I was looking at Delta-V budgets and noticed that it costs 4.1 km/s to get from LEO to L4/5, but just 3.2 km/s for LEO to geocentric C3 = 0 (escape velocity).

I find that counter-intuitive. Why does it cost less to escape Earth's gravity than it does to get to L4? Wouldn't getting to L4 be helped by the Moon's gravity? Is it because the budget for L4 includes slowing down?

  • $\begingroup$ Interesting; more delta-v to gain less potential energy... perhaps this is not absolutely true, and the problem is in the specific assumptions or routes used for this graphic. I think there have been one or two previous posts here calling this diagram... "less than ideal". $\endgroup$ – uhoh Jan 16 at 11:19
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    $\begingroup$ Getting to almost any high orbit takes more $\Delta v$ than C3=0. Think of it as getting on an escape trajectory is just 1/2 of a Hohmann transfer to another orbit. $\endgroup$ – asdfex Jan 16 at 12:08
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    $\begingroup$ To escape you just have to raise your apogee. To get to high orbit you also have to raise your perigee. Alternatively, you can get to L4/:5 more cheaply on a near-escape orbit, but you have to "stop" (match orbit) when you get there. $\endgroup$ – Steve Linton Jan 16 at 12:12
  • $\begingroup$ L4/L5 cost is very similar to L3, which has been answered here: space.stackexchange.com/questions/15469/… $\endgroup$ – asdfex Jan 16 at 12:13
  • $\begingroup$ @asdfex I don't really see an answer to "Why does it cost more...?" there. I think the answer is along the lines of (at)SteveLinton's comment above, but it will need some math to make it a real answer. $\endgroup$ – uhoh Jan 16 at 15:23

For the purpose of $\Delta v$ planning, an orbit at Earth-Moon L4 or L5 is about identical to just any orbit with a radius of 380,000 km. At L4/5 distance to Earth and Moon are identical and as the gravitational force changes with $r^2$, the pull of the Moon is just 0.01% of that of Earth at this point and can be neglected. It's sufficient to keep the orbit somewhat stable, but doesn't change the $\Delta v$ to get there.

As Steve Linton already commented - If you are in LEO and raise your speed (to be precise, energy) to C3=0, you have enough energy to escape Earths gravity well, but not enough energy to get into a high Earth orbit. If you check the chart, even going to GEO takes a lot more $\Delta v$ than reaching C3 = 0.

Adding 3.2 km/s (just short of C3=0 to avoid infinities) to your speed in LEO (200 km height) brings you to an orbit with a huge apogee (millions of kilometers), but the perigee is still in LEO. To get into a circular orbit in any height, you need to spend more fuel to raise the perigee to the same height, which will cost you another substantial amount of fuel. To get to L4/5 you have to spend 3.13 km/s in LEO, and another 0.83 km/s to raise the perigee.

And, indeed you are right, that L4 can be reached via a swing-by maneuver around the Moon, increasing the orbital speed at apogee and therefore decreasing the magnitude of the second burn (in the ideal case, by about 60%). In fact, the same could be used to get to L5: L5 is trailing the Moon by 60°, but at the same time can be thought of leading the Moon by 300°. I.e. you can "overtake" Moon, take the "long way" around the orbit and finally end up at L5. In both cases you need yet another, tiny, burn to adjust the orbital velocity to the right value once you reached L4 or L5.


tl;dr: The chart is outdated and unsupported and we are operating deep into three-body territory where Kepler orbits don't address the reality and talking about geocentric $C_3=0$ doesn't even make sense.

fyi I've just added a bounty to A spacecraft leaves Earth exactly with escape velocity V2 - what trajectory it will have in Solar System?

I find that counter-intuitive. Why does it cost less to escape Earth's gravity than it does to get to L4?

I'm not sure if this is really true or not, there are brute force powered ways to to get to Earth-Moon Lagrange points and there are more complicated 3-body techniques to get to, and move between Lagrange points, and a proper summary would have to address these and specify the orbital mechanics used to arrive at specific delta-v numbers.

This chart is problematic because it is not accompanied by any such explanation. The Wikipedia section on these(?) numbers laments this problem. From Delta-v budget; Earth-Moon space; high thrust:

The reference for most of the data[2] no longer works, and some things are not clear, such as why there is such a big difference between going from EML2 to LEO versus going from EML1 to LEO.

That plus what I wrote above:

...there are more complicated 3-body techniques...

Tells us that we have found ourselves deep three-body territory, and therein lies the rub!

The whole concept of $C_3$ is based on there being only a single central force, i.e. a Kepler orbit and that simply does not work well when:

  1. we are in cis-lunar space trying to settle down near Earth-Moon Lagrange points where our spacecraft plus Earth plus Moon is a proper three-body problem.
  2. we are working near escape from Earth's gravity (Earth's Hill sphere), where the spacecraft plus Earth plus Sun is a proper three-body problem.

It's time to stop writing prose answers that:

  1. assume that the chart is correct and needs to be defended and rationalized
  2. apply 2-body arguments to a pair of three-body problems suggesting this is a correct way to think.
  • $\begingroup$ What do you mean saying "The concept of C3 doesn't work well"? It's a reference point for calculations, like "LEO", just well-defined. $\endgroup$ – asdfex Jan 17 at 10:45
  • $\begingroup$ @asdfex It says "...doesn't work well when..." followed by an enumeration of two specific situations. It literally says what I mean. $\endgroup$ – uhoh Jan 17 at 12:02
  • $\begingroup$ @asdfex For example, we can talk about a 200 x 600 km orbit and that "works well" in that we know what it means. However even written to look like altitudes, it still has a +/- 5% ambiguity on minimum altitude depending on where periapsis falls. I would say that it works well in prose, but as an indicator of altitudes it doesn't work well. $\endgroup$ – uhoh Jan 17 at 12:04
  • $\begingroup$ Btw, this is not an answer to the question. It doesn't ask if the dv-plot is good or bad. It asks why going to L4 is more expensive than an escape trajectory. This is a fact, independent from any chart. $\endgroup$ – asdfex Jan 17 at 12:05
  • $\begingroup$ @asdfex Yes this is a good Stack Exchange answer to the OP's question. The premise of the question is that the chart is right and conveys undisputed facts. I've explained/advised why one should not simply believe and defend the chart and that one needs to go deeper before one can be certain which has the minimum delta-v when the type of trajectory and method of propulsion and number of impulses are unspecified. I then cited an independent source also describing problems with this chart. There are other answers here as well that point out problems with it. $\endgroup$ – uhoh Jan 17 at 12:11

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