Multiple revolution Lambert problem

Question

The Lambert problem solution determines the $\Delta V$ required to transfer from r1 to r2 in t time.

There are single and multiple revolution solutions of the problem. As I understand, the single revolution scenario assumes 2 impulses. What about the multiple revolutions?

BONUS

I'm using poliastro package on python. For single revolution the velocity is calculated as

(v0, v), = izzo.lambert(k, r0, r, tof, M=0)

However, for M=1 (which is, probably, 1 complete rotation), it's calculated as

(_, v_l), (_, v_r) = izzo.lambert(k, r0, r, tof, M=1)

What are v_l and v_r here?

Answer 1

Disclaimer: I'm the author of poliastro

No matter how many revolutions you consider, the Lambert solution gives always two velocities: departure and arrival. You could then consider that you need two impulses: one to attain the departure velocity, and one final impulse to fix the orbital velocity (because you arrived at the position you wanted, but maybe not with the velocity you desired).

Therefore, what we mean by multirevolution is the ability to go around the main attractor more times to lower the required cost in terms of impulses. To illustrate this, I represented possible arcs from Earth to Mars using 0, 1, or 2 revolutions:

The black trajectory ($M = 0$) goes towards the Sun, makes an almost complete revolution, and reaches Mars. What's more interesting is that the $M = 1$ and $M = 2$ trajectories go towards the Sun, make a full orbit, passes perihelion again, and then reach Mars. Their period is smaller and smaller, but by doing more revolutions they "wait for Mars" to arrive to the destination point.

There are two more things to note here: one is that there is no $M = 3$ solution. The $M = 2$ is already very close to the optimum, and if you try to find a solution with one more revolution, you will find out that there isn't.

On the other hand, did you notice that I wrote (a) in the multirevolution solutions? That's because there are two ways to get to Mars in these cases. You can visualize it by propagating both solutions a small amount of time:

What will happen is that one arc will reach Mars from the outside, and the other one will do from the inside.

Final note: This answer is incomplete because some of the orbits obtained for this particular case proved to be corner cases for poliastro propagators. Fixes are underway. Also, the original confusion comes probably from poor documentation and an unintuitive API. We will fix that as well.