Planets in Trojan points?

[granthutchison]

Ah-ha. Here we go: Laughlin and Chambers' Extrasolar Trojans: The viability and detectability of planets in the 1:1 resonance. All sorts of interesting things in this, but they've done the maths specifically relating to planets of non-zero mass in the L4 or L5 position.

Call the mass of our planet m1 and the mass of the trojan object m2; total system mass (primary+planet+trojan object) is M.
You'll recall from earlier in this thread that if m2 = 0, then the trojan position is stable if:

m1/M < (9 - root[69]/18 = 0.038520896...

Now, it transpires that if m1 = m2, this pair of mutual trojans is stable if:

(m1+m2)/M < (6 - 4*root[2])/9 = 0.038127305...

And for all intermediate values of m2 ( 0 < m2 < m1), the stability threshold for (m1+m2)/M lies between the two limits above.

Grant

[tony873004]

3.85% for a massless trojan and 3.81% for mutual mass trojans...
Now that's a thin line.

[wcomer]

Perhaps this has been mentioned in another thread on orbits about Libration Points; if so then my apologies.

Regarding libration orbits - it seems to me that as long as the orbital plane of the secondary mass is nearly 90 degrees to its axis of rotation, one can get a decent approximation to the libration point orbits. All you need is the orbital and rotation periods of the secondary mass, from which a suitable orbit about the secondary mass can be calculated for all five libration points. By treating the libration points as orbits about the secondary mass one can avoid the problems imposed by custom orbits of the secondary about the primary.

[tony873004]

Perhaps this has been mentioned in another thread on orbits about Libration Points; if so then my apologies.

Regarding libration orbits - it seems to me that as long as the orbital plane of the secondary mass is nearly 90 degrees to its axis of rotation, one can get a decent approximation to the libration point orbits. All you need is the orbital and rotation periods of the secondary mass, from which a suitable orbit about the secondary mass can be calculated for all five libration points. By treating the libration points as orbits about the secondary mass one can avoid the problems imposed by custom orbits of the secondary about the primary.

I'm not quite sure I understand you.
"...the orbital plane of the secondary mass is nearly 90 degrees to its axis of rotation...". Like the Earth's orbital plane is 23.5 degrees from its axis of rotation?

"...from which a suitable orbit about the secondary mass can be calculated for all five libration points..." Objects in the secondary's L1-L5 points don't orbit the secondary. Nor do they actually orbit the L point. They orbit the primary, but subtle forces from the secondary stabalize their positions relative to the secondary. In a rotating frame this makes them seem to orbit the L point.

[granthutchison]

By treating the libration points as orbits about the secondary mass one can avoid the problems imposed by custom orbits of the secondary about the primary.

It doesn't help, unfortunately. To keep their positions synched with the real libration points, your Celestia simulation would need to move the simulated libration points in scaled-down versions of the secondary's custom orbit - you just can't get away from the CustomOrbit/EllipticalOrbit mismatch.

Grant

[granthutchison]

Like the Earth's orbital plane is 23.5 degrees from its axis of rotation?

The Earth's orbital plane is 23.5 degrees from its equatorial plane - that may be feeding your confusion. An object with its axis of rotation 90 degrees from its orbital plane has an axial inclination of zero.

I believe wcomer is talking about the mechanics of simulating libration points in Celestia, rather than the behaviour of real libration points.

Grant

[wcomer]

Ok, I have no excuse to offer for the horridly wrong description from before. Axis of rotation is not relevant at all (darned tidally locked moon had me all mixed up )

I'm not certain how Celestia handles Keplerian orbits about secondary and tertiary bodies. I presume that the orientation of the ellipse is fixed with respect to the background stars while the focus of the ellipse moves with the center of the parent body.

If that assumption is correct then what I should have said is that: given a primary (the Sun), a secondary (the Earth), and a tertiary (the Moon) and if the orbital plane of the secondary and the tertiary are nearly the same (5 degrees), then we can approximate the libration point orbital mechanics with Keplerian orbits about the tertiary body even if the tertiary and secondary are using custom orbits. This assumes a bit about the dynamics of custom orbits which I'm not privy to, in particular they would need to have constant orbital periods; else, at some future time (eons from now) the libration points will stray into the secondary. Or, am I missing something in the geometry of the situation which prevents this approach from working?

In the case of the Earth-Moon system, the libration orbits would never be off by more than 5 degrees, which is not ideal but it is better an improvement over having them as Keplerian orbits about the Earth and wandering into the Moon.

[granthutchison]

The Earth-Moon system is a bad choice for this, since the Moon's orbit changes so quickly - the more slowly the CustomOrbit evolves, the longer a Keplerian approximation is useful.
But things would go wrong even if the CustomOrbit period remained constant - the movement of nodes and pericentre means that your Keplerian orbit starts to lollop around out of sync with the CustomOrbit. Unfortunately, even a small loss of sync between the two orbits will start wiggling the libration points visibly back and forth - for the Earth-Moon system, they're leaping around within a few months of setting the thing up

Grant

[Guest]

Thanks for the numbers, should be able to make some interesting systems...

Just wondering, does the article say anything about the situation where you have large masses in BOTH Trojan points? (I can't access it as I get a username/password dialog box).

~Chaos Syndrome

[granthutchison]

Just wondering, does the article say anything about the situation where you have large masses in BOTH Trojan points?

Whoooo. The four body problem? I think the only thing you can do for that is a simulation.

One idea I've just had for a system which I've never seen used before is to have a terrestrial planet in a horseshoe orbit with a brown dwarf, around a sunlike central star. The mechanics of the system would be the same as for Janus & Epimetheus, with the Earth-like planet and the brown dwarf swapping angular momentum back and forth and never passing each other. But because the brown dwarf is so much more massive than an Earth-like world, the brown dwarf's orbit barely shifts, whereas the Earth-like planet shifts dramatically inwards and outwards with each close encounter with the dwarf, experiencing large "seasonal" changes in both temperature and in year length. Life on the planet would have to evolve to cope with hot short seasons and long cold ones, alternating at multi-year intervals.

Grant

[wcomer]

Grant,

I don't see any reason why the four body libration points would not be tractable to the same techniques used by Laughlin and Chambers. Linear stability theory is rather straight forward, if mechanically a pain. It is, by definition, a linear method and hence doesn't fall victim to the sorts of analytic problems of nonlinear stability. The downside is that by throwing out higher order terms you cannot say much about the long term behavior (particularly as you move further away from the regions of equilibrium. The only real question is of which constraints to free up for linearization. Laughlin and Chambers introduce two degrees of freedom; this has the analytic convenience of producing a quartic characteristic equation for the eigenvalues. Obviously a higher order equation would likely require numerical, rather than closed form analytic, constraints on the masses which allow linear stability. Comparable generality for the four body case would require additional degrees of freedom, and hence, the zone of stability (in mass-space) likely would have to be solved numerically. There are plenty of less generalized two degrees of freedom possibilities for the four body case which could be analyized separately; the intersections of these zones of stability would likely be similar to the numerical results from allowing more degrees of freedom.

[granthutchison]

This is interesting. I was aware that in general four bodies require a numerical simulation to find out what's going to happen to them; and I don't know nearly enough about this stuff to have any sensible notion of what sort of restricted situations it might be appropriate and useful to examine ...
I'd certainly be interested to hear of any conclusions you can come up with!

Grant

[wcomer]

Grant,

In general four body does require numerical simulation. This is true for three body as well. I don't believe it is possible to say whether the L4 and L5 libration points are eternally stable to very small perturbations. It is possible to characterize how quickly L1, L2 and L3 orbits explode and pick up higher order terms. The same method applied to L4 and L5 shows that these systems have stable orbits but this doesn't mean they are stable forever. Rather, it means that the linear terms do not cause exploding orbits, but the higher order terms may even under very small perturbations (eternity is such a pain.) The interesting thing is that because these are conservative systems (and therefore time reversible), the orbits can never be sinks!

While on the subject I would point out that Laughlin and Chambers do not fully characterize the masses which have stable orbits. They are solving for an over constrained system. Three bodies have 9 (3x3) degrees of freedom. However, conservation of linear momentum (with translational independence) allows us to constrain 3 (1x3) degrees of freedom by six parameters. Similarly conservation of angular momentum allows us to reduce the system by 1.5 (3x.5) degrees of freedom. Conservation of energy reduces another .5 degrees of freedom. So the whole system can be reduced to 4 degrees of freedom. Or put differently we can write the entire system using 4 coordinates and 10 parameters. If we constrain motion to a plane then we end up with 3 = 6 (3x2) -2 (1x2) -.5 (1x.5) -.5 degrees of freedom and 6 parameters. But Laughlin and Chambers represent the system with only two coordinates; so they are over constrained for even the planar system. Therefore we can surmise that the mass limits they give are only part of the story, even in the planar case.

So having said that we can return to the question of eternal stability of L4 and L5. By now it should be clear that we have 4 degrees of freedom. This means that our characteristic equation is eighth order. So we must numerically solve for the eigenvalues. Suppose that for our three masses, the eigenvalues are all purely imaginary, then we have stability. But for how long. If we can add higher order terms then we get even higher order characteristic equations. If any of the roots have positive real components then we know we are unstable (but only for this particular set of masses) and we can determine how quickly the orbits explode. But if they are all imaginary then we have to go to the next higher order terms. The fundamental equations have infinite power series approximations, so this process only ends if you find positive real components. This approach only allows you to determine instability for trial masses and cannot determine eternal stability for trial masses. Other methods may provide more insight, but as far as I know it is an open question as to whether there exists any three body system which is eternally stable.

[granthutchison]

The idea of half a degree of freedom is probably too much for me to cope with ... But let me see if I can understand what you're saying by coming at it from another (simplistic!) direction. For the classic restricted three-body problem with a zero-mass "test object" in third position, am I right in thinking that two coordinates are acceptable, because m1 and m2 are constrained to maintain constant positions in the rotating frame, by virtue of the assumptions of circularity and coplanarity? But as a penalty for attributing mass to their third object, Laughlin and Chambers have lost that helpful constraint on the behaviour of the other two bodies?

Anyway, thanks for the insight into three-body "stability". The textbooks (and a lecturer almost 30 years ago ) disposed of the higher terms of the Taylor series with such brisk, reassuring confidence that I've never thought about them since.

Grant

[ajtribick]

Thanks for all the info everyone

Then again, planets in Lagrangian points could be a serious hazard, particularly if there are lots of violent impacts going on...

http://space.com/scienceastronomy/moon_formation_040621.html