Orbit reversal in a gravitational well A friend recently asked me this, and without wishing to spend a huge time thinking about it, I wondered if anyone knew the answer (or at least had an informed guess), or was familiar enough with these sorts of calculations to figure it out efficiently (rather than painfully, which is how I would end up doing it).

What is the most (energy) efficient way to reverse your orbital motion?

More specifically, assume you are in a non-relativistic regime in a spherical gravitational well, in some 'directed orbit' $O$. What is the most efficient way to end up in $-O$, the 'reversed orbit'?
Some fairly obvious suggestions include attempting to make the orbit highly eccentric and then maneuvering at apoapsis, or performing a gradual 180 degree plane-change. I have no immediate idea which is better, and whether there are other better techniques. Any insight appreciated!
Edit: Added emphasis above. I'm interested in whether there could be any other approach; I have no idea what it might look like, and suspect the above are the right possibilities to look at, but would like to be convinced about it.
 A: Right, well someone check my math but starting from equations 4.16-4.17 and 4.73 of the link Ben Crowell posted I work out that the delta-v of a boost/plane change/circularize relative to the direct plane change is
$$ \frac{\Delta v}{2v\sin(\frac{\theta}{2})} = \frac{\left(\sqrt{2}x-\sqrt{x(x+1)}\right)\csc\left(\frac{\theta}{2}\right)+\sqrt{2}}{\sqrt{x(x+1)}}, $$
where $x = r_a / r_p$ is the ratio of apoapsis to periapsis of the boosted orbit. If this awful function is less than 1 then the boosted maneuvre is more efficient than a direct plane change. So here is how it looks:

The dashed contour is the break even point, so a boosted maneuvre is always more efficient for a complete reversal. In fact the break even line asymptotes to an angle of $2\sin^{-1}\left(\sqrt{2}-1\right) \approx 49º$, so for any angle bigger than this you always do better by boosting the orbit first. The break even curve intersects $x=1$ for $\theta=2\sin^{-1}\left(\frac{1}{3}\right)\approx 39º$, so if the angle is between this and 49º you can do better by boosting, but do worse if you boost too much. If you are happy to aerobrake that would lower these thresholds even more.
If you just look at reversals $\theta=\pi$ the relative delta-v simplifies to
$$ \frac{\Delta v}{2v\sin(\frac{\theta}{2})} = \sqrt{\frac{2}{x}+2}-1, $$
which quickly asymptotes to $\sqrt{2}-1$, so you do better by boosting your orbit, but you don't do much better by boosting your orbit a lot (more than $x\sim$a few). So whether it is practical depends on other factors, like timing.
A: The approach to orbit plane change outlined by Michael -- bielliptic plane change -- is the solution for two-body problem (Fig. 1):

However, if the apoapsis of the transfer orbit is far enough then perturbations from other celestial bodies should be taken into account. So we need to consider (restricted) three-body (at least) problem. 
This was done in the paper "New class of optimal plane change maneuvers" by B.F. Villac and D.J. Scheeres (doi:10.2514/2.5109, pdf).
In essence, the timing of the first impulsive maneuver (the one changing circular orbit to elliptical) could be such that the plane change maneuver at apoapsis would not be needed -- the third body would perform all the work (see Fig. 2):

The paper outlines how to calculate such orbital transfer.
As for the savings that could be obtained by this maneuver, we read in the paper:

Savings on the order of 25% as compared with one-impulse maneuvers are obtained for 
  plane changes of ~60 deg, and the possibility of a reversal of the direction of motion 
  (plane changes of $\pm$180 deg) has been shown to exist using such an approach.
  Savings of more than 15% when compared to the classic results on parabolic transfers
  and 70% when compared one-impulse maneuvers are realized in these cases.

