Given two arbitrary elliptic orbits around an ideal single point mass, there will always exist a transfer with the minimal $\Delta v$ required.

It's easy to find an upper bound for this ideal transfer:

$$\Delta v \leq \sqrt{\frac{2}{r_{P1}}} - v_{P1} + \sqrt{\frac{2}{r_{P2}}} - v_{P2}$$

This is a general bi-elliptical transfer, and it's sometimes not possible to do better than that.

It sets a maximum $\Delta v$ required for arbitrary transfers. Are there any trivial lower bounds for the minimal $\Delta v$ required?

  • $\begingroup$ I'm no expert, but I would've thought the minimum delta V transfer around a non-gravitationally complex single mass in many cases would be your namesake. $\endgroup$
    – Ingolifs
    Sep 26, 2019 at 17:38
  • $\begingroup$ Only for planar, circular orbits. $\endgroup$ Sep 26, 2019 at 17:54
  • $\begingroup$ You could, perhaps, calculate all possible transfers at a specified interval then iterate through them using an algorithm to decide which is most efficient. That's kind of a cop-out though, I'm sure there's a mathematical way to do it. $\endgroup$ Sep 27, 2019 at 5:29

1 Answer 1


While it's a very loose lower bound, it does perhaps have some value to present one most trivial such bound.

With basis in the fact that under no circumstances there exists any more efficient way to increase apoapsis than a prograde burn at periapsis, the following bound exists:

$$\Delta v \geq \sqrt{\frac{2}{r_{P1}} - \frac{2}{r_{P1} + r_{A2}}} - \sqrt{\frac{2}{r_{P1}} - \frac{2}{r_{P1} + r_{A1}}}$$

(a simple vis-viva calculation)

Assuming $r_{A2} \geq r_{A1}$, which can be assumed without any loss of generality since the orbits could otherwise be swapped.

Inclination and argument of periapsis obviously add some non-zero cost on top of this, but it's nevertheless a lower bound.

  • $\begingroup$ Isn't the lower bound 0 (or infimum, if I got my nomenclature right), as "two arbitrary elliptic orbits" can be two co-planar elliptical orbits with the same orbital parameters expect e.g. infinitesimally small difference in $r_A$? I.e.: $\inf{\Delta v}_{r_{A_2} \to r_{A_1}} = 0$ Or am I missing something? $\endgroup$
    – Ludo
    Oct 5, 2020 at 15:40
  • $\begingroup$ @Ludo What you are missing is that the problem is really "two arbitrary orbits, given their parameters", not "two arbitrary orbits". Your example can be used as a sanity check for any bound, since plugging in the values should yield 0 in that case. $\endgroup$ Oct 5, 2020 at 15:43
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    $\begingroup$ Oh, you're looking for a lower bound as function of all the orbital parameters, not just a numeric answer. Interesting question, wish I had seen it before. $\endgroup$
    – Ludo
    Oct 5, 2020 at 15:45
  • $\begingroup$ Yes, that's exactly what I'm looking for. A function would be a nice way to think about it. (And this question is not resolved yet, by far) $\endgroup$ Oct 5, 2020 at 15:46
  • $\begingroup$ I'll put my thinking hat on. I was really into optimisation etc. in university, but that was 15 years ago. I might be a bit rusty... $\endgroup$
    – Ludo
    Oct 5, 2020 at 15:48

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