The L4 and L5 Lagrange points, 60 degrees in front of and behind the Moon in its orbit around Earth, orbit at the same speed as the Moon. But when a spacecraft flies to the Moon, the gravity of the Moon captures it, and then it only has to expend fuel to set up its orbit around the Moon the way it wants.

Though the gravity of the Earth and the Moon balance at the L4 and L5 points, there is nothing at those points to pull a spacecraft into orbit around them. So, does a vessel have to brake to set up an orbit around those points? How does the delta V to orbit those points compare to going from Earth into orbit around the Moon?

map of Lagrange points - L1 between primary and orbiting secondary, L2 on the far side of the secondary from the primary, L3 on the opposite side of the primary from the secondary, L4 60 degrees ahead of the secondary, L5 60 degrees behind.

Image courtesy NASA, originally posted here


2 Answers 2


According to Wikipedia, starting from LEO, it's almost exactly the same delta-V cost to enter low lunar orbit (~4.04km/s) as it is to reach the Earth-Moon L4/L5 (~3.99km/s), though the source for the information is fragmented. The braking costs are included:

Note that getting to one of the Lagrange points means not just getting to the right place but also adjusting the final velocity in order to stay there.

Your mileage may vary; another reference on Wikipedia gives ~4.1km/sec to EML4/L5 and ~3.9km/sec to lunar orbit.

The value for LEO to EML4/5 corresponds to that for a Hohmann transfer to the Moon's altitude, reasonably enough. Since that maneuver matches the moon's speed, but lunar orbital speed is around 1600 m/s moon-relative, it seems very counter-intuitive that the same ∆v could get you into orbit -- but in the case where you're going to the moon, the moon's gravity accelerates the approaching spacecraft by almost exactly the right amount to make up for it!


Enter perigee and apogee altitudes in my Hohmann Spreadsheet and the circularization burns show the delta V to get from circular orbit at perigee altitude to a circular orbit at apogee altitude. Here's a screenshot: enter image description here

So it takes about 3.9 km/s to get from LEO to L4 and L5. The moon is a lunar distance from either point so we don't get much help from the moon.

Using the Farquhar Route it takes about 3.5 km/s to get from LEO to EML2.

enter image description here

If you're willing to take a few months, LEO to EML2 is as little as 3.1 km/s. This route sends a payload to the edge of the Hill Sphere and uses the sun's influence to boost perigee to the EML2 neighborhood. See Hop's Route.

From EML2 it takes about .15 km/s to drop a rocket to a low altitude perilune (See the Farquhar illustration about). At perilune .61 km/s suffices to circularize. So LLO can be reached for as little as 3.1 + .15 + .61 which is about 3.9 km/s. Which is about the same as entering LLO directly from an earth to Moon Hohmann.

Here's delta V from EML2 to circular lunar orbits of various altitudes:
EML2 to 2000 km altitude .5 km/s
EML2 to 4000 km altitude .39 km/s
EML2 to 8000 km altitude .24 km/s

To get LEO to the above orbits add 3.1 or 3.5 km/s depending on whether you want use the 9-day Farquhar route or the longer route from the edge of earth's hill sphere and back.

  • $\begingroup$ Interesting - so, a similar approach could be used to reach L4 or L5 on return from the edge of the Hill Sphere, at i suppose a fairly similar delta V cost? Sounds very worth it, unless perhaps you had people on board. $\endgroup$
    – kim holder
    Apr 30, 2018 at 23:45
  • $\begingroup$ @kimholder 3.1 km/s can get a ship close to the edge of earth's Hill Sphere where (with correct timing) the sun's tidal influence can raise perigee to L4 altitude. This would still need about a .5 km/s circularization burn at L4 or L5 though. The moon's too far away to lend a hand with gravity. $\endgroup$
    – HopDavid
    May 2, 2018 at 0:14

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