# How deep is the force well of L4 and L5 Lagrangian Points of Earth-Sun set?

The Lagrangian Points are points in space, where the combination of gravitational pull of a set of two bodies and the centripetal force of orbiting one of them add up to zero. The special property of $L_4$ and $L_5$ points - in Earth-Sun system, located on Earth orbit a sixth of its length away from Earth, in leading and trailing direction respectively, is that they are actually stable - while, similarly to other Lagrangian Points they are local maxima of gravity field, unstable gravity-wise, the Coriolis Force of the set of the bodies creates local minimum there, overcoming the force of gravity locally and making them act similar to gravity wells; anything that gets sufficiently near and moving sufficiently slowly, will remain there indefinitely (or until impacted by a sufficiently fast meteorite).

That makes them very interesting from space exploration point of view, as potential points where e.g. deep space observatory unaffected by requirement of constant rotation around Earth could reside; also they are expected to have collected quite a few meteorites and shed some light on Earth history, and generally provide a very valuable point at stable distance from Earth and not just barely over its surface.

Now I'm aware the sum of forces keeping bodies inside $L_4$ and $L_5$ is fairly weak. I'm interested how weak it is are. Since they are just empty points in space, and not planets with own gravity, obviously quite a few measurements are not applicable, but I guess we could get something indicative, like what is the escape speed from these points' "force well", what is the centripetal acceleration at the "steepest part of their slopes" or such.

I'm quite interested if a space station located there would require active stabilizing to prevent escaping them, or opposite, they would cost significant extra fuel and energy just to escape them.

[see comments for discussion - this question used to ask about "gravity well" of these points, but it was shown they don't actually have any, being maxima like others; still, Coriolis Force seems to act like gravity for all practical purposes there, and I'd be quite interested in learning how strong it is there.]

• the concept of the interplanetary highway also relies on the gravity wells of lagrange points, see if you can find somthing linked from the corresponding wikipedia article
– mart
Commented Jul 17, 2013 at 15:49
• Commented Jul 17, 2013 at 15:56
• @Manishearth: Interestingly, answers from your link suggest that the force that makes these points stable is the Coriolis force - while gravity is there at equilibrium (doesn't disturb) it's the Coriolis' force that prevents objects from leaving the places. That would mean there is no g, the gravitational acceleration there, but there is a viable substitute, a measurable centripetal acceleration coming from the Coriolis force. That's not a gravity well, but a measurable force well nevertheless.
– SF.
Commented Jul 17, 2013 at 16:37
• The relativly low escape velocity from a Lagrange point is used in the comncept of the interplanetary transport network, look at the articles source there. Vaguely related question
– mart
Commented Feb 18, 2014 at 13:54
• I think "and the centripetal force of orbiting one of them add up to zero" should read "and the centrifugal force of orbiting both of them add up to zero." Lagrange points are usually described in a rotating frame of reference where the non-existent centrifugal force exists. Lagrange points orbit both bodies, in the same direction, and with the same period. Commented Jan 8, 2022 at 22:07

L4 and L5 are stable in an ideal circular 3 body scenario where the central body is 26 times more massive than orbiting body. But that isn't a very accurate model for the real world. While the sun is a lot more massive than 26 times earth's mass, there are more than 3 bodies exerting an influence.

When I included the Sun's influence in my orbit sims, the Earth-Moon L4 and L5 were destabilized by the Sun's influence. I haven't looked at Sun-Earth L4 and L5, but I would expect Venus' influence to destabilize Earth trojans. Venus would come within 0.28 AU of an earth trojan each 1.6 years (Venus Earth synodic period). Jupiter also exerts a substantial tug.

Jupiter has a bunch of trojans. But, outside of the Sun, Jupiter is by far the biggest frog in the pond. Jupiter's trojans are less vulnerable to perturbations by the weaker planets.

So I would venture to guess the escape velocity from Sun-Earth trojan regions is 0 km/s.

Edit: Did some Googling, there's an Earth trojan 2010 TK7. But it's not very stable. Wikipedia says:

2010 TK7's orbit has a chaotic character, making long-range predictions difficult. Prior to A.D. 500, it may have been oscillating about the L5 Lagrangian point (60 degrees behind Earth), before jumping to L4 via L3. Short-term unstable libration about L3, and transitions to horseshoe orbits are also possible.

• Rereading this answer, the opening should read to specify that L4 and L5 are (linearly) stable in the ideal CR3BP for mass ratios below the critical mass ($\mu < 0.038$), which is valid for the Earth-Moon system. Commented Feb 13, 2015 at 3:49
• I hadn't even thought about μ in this question's context since the sun's so much larger than earth. But you're right -- I attempted to correct my answer. I'm not sure what you mean by (linearly) stable. Commented Feb 13, 2015 at 16:34
• @jah138 maybe you could answer one of my questions: astronomy.stackexchange.com/questions/3595/… Commented Feb 13, 2015 at 16:39

Probably the best way to think about gravity wells is to look at the delta-V required (proportional to energy required for a given mass) to move from one point to another in the well. This graphic is particularly good. From it, you can calculate the delta-V to "get around the neighborhood."

• If that graph contained L4/5 -> L3 transition, it would pretty much answer my question.
– SF.
Commented Jul 17, 2013 at 20:36
• It's not entirely infinitesimal - since L4/5 are both "stable equilibrium" so they need some escape velocity.
– SF.
Commented Jul 17, 2013 at 21:53
• Yeah @SF. I stand corrected. There is certainly non-zero energy required.
– Erik
Commented Jul 17, 2013 at 22:00
• This is actually a very interesting question. It turns out that L4/L5 are stable due to coriolis forces not due to the potential energy surface. They are actually local maxima on the potential energy surface. See this: physics.stackexchange.com/questions/36092/…
– Erik
Commented Jul 17, 2013 at 22:08
• See comments under my question, above ;) I agree though, this is interesting - energy of escaping the Coriolis Force well?
– SF.
Commented Jul 17, 2013 at 22:44