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Problem: Given two orbits (e.g. in the 5 keplerian elements excluding true anomaly) A and B calculate the $\Delta v$ necessary to perform a transfer from A to B using instantaneous burns.

Simple question, but the answer seems difficult to me. I could not find a solution to this problem in my searches. Could someone please point me in a good direction or even give a comprehensive answer?

Turakar
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  • You are right, the answer is difficult and not simple. But luckily that question does not ask for the optimal transfer so you can do it in the simplest way to calculate possible, which I assume is two steps; 1) plane change maneuver, 2) Hohmann or similar co-planar circular orbit transfer. "Could someone please point me in a good direction or even give a comprehensive answer?" Yes, search this site for similar and related questions and read their answers! – uhoh Nov 03 '20 at 15:21
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    There is a reason interplanetary trajectory experts are few and far between... – Carlos N Nov 03 '20 at 20:28
  • @CarlosN actually all the ones we've ever known of have always remained within a 0.00009 AU diameter sphere. – uhoh Nov 05 '20 at 22:57

2 Answers2

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This is indeed a difficult question. On this site, you can for instance see that even my quest to resolve the co-planar case hasn't been particularly resolved, and yet even more restricted forms like optimal inclination change between circular orbits still quickly grow complicated.

However, that is for optimal transfers. There exists a simple strategy that is sufficient:

  1. Starting from your original orbit, accelerate up to very nearly escape velocity. For your instantaneous case, that means a burn at periapsis.

  2. At a very great distance, perform inclination, longitude and periapsis changes, since that costs ~0 $\Delta v$ at that distance.

  3. Fall back towards the parent body, breaking into the target orbit at periapsis.

Only the burns at 1) and 3) have a cost larger than 0. Those can be calculated from the vis-viva equation for velocity in elliptical orbits:

$$v = \sqrt{GM\left(\frac{2}{r} - \frac{1}{a}\right)}$$

This strategy is not generally optimal (rather poor for low relative inclination), but it will always work, and is sometimes even optimal.

This can be regarded as an upper bound. Asking for lower bounds hasn't been particularly successful, although there's a quite simple one.

SE - stop firing the good guys
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This is to some extent equivalent to Lambert's problem. Namely, if you pick any point along the two orbits you can draw transfer orbits between. This can be constrained to one orbit if you for example also specify the true anomaly when departing or arriving (if the two points and the celestial body do lie on one line). You could also specify the transfer time, but this can yield multiple solutions. The ∆v can be calculated by adding the difference in norm of the difference in velocity at the two points.

It can be noted that this only considers transfers with only two (instantaneous) velocity changes. Therefore, this method as described can not obtain things like the bi-elliptic transfer or mid-transfer plane change. You could of course expand the method and add an additional velocity change, for example by picking a additional point in space where you would perform another burn. This would add four additional degrees of freedom (three for the point and one for true anomaly/travel time between points).

Lambert's problem is also used to generate pock chop plots. Though, that problem has only two degrees of freedom, since the arriving point is a function of the departure point and travel time

In general there not an analytical solution for the transfer that minimizes the total ∆v, even not for the pock chop plot case. I think that such problem is not convex in which case there is also no guarantee that numerical methods can approximate the optimal transfer.

fibonatic
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