[SPACE] Rules considering Lagrange Points in Advanced Worldbuilding

[Sabersonic]

For my first thread on this board, I have a few inquiries related to the Advanced Worldbuilding in the creation of a customized star systems, or rather the lack of information for details that I require. Mainly Lagrangian/Liberation/Trojan points.

For Lagrangian Points subject, well I wanted to create a habitable moon of a Gas Giant that had Co-Orbital Trojan Moons along its orbit at the L4 and L5 locations after recalling and looking up information on natural occurring Trojan Moons of the Saturnian Moons Tethys and Dione as explained in this Wikipedia Article. Since my hardcover copy of GURPS Space 4th Edition did not have the information I sought, I attempted to search for the information through the forum. Although I have found some interesting notes about Lagrange Points such as

Originally Posted by dataweaver

(maximum mass for satellite with stable trojan points = 4% of primary's mass)

Originally Posted by Brett

The L4 and L5 points are only stable if the ratio of the masses of the two stars is greater than about 25.

However, I did not find the answers to my questions in determining if a Planet-Moon system could have Lagrangian Points or not and the maximum mass/moon type/moonlet that could occupy that zone. Is there a suggested dice formula to determine if such a system could even have Lagrangian points if they meet certain requirements or do I just dictate that those two bodies have those points and leave it as that?

[Diomedes]

Originally Posted by Sabersonic

However, I did not find the answers to my questions in determining if a Planet-Moon system could have Lagrangian Points or not and the maximum mass/moon type/moonlet that could occupy that zone. Is there a suggested dice formula to determine if such a system could even have Lagrangian points if they meet certain requirements or do I just dictate that those two bodies have those points and leave it as that?

They'll have the first three points, and the other two if the ratio of the masses are high enough (the aforementioned 25:1). As to the maximum mass that can be placed there, probably considerably smaller than either the primary or the satellite, to avoid disturbing their orbits.

[teviet]

Originally Posted by Diomedes

They'll have the first three points, and the other two if the ratio of the masses are high enough (the aforementioned 25:1). As to the maximum mass that can be placed there, probably considerably smaller than either the primary or the satellite, to avoid disturbing their orbits.

My guess is that the tertiary mass would not matter so much for stability, as long as both the secondary and tertiary were less than 1/25 of the primary. But this is a guess (based on a vague notion of how the rotating potential will change as the tertiary mass increases), not an analysis.

Nonetheless, among the many Trojan points in our Solar system, and the many bodies orbiting them, none have accumulated into a major-moon-sized mass, suggesting that such objects would be quite rare.

TeV

[malloyd]

Originally Posted by teviet

My guess is that the tertiary mass would not matter so much for stability, as long as both the secondary and tertiary were less than 1/25 of the primary. But this is a guess (based on a vague notion of how the rotating potential will change as the tertiary mass increases), not an analysis.

Since two such moons would be in *each other's* L4/L5 points (one is 60 degrees ahead of the other, hence the second is 60 degrees behind the first) there shouldn't be any mass ratio considerations, or you simply redefine which is in the point. Though possibly the combined mass of the two will need to be sufficiently low compared to the primary instead of just the individual masses.

Nonetheless, among the many Trojan points in our Solar system, and the many bodies orbiting them, none have accumulated into a major-moon-sized mass, suggesting that such objects would be quite rare.

You really would expect that. In order to move into the point and stay there something would have to be in an orbit that passed nearby without much energy difference, which means on anything like a geologic timescale it also passed near the main body lots of times too, not a recipe for staying in that orbit.

[Agemegos]

Strictly speaking the stability limits of the Trojan points are unknown, since the Lagrangian solution holds only for test masses (i.e. if the mass of the third body is ignored). No analytical solution exists for that case of the three-body problem. As for numerical simulations, I understand that they always lead to a collision or the expulsion of the smaller body for any mass ratio that is tried.

Obviously the orbits are close enough to stable for really large mass ratios, since there are Trojan asteroids and analogues in the orbit of Saturn. But I don't know of any Trojan-analogues in the orbit of Earth, which suggests that the ratio of masses of Earth to an asteroid visible at 1 AU is not enough. One of the leading theories for the formation of the Moon is that an object about the size of Mars formed in one of Earth's Trojan points, where its orbit was unstable and led in time (only 20–30 million years) to a collision.

I can't be quite definite, but I would have to guess that having a planet in the Trojan point of another planet, even a gas giant, is a space opera conceit rather than a hard SF plausibility.

[teviet]

Originally Posted by Brett

Strictly speaking the stability limits of the Trojan points are unknown, since the Lagrangian solution holds only for test masses (i.e. if the mass of the third body is ignored). No analytical solution exists for that case of the three-body problem.

No, the Lagrange solutions are equilibrium states for arbitrary masses. Their stability against small perturbations can be calculated analytically. (I haven't actually done it but the method is straightforward.) Klemperer rosettes are a similar N-body equilibrium state with nonzero masses.

But these are special cases that are either highly symmetric or stationary in the rotating reference frame. You are correct that the general (non-equilibrium) three-body problem has no analytic solution.

TeV

[Agemegos]

Originally Posted by teviet

No, the Lagrange solutions are equilibrium states for arbitrary masses.

Can you point me in the direction of a citation for that statement? Every discussion of the Lagrange Points I have ever come across has described them as solutions to the restricted three-body problem in which the third mass is negligible. If there is a more general result I would be interested in hearing about it.

[teviet]

Originally Posted by Brett

Can you point me in the direction of a citation for that statement? Every discussion of the Lagrange Points I have ever come across has described them as solutions to the restricted three-body problem in which the third mass is negligible. If there is a more general result I would be interested in hearing about it.

Hmm, I saw it as a problem in a classical mechanics textbook somewhere, but I can't remember which one. How about I just give you the proof?

Theorem: The Trojan orbital configuration is an equilibrium for arbitrary masses.

Summary: The Trojan orbital configuration consists of three arbitrary masses at the vertecies of an equilateral triangle, rotating about their common centre of mass. This is an equilibrium configuration if the centripetal acceleration equals the gravitational acceleration for every mass. I will show that (a) the gravitational force on any mass is always directed towards the centre of rotation, and (b) there is a single rotation rate Omega that balances the centripetal and gravitational accelerations.

Proof: (a) Consider three masses m1, m2, m3 at the vertecies of an equilateral triangle of side R. Without loss of generality, choose m1 to be momentarily at the origin of the coordinate system, and m2 and m3 to be at positions r2 and r3 respectively, where |r2|=|r3|=R.

The centre of mass of the configuration is:

rc = ( m2*r2 + m3*r3 )/M

where M = m1 + m2 + m3 is the total mass of the system. The acceleration of m1 due to the gravity of m2 and m3 is:

a1 = G*m2*r2/|r2|^3 + G*m3*r3/|r3|^3
= G*( m2*r2 + m3*r3 )/R^3
= ( G*M/R^3 )*rc

Thus m1 accelerates towards the centre of mass.

(b) Set the configuration spinning about its centre of mass with an angular speed Omega, and require that the centripetal acceleration rc*Omega^2 equal the gravitational acceleration a1. This gives:

Omega^2 = GM/R^3 .

The same analysis carried out for either of the other masses would give the same value of Omega. Thus when the configuration is rotating at this rate, all masses are in equilibrium.

Note: This only proves that the Trojan configuration is an equilibrium configuration for any set of masses, but says nothing about the stability of the configuration (i.e. whether small perturbations to the configuration will remain small or will grow over time). As has been pointed out, the stability of the Trojan configuration does depend on the masses of the objects.

A key to this proof is that each mass is exactly the same distance from every other mass and they lie in a common plane. It will not generalize to systems with more than three masses: rosettes have further restrictions on the symmetry of the mass distribution.

TeV

[Agemegos]

Originally Posted by teviet

Hmm, I saw it as a problem in a classical mechanics textbook somewhere, but I can't remember which one. How about I just give you the proof?

Even better!

Theorem: The Trojan orbital configuration is an equilibrium for arbitrary masses.

Proof looks good to me. Do you know why it is so often written that the result only holds if the third mass is light? Is it that the configuration was discovered (by Lagrange) by considering a test mass?

Note: This only proves that the Trojan configuration is an equilibrium configuration for any set of masses, but says nothing about the stability of the configuration (i.e. whether small perturbations to the configuration will remain small or will grow over time). As has been pointed out, the stability of the Trojan configuration does depend on the masses of the objects.

I see (above) that you don't happen to know the stability limits. Do you happen to know whether the standard result for the stability limit does depend on an assumption that the third mass is arbitrarily light?

In reading about the Giant Impact Hypothesis I often come across statements that Theia's orbit in one of the Earth's Trojan points became unstable once it accumulated (through accretion) more than a limiting mass (eg http://www.search.com/reference/Giant_impact_hypothesis). Earth is supposed to have been nearly fully-formed, and Theia about the mass of Mars.

[teviet]

Originally Posted by Brett

Do you know why the statement that the result only holds if the third mass is light is so common? Is it that the configuration was discovered (by Lagrange) by considering a test mass?

Possibly. Myself, I don't recall encountering that statement before. But certainly the general 3-body problem in which the masses are not in a fixed co-rotating configuration is a lot harder (in fact impossible) to solve analytically when the third mass is nonzero.

Originally Posted by Brett

I see (above) that you don't happen to know the stability limits. Do you happen to know whether the standard result for the stability limit does depend on an assumption that the third mass is arbitrarily light?

It does. I'm not sure how much. I was intrigued enough to take a look at it, but all I concluded is that I am gradually losing the ability to perform basic algebra without Mathematica (it's only a 4d eigenvalue problem, how hard can it be?!?).

TeV