Para-Lagrange Points?

After wondering about a apparently odd feature of Elite 'planetodynamics' and pondering the writeup on the Lagrange Point Wiki page

... points of equilibrium for small-mass objects under the gravitational influence of two massive orbiting bodies.

and relating that to Elite Dangerous, it seems there might be an additional kind of equilibrium point in the Elite Dangerous universe.

I'm thinking of the 'center' of a pair of bodies orbiting each other, i.e. orbiting a barycenter located at the center of their respective orbits, yet about 'equally far' from the bodies themselves. This is close to a Lagrange L1 point, but is 'stationary' in relation to the two bodies, while L1 orbits along with the minor body. (When I try to think of what locations of L2--L5 might be in such binary systems, my mind squeals in distress ...)

I suspect this is highly unlikely in the real universe, but there are certainly many examples of binary planets in Elite.

Anyone know if this kind of 'para-langrangian' system has been discussed anywhere?
 
When I try to think of what locations of L2--L5 might be in such binary systems, my mind squeals in distress ...)
Relax the calcualtions are the same for binary objects as singular objects. You just use the mass of both planets at the barycenter and then it calculates out the same way. At the distances involved since they're locked together you treat them as a single body just like you a planet and it's moons are treated as a single body.
 
Relax the calcualtions are the same for binary objects as singular objects. You just use the mass of both planets at the barycenter and then it calculates out the same way. At the distances involved since they're locked together you treat them as a single body just like you a planet and it's moons are treated as a single body.

I see I was unclear. I'll try to rephrase: "where are the Lagrange points in a 2-body system in which the bodies are double-planets as in Elite Dangerous?" (The system map brackets the bodies.) The 'normal' IRL situation is that the two bodies are so different that the minor orbits the major. That's the kind of situation we have in the Jupiter/Sun system. But I'm asking about a system in which the bodies are almost the same: approximately the same size, approximately the same gravity and velocity. Instead of a minor/major we get a system of two equals orbiting ... each other.

In the Sol system, Charon/Pluto are an example of what I'm thinking of (though I would prefer the bodies to be more similar). Note that the system map 'brackets' the bodies

The Charon/Pluto system is not quite what I'm thinking of, but it is probably the closest we have in-real-life. ED models them as two bodies orbiting a common barycenter, which in turn orbits Sol. (I would prefer a system with 'orbits' fairly close in size and extent, but ... ) In LHS 2938, we have 1/2 (planets) as well as 8C/8D (moons) showing the same relation.

L1 (normally following the orbit of the minor body) still remains between B1/B2, but will as far as I can see coincide with the barycenter of the B1/B2 system. (I am assuming that Elite Dangerous models the orbital situation correctly, with both bodies locked in orbits 'opposite' each other.)

But is there an L2 or an L3 outside the bodies? And are L4 and L5 still sufficiently stable to allow debris to collect ('trojans')?
 
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but will as far as I can see coincide with the barycenter of the B1/B2 system.
The barycenter will function as a lagrange point because it's the centre of the two bodies but L1 is calculated from the barycenter as if the two bodies were one. You cannot calculate them independently as their rotation around each other with destablisie the L points each has as an indepenendent body except for the barycenter. Their combined mass and barycenter pushes the points further and you'll get some vague form of stability that should be able to function as a normal lagrange cloud.

I haven't gone looking but I'm not aware of ED ever placing a lagrange cloud at the barycentre. I'm not sure it technically is one since it's it's own thing.
 
I haven't gone looking but I'm not aware of ED ever placing a lagrange cloud at the barycentre. I'm not sure it technically is one since it's it's own thing.

Something like that (at least for L1) would be the next question in the series, but that moves into tinfoil territory, so I'm not going to ask it in Exploration.
 
And are L4 and L5 still sufficiently stable to allow debris to collect ('trojans')?

Do you mean something like that?
it's a triple, my only find of an ahead and following Trojan so far
20250711192237_1.jpg

sdd en.png


the barycenter of 16 +17 has the same semiMajorAxis as 18 and 19
sdd b.png
 
No ... looks like a nice find, though.

I'm concerned with the three-body problem, of which the Langrange points are a special solution in that they stay in a fixed position relative the other two.

And I'm concerned with the special case where the two main bodies form the typical Elite double-planet, i.e. they rotate around each other. In Sol, the Charon/Pluto pair are fairly close to that. I don't include Sol in the equation, I'm only interested in what the L points the Charon/Pluto pair (or similar in other systems) may exhibit on its own.

The Charon/Pluto L1 must be their barycenter.

Normally, the L2 point is on the line through the two main bodies, but beyond the smaller of them. But ... in the situation I'm thinking of, none of the two bodies can be said to be the smaller. I'm guessing that any L2 point is either inside the minor body (in Charon/Pluto it would be inside Charon), or very close to its surface, but if the bodies are equal in gravitational strength, it's not as clear where the locus is (addded: as none of the bodies is the smaller).

L3 would also be on that line, but beyond the larger of them. But again, if neither is larger ... (In a perfecttly balanced system, the orbits of the two bodies would probably coincide.)

We probably have L2 split between the two bodies, together with L3 similarly split, and ... intuitively ... colocated with both L2 in body1 and L2 in body2.
(A split L2 and L3? Weirdly interesting. But not something I associate with Lagrange points ... so we may have a special-special solution.)

L4 and L5 ... I leave that for now. I'm going to use that instead of counting sheep.
 
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The Charon/Pluto L1 must be their barycenter.

Calculation of the L4 and L5 Lagrange points don't care about how big the objects are relative to each other. The L4 and L5 points are always at the vertices of equilateral triangles, no matter what the masses of the 2 objects are. The relative masses will have an impact on exactly how stable an object is when placed at the Lagrange point, but not the location of the Lagrange point itself.

The other three Lagrange points - L1, 2 and 3 - do all require a comparison of the relative masses. But none of them are located at the barycentre. Consider: the barycentre is always located closer to the more massive of the two bodies - the L1 (gravitational equivalence point) must logically be closer to the smaller object, since a third body needs to be closer to the smaller object in order to feel the same gravitational attraction as the larger one. An object placed at the barycentre is always much closer to the larger object and will immediately start falling toward the larger object. This is why we never find objects either in ED or in real life sitting at the barycentre point of two other objects.

The L1 point will only be at the barycentre in the very special circumstance of two objects with exactly the same mass are orbiting each other. For unequal masses, the barycentre will always be closer to the larger mass, and the L1 point closer to the smaller mass. Thus, the Earth-Sun L1 point is quite close to Earth, and the Earth-Moon L1 point is quite close to the Moon. The Earth-Sun barycentre, on the other hand, is so close to the Sun it's buried deep inside it, likewise the Earth-Moon barycentre is buried deep inside the Earth.

The equation for calculating the location of the L1 point is: [the distance from the smaller planet to the L1 point] equals approximately [the distance between the two planets] x the cube root of ([mass of smaller planet] / (3 x [mass of larger planet])). L2 is located the same distance from the smaller object, but on the opposite side of the smaller object. L3 is essentially on the orbital path of the smaller object, but on the opposite side, very close to the larger object (perhaps buried inside the larger object, if the masses are nearly equal).

Note that all Lagrange point equations assume that the co-orbits of the two bodies are essentially circular, or near enough to circular to make no practical difference. If the orbits are noticeably elliptical, then that distance between the two objects, at the core of all the calculations, is constantly changing and there can be no fixed stable Lagrange points.
 
Calculation of the L4 and L5 Lagrange points don't care about how big the objects are relative to each other. The L4 and L5 points are always at the vertices of equilateral triangles, no matter what the masses of the 2 objects are. The relative masses will have an impact on exactly how stable an object is when placed at the Lagrange point, but not the location of the Lagrange point itself.
True up to a limit. Unfortunately "how stable it is" does indeed depend on the ratio of body sizes, and once that is lower than 24.96:1 it becomes metastable rather than stable.

For comparison Earth:Moon is about 83:1 and Pluto:Charon is 8.2:1 - so Earth-Moon has a stable island at L4/5 and indeed has asteroids sat there; Pluto:Charon does not. A true binary planet we like to get excited about in Canonn has a ratio of 1:1 and definitely does not have a stable L4 point.

The L4/L5 effect arises because the barycentre is a little towards Body 2 from Body 1 - which is not the same thing as being towards the L4/L5 point - so there's an offset in play. Beyond that 24.96 ratio I assume what happens is the barycentre is so far above Body 1 the angle of dangle of that gravity starts pulling in the wrong direction to oppose anything and instead starts reinforcing any drift. I would not like to sit down and work out why it's 24.96, I am scared of arctangents.

The barycentre's position doesn't make a difference to Body 2 (or indeed the L2 point behind it) because in that two-body system there's only one dimension you care about, the barycentre is just moving closer or further away and that just alters what the stable orbital speed happens to be and how far along the same axis L1/L2/L3 will be. But for an observer at L4 that barycentre is indeed moving across the sky as the mass ratios change (and a little bit closer or farther)

When you perturb Body 3 at the L4/L5 point, there's a bunch of complex orbital mechanics that determine what happens next (we are in orbit here so it's not as simple as objects going in straight lines in the direction you boot them) but in an L4/L5 system the effective direction of gravity - ie towards that barycentre - opposes the perturbation in useful ways, so when you add up apparent centrifugal, apparent Coriolis, and true gravity from Body 1 plus Body 2 they cancel out.

That extra handy offset vector is not available at L1/L2/L3 because everything's in a straight line, so that's why those points are saddles, not wells.

(Apparent centrifugal and apparent Coriolis are two things that happen because the satellite at L4 IS orbiting something but it is ALSO an object moving in Galilean space and has Galilean inertia, but Newtonian gravity is bending that into a curve; both centrifugal and Coriolis are mathematical conveniences to stop you having to remember mass-inertia is a thing and also that you've been messing around with coordinates a lot; they are both conveniences to allow you to think in points and lines and forget dealing with curves. There's another one called the Euler force which we can fortunately ignore because it arises from angular acceleration, which we don't have here, nobody said nuttin' about spinning these bad boys faster.)

Note that all Lagrange point equations assume that the co-orbits of the two bodies are essentially circular, or near enough to circular to make no practical difference. If the orbits are noticeably elliptical, then that distance between the two objects, at the core of all the calculations, is constantly changing and there can be no fixed stable Lagrange points.
I hate to break this to you but what you get instead is Lagrange orbits.

As Body 1 and Body 2 orbit each other, they still have a stable point at L4, it's just that exact position will move around as the bodies move around. So what you can do is calculate that emergent orbit which is "around" an average L4, and put your spacecraft into an unpowered orbit the same way you would inject it into any other orbit. It's just a bloody weird shape, usually called "kidney bean" and at the centre of it is... nothing.

Either way though the major advantage of using these points is that you can station-keep using very little propellant and planning for them doesn't tend to rely on any assumption that it would stay there forever if parked. Sussing out the kidney bean orbit is the lowest-energy way to do it.
 
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