2-sphere.

If $S_E$ is topologically spherical, then vector fields on $S_E$ are subject to simple topological analysis, including the famous "hedgehog theorem" stating that there are no non-vanishing continuous vector fields on $S_E$.

Edit:

I no longer believe it is reasonable to think in terms of perihelion and aphelion, since that assumes an orbit in the first place. We can still think in terms of tangents at some equipotential surface, but have to allow for the full range of kinetic energies (speeds). We are thus characterizing a set of geodesics by a point on the sphere plus a tangent vector, rather than just a point on the unit circle.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ It would be helpful to expand a little further on this. How does this relate to stability under perturbations $\endgroup$
    – Steve Linton
    Commented Apr 24, 2018 at 19:01
  • $\begingroup$ Thanks for your answer, I will give it a thorough read in a few hours. $\endgroup$
    – uhoh
    Commented Apr 24, 2018 at 22:49
  • $\begingroup$ @Steve Linton This is a different topological perspective than stability under perturbations, in which I am hinting there might be some simple hedgehog theorem trick. If hedgehog doesn't work then perturbations seem the most promising avenue. $\endgroup$
    – Brian B
    Commented Apr 25, 2018 at 0:31
  • $\begingroup$ Okay, well it's been longer than a few hours now, but I'm almost there. I'll have a look when it's quieter this evening. "Hairy ball theorem" is the answer to my question Are there necessarily always at least two points where the Earth's magnetic field is vertical? and I'm guessing that that's simply a spherical hedgehog? $\endgroup$
    – uhoh
    Commented Oct 10, 2018 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.