This answer to How much mass can be put in an L4 or L5 and it still maintain reasonable stability? has left me with the gedankenexperiment of the Earth in a circular orbit around the Sun, and the Moon placed somewhere near where the Sun-Earth Lagrange point would normally be.

Question: Could the Moon placed near the Sun-Earth L1 point remain in a heliocentric 1:1 resonant orbit with Earth? If Earth were still at 1 AU, what would the Moon-Earth distance be?

  • 1
    $\begingroup$ L1 is unstable, so the Moon would soon drift away from it because of perturbations from the other planets. $\endgroup$
    – PM 2Ring
    Commented Oct 2, 2021 at 8:10
  • $\begingroup$ @PM2Ring since the Moon has substantial mass, there is no longer any such thing as L1. The whole point of my linked answer is that this is not the restricted three-body problem of Lagrange any more. An answer here will need to be more than a one-liner. It will likely need some combination of math and source-citing. $\endgroup$
    – uhoh
    Commented Oct 2, 2021 at 8:18
  • $\begingroup$ @uhoh, PM 2Ring is correct. This question is nonsense, and your comment says it's nonsense. The question asks "Could the Moon placed near the Sun-Earth L1 point remain in a heliocentric 1:1 resonant orbit with Earth?" and your comment says "since the Moon has substantial mass, there is no longer any such thing as L1." So which is it? $\endgroup$ Commented Oct 2, 2021 at 9:54
  • $\begingroup$ @uhoh Lagrange's solution was for arbitrary masses, not just the restricted 3-body problem, so it's perfectly sensible to talk about L1, L2, etc. in such cases. It was shown in the 19th Century that the L1, L2, and L3 points are always unstable, while the L4 and L5 points are stable when the sum of the masses of the smaller objects relative to the total mass of all three is small enough. $\endgroup$ Commented Oct 2, 2021 at 9:58
  • 1
    $\begingroup$ @DavidHammen no the question is not nonsense. If you don't understand the question you can ask for clarification in a comment, rather than use it for disparagement. The Sun-Earth L1 point is a defined concept. It's a place about 1.5 million km from the Earth towards the Sun. It is no longer a useful concept when the massive Moon is there, but we can still understand what area that "near the Sun-Earth L1 point" refers to. $\endgroup$
    – uhoh
    Commented Oct 2, 2021 at 14:31

1 Answer 1


Most discussions of Lagrange points use the simplification of the circular restricted three-body problem, where two objects (e.g., the Sun and the Earth) are massive and in fixed circular orbits about the center of mass, and we are interested in the possible motions of a third particle with effectively infinitesimal mass (e.g., a spacecraft or asteroid). This is simple and easy to understand, and is a good approximation for things like satellite orbits, Trojan asteroids in the Sun-Jupiter system, and the like. But the underlying basis of this was actually worked out for the general three-body problem, where all three objects have nonzero mass (and orbits can be elliptical).

The start of this was work by Euler in 1760s, who described a family of "collinear" exact solutions of Newton's equations where all three objects had orbits that kept them perfectly lined up. Here's an example figure, showing an elliptical-orbit solution for the case where $m_1 : m_2 : m_3 = 1 : 2 : 3$

enter image description here (From Hestenes, New Foundations for Classical Mechanics)

In the 1770s, Lagrange found an additional set of exact solutions, where the three objects are always arranged in a equilateral triangle (the orientation and possibly the size of the triangle changes with time, but the angles between the bodies are always 60$^{\circ}$). Here's an elliptical-orbit example for the same mass ratios as the previous figure

enter image description here (From Hestenes, New Foundations for Classical Mechanics)

In the figure, $m_3 > m_2 > m_1$, so the circular-restricted version of this would have the primary (e.g., the Sun) as $m_3$, with a nearly zero-sized orbit and the secondary (e.g., the Earth) as $m_2$ with a circular orbit. $m_1$ (the test particle) would be located either 60$^{\circ}$ ahead or 60$^{\circ}$ behind $m_2$, in what are conventionally called the L4 and L5 positions.

Work in the 19th Century by Liouville (1842), Gascheau (1843) and Routh (1875) demonstrated that Lagange's triangle solution (i.e., the L4 and L5 positions) was stable only if the sum of the masses of the two smaller objects was less than a small fraction of the total mass, and that Euler's collinear solution (corresponding to L1, L2, and L3) was always unstable. (See, e.g., the discussion in the Introduction of Leleu et al. (2018).)

So while it is in principle possible to have the Moon at L1, L2, or L3 -- that is, a collinear arrangement where the Sun, Moon, and Earth are always on a line -- this is always unstable, and so you could never get it to work in practice.

I haven't been able to find links to Liouville (1842) or Gascheu (1843), but here are the references:

Liouville, J. (1842) "Sur un cas particulier du problème des trois corps." C. R. Acad. Sci. Paris 14: 503.

Gascheau, M. (1843) "Examen d’une classe d’équations différentielles et applicationa un cas particulier du probleme des trois corps." Comptes Rendus 16.7: 393.

  • $\begingroup$ This is excellent, thank you! I will do some more reading first, and then see if I can formulate a question something along the lines of "Are 1:1 resonant orbits around a more massive body possible? (other rational numbers are) If so, under what conditions?" But I'll have to think about it first. $\endgroup$
    – uhoh
    Commented Oct 2, 2021 at 14:37
  • $\begingroup$ Aren't Trojan asteroids examples of "1:1 resonant orbits around a more massive body"? $\endgroup$ Commented Oct 2, 2021 at 14:43
  • $\begingroup$ Yes, once you move off of the zero-velocity potential there are all kinds of stable orbits in CR3BP. I don't recall if those L-points themselves are stable or not, but there's a whole zooniverse of 3-body orbits if the third is massless. scroll down But for the whole Trojan asteroid analogy we think of the third body as massless. When the two are of comparable masses (as in two moons being resonance around a planet) the situation seems different. $\endgroup$
    – uhoh
    Commented Oct 2, 2021 at 14:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .