Given two arbitrary orbits around a point mass, there exists some optimal transfer between them in terms of delta-v.
What's the highest number of impulses such a transfer could require? (That is, I'm asking for a specific quantity of solutions to a variant of Lambert's problem.)

"Optimal" in a mathematical sense. Burns not being perfectly impulsive, transfers taking unreasonable amounts of time being undesirable, perturbations, three-body effects and so on can all be ignored.

The number is obviously larger than 1, since not all orbits share a common point.

If all optimal planar transfers are bi-tangential orbits, the answer is 2 for planar orbits.

The number is larger than 2, since solutions with 3 impulses are better for some types of transfers.

An infinite apoapsis generalised bi-elliptic transfer, which is sometimes optimal, has two non-zero impulses and two zero-impulse manoeuvers. Whether this counts as 2 or 4 impulses is less important since: 1) There can at most be 2 zero-impulse manoeuvers in any optimal transfer, and 2) Any optimal transfer containing a zero-impulse manoeuver can at most have 2 non-zero impulses.

Does an optimal transfer requiring 4 or more non-zero impulses exist?

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    $\begingroup$ I was thinking a bit about adding a bounty to your question (because I have a hunch this will turn out to be really interesting) but I don't know if it interferes with your timing or intent. How would you feel about that? $\endgroup$ – uhoh Nov 16 '20 at 0:15
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    $\begingroup$ @uhoh Sure, go ahead. I too would like to know the answer to this. (Even though bounties haven't proved very effective for this kind of question in the past). $\endgroup$ – SE - stop firing the good guys Nov 16 '20 at 1:03
  • $\begingroup$ This is quite interesting. For ill-conditioned transfers (practically everything with 30+ degrees inclination change; generally vast majority), the answer is "4" (1. raise apoapsis to near-infinity, 2. circularize+plane change to enter a slow orbit to a point co-axial with target apoapsis, 3. drop periapsis + plane change for target orbit, 4. drop apoapsis.) Considering the "mathematical perfection" cost of 2. and 3. is infinitesimally small, (and transfer time between them is infinite), but the whole thing costs $\sqrt{2} ( v_{Pe1} + v_{Pe2} ) + \epsilon $ $\endgroup$ – SF. Nov 16 '20 at 8:49
  • $\begingroup$ This is quite a bit, and there will be a margin of degenerate transfers that are not-quite-as-ill-conditioned, that can be done on less delta-V but on more burns. $\endgroup$ – SF. Nov 16 '20 at 8:53

For coplanar orbits, a bi-elliptical transfer is more efficient than an Hohmann transfer when the ratio of the initial and final radii is greater than 15.58. When the ratio is less than 11.94, an Hohmann transfer is more efficient. (Thanks to notovny for correcting me.)

A bielliptic transfer is effectively two subsequent Hohmann transfers. Section 6.3.2 of "Fundamentals of Astrodynamics" by Vallado (p. 328 in the 4th edition) compares Hohmann transfers to the bi-elliptical transfer. In a bi-elliptical transfer, you will need three burns: one to depart the initial orbit onto an elliptical orbit (you must depart when your flight path angle is zero), then perform an apogee burn on the elliptical orbit, and finally perform a final maneuver on the destination orbit, also where you should get a flight path angle of zero.

For any other transfer, it really depends on the problem you are trying to solve, and the variables of the problem (e.g. how many times can you reignite the engine, what will be the errors in the thruster performance, where are the ground stations placed for navigation, etc.).

For example, for interplanetary or lunar missions, one would set up the problem to assume 4 to 8 control points, i.e. positions in the trajectory where you should place a maneuver. One would rarely place more than 8 control points. Each control point is assumed to be a point in the trajectory where a maneuver will be executed, and those require some operational overhead. As such, we ensure there is some time between each potential maneuver. For example, before a maneuver, it is important to have very good knowledge of the position and velocity of the spacecraft before the maneuver (i.e. a good navigation solution), and be able to continue tracking the spacecraft soon after the maneuver. In short, the fewer the maneuvers, the easier it is to fly the spacecraft. So there's a trade off between the fuel savings and the overhead needed for each maneuver.

Moreover, optimizers (like SNOPT) would be used to optimize the placement of these control nodes and the optimizer will try to minimize the delta-V at each node. This approach is called "multiple shooting" and is used for Ballistic Lunar Transfers to libration point orbits. The optimizer may show that some of the control nodes have extremely small delta-Vs (e.g. less than a few millimeters per second), and in which case, you can omit that maneuver, and rerun the optimization problem.

A similar approach would be done for Earth orbits on different planes. As you also correctly stated, one would generally start with a Lambert solution for a first level approximation. Then, you would place the control points at different positions and let the optimizer find the best solution.

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    $\begingroup$ This is a really interesting answer and I'll bet there's a question about three-body orbits that would suit it better. The current question post begins: "Given two arbitrary orbits around a point mass..." so I think at least the intent is to ask about transfers between Keplerian orbits. It doesn't specify anything else about those orbits, so while "planar" is mentioned in an example of the counting procedure, my guess is that the maximum number will be between non-coplanar elliptical orbits that are either awkwardly asynchronous or "maliciously synchronized" to make it hard. $\endgroup$ – uhoh Nov 15 '20 at 1:35
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    $\begingroup$ While interesting about trajectory optimisation in general, this doesn't answer my question. $\endgroup$ – SE - stop firing the good guys Nov 15 '20 at 12:52
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    $\begingroup$ "It really depends on the problem you are trying to solve". Optimal delta-v transfers between two orbits around a point mass. $\endgroup$ – SE - stop firing the good guys Nov 15 '20 at 12:54
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    $\begingroup$ @ChrisR: No mission profile ever involves maneuvers that would lean on delta-v equivalent of turning inclination of LEO by 180 degrees ("""optimally"""). There's no point bringing real-life missions to this discussion because these almost-worst-case scenarios are already so bad nobody even takes them into consideration. $\endgroup$ – SF. Nov 16 '20 at 21:36
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    $\begingroup$ @ChrisR I see you're confused because you take a lot of engineer's approach, not mathematician's, considering factors which are non-factor in asker's conditions (like delay between maneuvers). For example, " in which case, you can omit that maneuver," - if you see my comment about all the "worst scenario" transfers, the solution consists of two large burns and two minuscule ones, and the minuscule ones are critically important. It also has completely impractical duration, which again is non-factor. Or consider a delta-v optimal bielliptic transfer - transfer orbit being infinite is OK. $\endgroup$ – SF. Nov 17 '20 at 10:29

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