What are the lowest delta-v trajectories a spacecraft could take from Earth to EML-1? (Earth Moon Lagrange point 1). provided the payload is massive. ( the amount of time it takes is not an issue)

  • $\begingroup$ I've made some edits to your question. The word "best" is very problematic in Stack Exchange so it's "best" to avoid it, since it could be interpreted as asking for an opinion. Of course you did say lowest delta-v, so it's "best" to drop "best" and just ask only for lowest delta-v. I noticed that you are asking from Earth, and have added the launch-trajectories tag. It might be easier for an answer to separate out the delta-v required starting from LEO (low earth orbit) and then just describe the aspects of the Earth orbit that would minimize the delta-v that the spacecraft itself needs. $\endgroup$
    – uhoh
    Mar 11, 2018 at 6:49
  • $\begingroup$ But if you really want to know the lowest delta-v starting from Earth's surface at launch (if this is a very massive payload and not just a communications link), maybe it's better to mention that more clearly. $\endgroup$
    – uhoh
    Mar 11, 2018 at 6:51
  • $\begingroup$ New edits look good. $\endgroup$
    – uhoh
    Mar 11, 2018 at 14:12

1 Answer 1


In fact it could be arbitrarily low. But wait before you rejoice.

The Moon has an elliptic trajectory. This elliptic motion perturbs any orbit of a satellite in LEO. If you first assume your satellite is in the same plane as the Moon and describe the system as a time dependent Hamiltonian system with two degrees of freedom, a process called Arnold diffusion will occur.This implies that with time, any orbit (outside zero set measure) can be attained without any propellant. However, this effect is much much smaller than orbital precession (it corresponds to semi-major axis perturbations). Roughly, the idea would be to place the satellite in a resonant orbit with the Moon, and jump between resonances if necessary (this can require arbitrarily small quantities of propellant). The problem: Arnold diffusion will work in time scales $\exp(\exp(\frac{1}{v}))$, with v (up to some constants) the $\Delta V$ necessary. This will typically be larger than the age of the universe.

To obtain "reasonable" transfer times, you cannot use this effect, and so you will have to use thrust to change the value of the Jacobi integral. Bad news, its maximum is EML-1... You can then just optimize your transfer time by arriving at EML-1 when it's closest. sparing probably a few 10's of m/s $\Delta V$.

  • 1
    $\begingroup$ Welcome to Space Exploration SE. While I'm not familiar with Arnold diffusion or Jacobi integral, I generally understood what you are explaining here. The one part that I didn't understand was "Bad news, its maximum is EML-1." I thought EML-1 was a mathematically defined point or maybe even orbits about that point. Does EML-1 have a value that can be quantified that can be related to thrust? $\endgroup$
    – Bob516
    Dec 31, 2019 at 13:49
  • 1
    $\begingroup$ @Bob516 it's possible that we can just insert the word "at" or "located at" before "EML-1". see this section for example. I think the "bad news" spoken of here is that L1 and L2 are maxima of the Jacobi integral (think "energy") so no matter what tricks are available, you'll still need to do some work to get there; we can't just "fall into" them. I'm sure user34417 can explain this better, I'm just speculating. $\endgroup$
    – uhoh
    Dec 31, 2019 at 14:49
  • $\begingroup$ @user34417 Welcome to Space! There are several questions tagged three-body here, more tagged halo-orbit and a zillion tagged orbital-mechanics $\endgroup$
    – uhoh
    Dec 31, 2019 at 14:56
  • $\begingroup$ Yes the Lagrange points are the critical points of the Jacobi integral. As the function is in practice (over reasonnable times) conserved, you need to apply work to change its value no matter what trajectory you use. Arnold diffusion is much more efficient near separatrices, but these are the levels of Jacobi integral of L1,L2,L3 anyway. $\endgroup$
    – user34417
    Jan 2, 2020 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.