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The assumptions about the Lagrangian points being stable (...in the traditional meaning of the word: not moving around; they are unstable in the mathematical sense, except for arguably L4, L5) that I've commonly encountered are:

  • a pure 2-body system, with any satellites that sit at these points being of negligible mass.

  • the central body is vastly more massive than the orbiting one

  • the orbiting one moves in a circular orbit.

I wonder though, do they remain reasonably stable - allow for a satellite to sit there with only minimum of station-keeping, in cases of:

  1. the planet's orbit being eccentric, not a perfect circle
  2. The two massive bodies being of comparable mass, orbiting a common barycenter (though both in circular orbits). Specifically, how would they morph/move?

bonus: what about a ternary system, where a distant planet orbits something like a contact-binary star? Does the idea of Lagrangian points assume uniformly round celestial bodies?

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    $\begingroup$ For 2), there's this funny looking constant. L4 and L5 are stable if m1/m2 is greater than (25 +3*sqrt(69))/2, approximately 24.96. As for how they move, L4/L5 and and the two bodies always forms a equilateral triangle $\endgroup$ Sep 10 '18 at 10:02
  • $\begingroup$ apparently @DavidHammen knows just enough about things like this to be dangerous. $\endgroup$
    – uhoh
    Sep 10 '18 at 11:17
  • $\begingroup$ Somewhere there is also an answer or a comment by @DavidHammen about the stability of a point midway between two stars, perhaps in Astronomy SE, and there might be something there that relates to Lagrange points. dunno... $\endgroup$
    – uhoh
    Sep 10 '18 at 11:42
  • $\begingroup$ @uhoh: I'm quite sure for circular orbits, similar masses still have L1, L2, L3 of old, but as masses change, L1 shift towards the body that gets lighter, and L2 of one body becomes L1 of the other. Can't quite picture how L4 and L5 would behave. $\endgroup$
    – SF.
    Sep 10 '18 at 11:46
  • $\begingroup$ As am I. There are a lot of questions in your question! I'll add another answer to my answer. $\endgroup$
    – uhoh
    Sep 10 '18 at 11:48
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  1. the planet's orbit being eccentric, not a perfect circle

Apparently @DavidHammen knows just enough about things like this to be dangerous. I can't speak to the topic mathematically, but in the figure the little blue cigar is the Earth oscillating towards and away from the Sun as it moves in its slightly elliptical orbit, and the cool looking orbit is in this case SOHO, who's been in a halo orbit for more than 20 years, faithfully executing it's station keeping maneuvers towards or away from the Sun, except when it didn't (see Roberts 2002 linked there).

The near-rectilinar halo orbit of the proposed, future stargate err... ,wormhole, err... gateway will be a (near-rectilinear) halo orbit (1, 2 and answers therein) around the Earth-Moon L1 and/or L2, and the eccentricity of that is much larger than that of Earth's heliocentric orbit ($\epsilon$=0.055 vs 0.017).

I always get my science fictions mixed up. (Space Force!, Rocket Racing!, etc.)

Also, Chang'e-4's radio link to the Earth Queqiao relay satellite will be a the Earth-Moon L2.

So for significantly elliptical systems, yes. For very significantly elliptical systems, someone else will have to answer about the ER3BP, and of course, someone has already asked Are there any natural circular orbits? to begin with.

  1. The two massive bodies being of comparable mass, orbiting a common barycenter (though both in circular orbits). Specifically, how would they morph/move?

In this answer I show just how to calculate the positions of L1 and L2 for two masses. I'll try to make a plot of how the two points move as the ratio of the two masses changes, that will take a half-day, need to get a stable 10-20 before looking at the stable points.

GIF: SOHO orbit, data from Horizons, plot from here

enter image description here

below left: Top-down view, eleven years. right: View from the side, one year. (Sun to the left) James Webb Telescope prototype orbit, data from Horizons, plot from here

image image

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$\newcommand{\r}{\mathbf{r}}$ $\newcommand{\i}{\mathbf{i}}$

Lagrange's set of solutions for the three-body problem is more general than just the points, which are special cases of a broader family of possibilities. The conditions under which they exist are very restrictive, but less so than the standard assumptions you list. Everything here is for the full three-body problem (in which all three masses are non-negligible), not the restricted three-body problem (in which one mass is considered small enough to have negligible effect on the motion of the larger two). These solutions include a form of arbitrary elliptical motion, which I will attempt to describe. The text I am following is Richard Battin's An Introduction to the Mathematics and Methods of Astrodynamics, chapter 8. One possible free alternative to buying the book is to use these lecture notes from MIT Open Course Ware, as used by Battin himself to teach out of his own book, but they offer many fewer words of explanation between the equations, and stop before reaching the final elliptical form I'll reference.

Lagrange's solutions are exact only in the case of there being three bodies in the same plane, all orbiting their common center of mass. Any three bodies are always instantaneously in the same plane, no matter what else is going on, but these solutions all require that the three bodies of interest all stay in that one same plane for all times. They are "points" only from the perspective of a coordinate system rotating with one particular value of constant angular velocity ($\omega$) around the center of mass of the three bodies. The three bodies must be arranged either all along a line (giving L1, L2, and L3), or at the corners of an equilateral triangle (giving L4 and L5). There are lots of other strange and interesting things that can happen in multi-body systems, but these are the only ones which can be described exactly in this relatively simple form.

In the triangular solutions, the triangle must be exactly equilateral, so the bodies must always be equidistant from each other, but the three masses can have any values. If the three masses are the same, the center of mass of the triangle is equidistant from the three equal masses, so the three all rotate in the same circular orbit around it, spaced by 120 degrees. When the masses are unequal, the center of mass (barycenter) moves to a point other than the geometrical center (centroid), so the distances from the three masses to the center of mass are no longer equal, but the distances from the three masses to each other must remain equal. The positions of the bodies relative to the center of mass are by definition $m_1 \r_1 + m_2 \r_2 + m_3 \r_3 = \mathbf{0}$; if the vectors happen to be arranged so that the triangle's three sides ($\r_1-\r_2$, $\r_2-\r_3$, and $\r_3-\r_1$) are all of equal length, then the conditions might be satisfied. The special orbits exist if the velocities of the three objects also happen to work out so that the equilateral shape is maintained (more on this further down). In the special orbits, the heaviest body moves in the smallest circle (since the center of mass is closest to the heaviest thing), and the lightest moves in the largest circle, but all three circular orbits have the same angular velocity. This is contrary to our usual experience, where the farther away a body is, the slower its angular velocity; but that's an expectation derived from Kepler's third law ($\omega^2 = GM/r^3$), and Kepler's "laws" only apply exactly in a two-body system. The corrected version, which applies exactly in the three body equilateral triangle case (the straight line version is much messier), is $\omega^2 = G(m_1 + m_2 + m_3)/r^3$, where $r$ is the distance from the three masses to each other.

When all three masses are in a line, the equations describing the allowed conditions get more complicated. Since the center of mass must also lie on the line segment connecting the three masses, two of the masses will be on one side of it, rotating exactly in phase with each other, while the third rotates exactly 180 degrees out of phase with the other two (in the degenerate case, if the mass ratios are arranged just right, the middle body might be located at the center of mass, in which case its "orbit" has zero radius, with the other two exactly balanced to circle around it). Again, each orbit is a perfect circle when viewed with respect to the center of mass, but if you try to plot them as paths around each other, they aren't circles anymore. Numbering the bodies in some order, we write the coordinates of those bodies as multiples of some arbitrary vector $\i$ from the center of mass along the line, as $\r_1=r\i$, $\r_2=(r+\xi)\i$, and $\r_3=(r+\xi+\xi\chi)\i$, and attempt to solve for $\xi$ and $\chi$ in terms of the masses and $r$ to find out where on the line the three must be placed. Some algebra results in an equation for the allowed values of $\chi$, which is $$(m_1 + m_2)\chi^5 + (3m_1 + 2m_2)\chi^4 + (3m_1 + m_2)\chi^3 - (m_2 + 3m_3)\chi^2 - (2m_2 + 3m_3)\chi - (m_2 + m_3) = 0$$ This, and corresponding equations for $\xi$ and $\omega$, are the source of the "funny constants" found in formulas for L1, L2 and L3. Battin's comment is "The condition equation has one and only one positive root as can be seen from the fact that the coefficients change sign only once. However, a total of three different straight line solutions exist since two more can be obtained by a cyclic permutation of the order of the masses."

There is a way to get ellipses back into this, if we consider maximizing the generality of what has come so far. In both the triangular and linear cases, the allowed shape of the positions giving rise to exact solutions is tightly constrained by the ratios of distances, but there is also an overall scale factor we haven't played with yet. That is, for any three masses whose positions $\r_1$, $\r_2$, and $\r_3$ satisfy one of the cases described so far, for an arbitrary scale factor $\rho$, the positions $\rho\r_1$, $\rho\r_2$, and $\rho\r_3$ also satisfy the same case, though the rotational velocity $\omega$ has to decrease as $\rho$ increases. This even works dynamically, where the scale factor varies with time, under certain conditions. In particular, if we now consider a coordinate frame rotating with a variable angular rate $\omega=d\theta/dt$, the consistency conditions on $\rho(t)$ and $\theta(t)$ turn out to be identical to the standard Keplerian two-body solution! The three orbits of the three masses around their common center of mass can have any eccentricity you like, so long as the three orbits all share the same eccentricity, and are phased so that the equilateral triangle or collinear setup grow and shrink together as they rotate together, via the familiar $$\frac{d}{dt} \left( \rho^2 \frac{d\theta}{dt} \right) = 0 \,\,\, \mathrm{and}\,\,\, \frac{d^2\rho}{dt^2} - \rho\left(\frac{d\theta}{dt}\right)^2 = -\frac{\mu}{\rho^2}$$ For more than three bodies, all of these equations are no longer correct; but if the masses of everything else are very small compared to the big three, then they will remain approximately correct, at least for the big three themselves. This is essentially the same statement as saying Kepler's laws, which are exactly true only in the case of just two point masses, will remain approximately correct, as a variety of small correction terms are added. I don't know of any actual three-star systems which have anything close to the structure necessary for three-body exact Langrange solutions to describe them at all well, but if you're looking to make a science fiction story about Lagrange three-body orbits, it would be a fun way to set up the three stars, and let things of relatively negligible mass move in their gravity field.

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  • $\begingroup$ Another possible free alternative to buying the book is to borrow it from a library (remember those?), or to spend a day at a library and read it there. :-) Wonderful answer! In the spirit of the question as asked, is it possible to add some definitive yes/no's to the following? Starting from the standard CR3BP then expanding more slowly in concept space, do "Lagrange's set of solutions for the three-body problem" allow equilibrium points for 1) more equal masses rather than one being much larger than the other? 2) for the elliptical restricted three body problem? $\endgroup$
    – uhoh
    Mar 2 at 0:45

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