For what I know, L4 and L5 are high points. Although, they are stable points. Why? I thought it was the same being a high point and being a unstable point.


2 Answers 2


It does indeed seem counterintuitive that $L_4$ and $L_5$ would be at the same time both high points of potential as well as stable points in the system. In fact, a quick look at an example contour plot with all five Lagrangian points demonstrated would also suggest that $L_4$ and $L_5$ would be unstable:

From Wikipedia: Lagrangian point

In the picture you can see that $L_1 - L_3$ are placed in saddle nodes, and intuitively one concludes that the smallest perturbation would set them "falling" in either direction, i.e. they need station keeping.

However, $L_4$ and $L_5$ would still seem to be unstable, so what gives? What makes $L_4$ and $L_5$ stable is that each of them is located equidistant from both of the masses. This leads to the gravitational forces from each of the bodies towards $L_4$ and $L_5$ to be in the same ratio as the two bodies' masses, hence the resultant acceleration points towards the barycentre of the system (which is also the centre of rotation for a three-body system).

This resultant acceleration, and by extension force, leads to the object in either $L_4$ or $L_5$ to in fact have just the correct amount of acceleration / correct force towards the system's barycentre so that it will not fall out of orbit (it "falls" towards the barycentre as the system rotates, like Earth similarly falls towards Sun throughout its orbit), as counterintuitive as it sounds at first. (Furthermore, this effect is not only limited to bodies of negligible mass, but is independent of mass, which is why you can have more massive bodies orbiting in $L_4$ and $L_5$ like Jupiter's trojans, as well as smaller, artificial satellites.)

Source: Lagrangian point - Wikipedia

  • $\begingroup$ Ok, I think I need to think about it. I don't understand why, if a body is a little closer to the two massive bodies, it doesn't tends to leave L4 or L5 but it goes back and orbits the point. I guess I must give it a nap (spanish way). Thanks for your answer. $\endgroup$
    – Felix L.
    Commented Aug 22, 2015 at 22:07
  • 1
    $\begingroup$ @Krotanix No problem. I actually now found this Physics SE answer that describes why a body tends to orbit $L_4$ or $L_5$: when a body is perturbed away (at a low velocity) from the $L_4$ or $L_5$, the Coriolis force created by the rotating system curves the body's trajectory towards the Lagrangian point, resulting in the body doing this repeatedly and leading to it orbiting the Lagrangian point. If the body's velocity is greater than a certain threshold, it "escapes" Coriolis force and removes itself from its orbit around the Lagrangian point. $\endgroup$
    – V-J
    Commented Aug 22, 2015 at 22:25
  • $\begingroup$ Wow, thank you very much. You made my day :) Thank you! $\endgroup$
    – Felix L.
    Commented Aug 23, 2015 at 0:14

Strictly speaking, the premise of your question is false. The L4 and L5 points in a general system are not necessarily stable. Whether they are stable or not depends on the ratio of masses in the system. It does turn out that in many systems of interest, the ratios are such that the these points are stable.

When they are stable, it's basically because the Coriolis force balances the gravitational force.

See, for example, www.physics.montana.edu/faculty/cornish/lagrange.html

  • $\begingroup$ Thanks. I've read some papers and for what I've seen, it can also be proved that these points are stable (or not) by the sign of their eigenvalues. Seen in: Hirsch, Smale, Devaney. Differential equations, dynamical systems, and an introduction to chaos (2ed., Elsevier, 2004)(ISBN 0123497035)(432s)_PD_ $\endgroup$
    – Felix L.
    Commented Aug 24, 2015 at 16:22
  • $\begingroup$ @Krotanix Yes, what you said is consistent with what I said. Ultimately the sign of the relevant eigenvalues depends on the ratio of masses of the two main bodies. The link that I referenced (among others that you can easily find with a search engine), address that. $\endgroup$
    – Brick
    Commented Aug 24, 2015 at 16:24

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