Wikipedia gives 0.64 or 0.65 km/s for going from LLO to EML-1 or back. I would like to know what the delta v of the specific burns involved are, and if my initial calculations are close, I don't see how their number can be correct.

Here is Wikipedia article about delta-v budget.

It's complicated because it's a full 3-body problem, so I'm left trying to approximate it with a simplified model in the Moon's sphere of influence. LLO I have taken to be at 110 km, at 1.629 km/s. I have 61,000 km for the distance from the Moon to the EML-1 (from the approximation $R \cdot \sqrt{\frac{m_1}{3 \cdot m_2}}$).

First burn, I consider getting into an orbit with apogee at 61,000 km, which is at the radius of EML-1. Perigee for the elliptical orbit would be 2.27 km/s, with the difference from LLO being 0.641 km/s, which is our first burn.

NOTE: the numbers here are wrong staring with 0.714 km/s; see my answer for details.

Second burn, I calculate the speed at apogee of that elliptical orbit to be 0.068. However, EML-1 is circling around the Moon, which I would calculate by:

$$\frac{2 \cdot \pi \cdot R}{30 \cdot 24 \cdot 60^{2}}$$

if R is the 61,000 km distance from the Moon. This gives 0.714 km/s. If the orbit is going in the "right" direction, then I subtract the apogee speed from this to get 0.641 km/s for the second burn.

If you added these, it's 1.355 km/s total, and both burns are of about equal magnitude. I don't think that's right. These calculations are probably correct for the equations I'm using, but I strongly suspect that the underlying assumptions are highly flawed. Perhaps Earth's gravity has a major effect in the mechanics? In other cases, the EML-1 and EML-2 points are touted as extremely low delta v points to reach, but these numbers would seem to somewhat undermine that, if the 0.641 km/s burn is going "out of the way". But I'm not sure.

Does anyone have any serious numbers for the transfer from LLO to EML-1?

  • 2
    $\begingroup$ EML-1 or EML-2? $\endgroup$
    – user
    Commented Sep 15, 2016 at 12:36
  • $\begingroup$ My interest was in EML-1, the near-side of the moon. Similar statements might apply for EML-2? Since that point is further from Earth, it would probably follow that the circularizing burn upon arrival at the point would be greater, according to the approximation I used, but this can't possibly be right either. Other people are surely interested in EML-2, but my interest is in EML-1. $\endgroup$
    – AlanSE
    Commented Sep 15, 2016 at 14:28
  • $\begingroup$ Your question title says EML-1 to LLO, but the very first sentence says LLO to EML-2. In the second to last paragraph you talk about both EML-1 and EML-2. Then at the end it's back to LLO to EML-1. $\endgroup$
    – user
    Commented Sep 15, 2016 at 19:11
  • $\begingroup$ That was a pure error. I didn't previously notice that. Edited the incorrect occurrence. $\endgroup$
    – AlanSE
    Commented Sep 15, 2016 at 23:38
  • $\begingroup$ FWIW, the mean distance from the centre of the Moon to EML-1 is ~58,000 km, but it varies between 54,000 km & almost 62,000 km. See i.sstatic.net/Lnqjr.png & astronomy.stackexchange.com/a/49269/16685 $\endgroup$
    – PM 2Ring
    Commented May 27, 2022 at 17:36

2 Answers 2


I realized an error in the math in my question that is significant enough to leave as a formal answer. The error occurred here, and I'll point it out in bold.

However, EML-1 is circling around the moon, which I would calculate by 2*pi()*R/(30*24*60^2), if R is the 61,000 km distance from the moon. This gives 0.714 km/s. If the orbit is going in the "right" direction, then I subtract the apogee speed from this to get 0.641 km/s for the second burn.

The distance from the center of the moon to EML-1 is about 61,000 km, but the calculation erroneously plugged in 322,000 km, which is the distance from EML-1 to the center of Earth. If you plug in the intended (and more logical) number, you come to the conclusion that EML-1 is moving at about 150 m/s relative to the moon.

This is a very low number, but reasonable considering the circumstances. EML-1 is much closer to the moon, so if we're considering moon-based SOI calculations, it'll be notably lower than Earth-based SOI calculations. It is, at 150 m/s versus 700 m/s.

If the orbit direction is the same as the rotation of EML-1 about the moon, then we subtract 150 m/s - 68 m/s = about 81 m/s (rounding after the fact here) in order to get the burn necessary to match the velocity vector of EML-1 for a projectile in transfer from LLO. As km/s, that is 0.081, so it would add fairly marginally to a 0.6 km/s burn at a perigee at LLO altitude, for 0.681 km/s total. The altitude of LLO is also a fungible number which can account for the range given in the slide referenced in the other answer, in additional to a possibility of some marginal gravity drag at LLO departure as well.

Let me embellish on another detail that I blew through in the question. Should it be 150 - 68 or 150 + 68? What direction do LLO orbits typically go in (for planned space missions)? Probably in the direction of the moon's rotation. So let me go point-by-point:

  • Sun rises in the East, so Earth rotates counter-clockwise if viewed from N. Hemisphere
  • Moon orbits in same direction as the Earth rotates, so counter-clockwise
  • Moon is tidally locked, so it rotates counter-clockwise as well
  • EML-1 rotates counter-clockwise about the moon by similar statements

Thus, it should be subtracted, instead of added. Possibility still remains for a 0.13 km/s modification (increase) if the mission is retrograde in LLO, which is admittedly unlikely. That could be why the range goes from 0.6 km/s to 0.8 km/ in the referenced slide, although I find little reason to ever assume the higher number.

The validity of the approximation is still highly questionable. Perhaps this the number here is still a pretty good ballpark. On Hop's blog, several mentions are given to a approx 0.5 km/s burn to drop from EML-1 into an Earth grazing orbit (not so different from 0.7). That should be a 3-body problem, but you can find that the Earth SOI approximation is very acceptable as a ballpark estimation. If the moon-side follows the same pattern, then we would conclude that 80 m/s is relatively close to the true answer.


This is not a complete answer, but it is a start. I think you may not be able to arrive at correct numbers for a full 3BP maneuver doing piecemeal transfer orbit calculation. Your arithmetic may not ever add up to the real values that way.

The value of 0.65km/s from a halo orbit around Earth-Moon L1 to LLO can be found on this slide from the ULA presentation "Transportation Enabling a Robust Cislunar Space Economy".

There is also a lot to read in Zazzera, Topputo, and Massari's "Assessment of Mission Design Including Utilization of Libration Points and Weak Stability Boundaries" and Koon, Lo, Marsden and Ross’ (KoLoMaRo’s) ”Dynamical Systems, the Three Body Problem, and Space Mission Design" as well.

Here is just one example of getting from LEO to the EML1 halo: Zanzottera, Mingotti, Castelli and Dellnitz's "Earth-to-Halo Transfers in the Sun–Earth–Moon Scenario"

Bear in mind: the ellipses (and for that matter the distances) in this slide are schematic only. Those are not what the orbits actually look like.

enter image description here

I haven't found an independent source for L2, but but at least the article in the Web Archive cited by Wikipedia seems to treat L1 and L2 as roughly equivalently difficult (or easy) to access. Wikipedia's tabular values for high thrust:

enter image description here

...and low thrust:

enter image description here

...are all in the 0.6 to 0.8 km/sec range.

  • 1
    $\begingroup$ Thanks, your answer helped contextualize some. I can see how the network additions add up, and the slide shows clearly that they meant to use that number. I browsed through the pdfs, but I'm not hopeful about any of them giving the number I wanted (the delta v going from EML-1 to the transfer orbit). I do see, however, that the number of 0.64, 0.60 or 0.80 is almost exactly the escape velocity from the moon's SOI. The implication is that the burn to match EML-1 velocity when you get there is "essentially zero". I'm willing to accept that, but I still seek a more scientific confirmation. $\endgroup$
    – AlanSE
    Commented Sep 18, 2016 at 10:58
  • $\begingroup$ Yep, like I say "This is not a complete answer, but it is a start" - I am hoping someone more familliar with this will chime in. @hopdavid 's blog - including this entry linked in this answer has some interesting reading (so does this one as well). $\endgroup$
    – uhoh
    Commented Sep 18, 2016 at 11:15

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.