# Thread: Trojan point stabiility (redux)

1. ## Trojan point stabiility (redux)

We'd just started to get somewhere interesting in the "Trojan" thread, concerning the stability of the Trojan points for non-zero masses, and then it went to ATM in a handcart.
I've therefore sprung the following loose from the original thread, in order to start a new thread on the topic.

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With reference to StupendousMan's post in that other thread:

Originally Posted by StupendousMan
Tony's excellent work prompted me to do what I should have done a few days ago: get up and walk to the library. I scanned through a book called "The Three-Body Problem" by Christian Marchal, published by Elsevier in 1990. Chapter 8 in that book discusses "Simple Solutions of the Three-Body Problem", one of which is the system Tony proposes: a very massive star (the Sun) and two much less massive bodies (Earth-1 and Earth-2), with the two small bodies orbiting the massive one in circular orbits of the same radius, separated by 60 degrees.

Marchal discusses the stability of these systems at some length, using first-order and higher-order analysis. He provides references to papers by C. Richa (thesis from University of Pierre and Marie Curie, 15 Oct 1980) and himself (seminar at the Bureau of Longitudes, 7 Mar 1968). In his discussion, he defines a couple of parameters which involves the relative masses of the three bodies:

N = sqrt( m1^2 + m2^2 + m3^3 - m1m2 - m1m3 - m2m3) / (m1 + m2 + m3)

which, in our case of m2 = m3, simplifies to (please check my algebra here!)

N = (m1 - m2) / (m1 + 2m2)

and the parameter

R = ( 3 - sqrt(12 N^2 - 3) ) / 6

Now, in the case Tony has suggested, in which m1 = Sun's mass and m2 = m3 = Earth's mass, the parameter N is a fraction very close to 1 (I find N=0.999994), and the parameter R is a number very close to zero (I find R=4 x 10^(-6)). Again, check my arithmetic!

The payoff for computing these values is that Marchal presents a nice graph (his Figure 13 on page 49) showing zones of stability for this "Trojan" arrangement. For circular orbits, there are two zones in which the orbit can be proved stable:

0 <= R < 0.02860...
and
0.02860... < R <= 0.03852....

Note that Tony's system falls well within this first zone of stability. In fact, this suggests a test: the arrangement should still be stable until the two "Earth-like" planets grow in mass so that R = 0.03852, which means N = 0.9428 or so. And that, in turn, if I can do algebra again, means that the system should be stable until each of the two "Earth-like" planets has a mass of m1*(0.01982). In other words, until the two Earth-like planets added together have a mass of about 4 percent of Sun.

Sorry for being so forceful in my incorrectness earlier :-(

Tony, thanks for sticking to your guns and helping me to learn something new today ....
Marchal actually gives a precise value for the parameter R at which circular motion is unstable - it's (3-sqrt(8))/6. That corresponds to his parameter N = sqrt(11/12). So in Marchal's analysis there's no range of unstable conditions in this vicinity - just a single point. Of course in the real world there are always perturbations, so we can expect to see unstable behaviour arising in the vicinity of this point in parameter space, but I'm not clear why Tony's simulation should show a range of unstable values.

For reference, here's the relevant page (click to enlarge):
Attachment 22878

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Grant Hutchison
Last edited by grant hutchison; 2018-Jan-07 at 03:10 AM. Reason: fixed brackets

2. And as a followup, in answer to a question George asked in the original thread, my understanding is that the boundary between the unstable and stable cases in Marchal's diagram (the attachment above) is determined by the onset of resonances between the orbital period of the whole system, and the period of oscillation around the Trojan point. When such resonances occur, they drive the oscillation towards instability. Such a resonance occurs in the vicinity of Marchal's R = (3-sqrt(8))/6 [I see I placed the brackets wrongly in my orginal post], and extends its effects as eccentricity increases, but should only represent a single point on the R axis when e=0.

Grant Hutchison

3. Thanks Grant. I am curious if there is a somewhat simple physical explanation why there is a sweet, or perhaps sour, spot for that R value, where even a circular orbit is unstable? It is the anti-Goldylock's spot where hotter or colder is better (stable at least for a limited range if the Trojan is more massive from this mass ratio sour spot).

Originally Posted by StupendousMan
N = sqrt( m1^2 + m2^2 + m3^3 - m1m2 - m1m3 - m2m3) / (m1 + m2 + m3)

which, in our case of m2 = m3, simplifies to (please check my algebra here!)

N = (m1 - m2) / (m1 + 2m2)
No error, just an understanble typo (bold -- m3^2). Your algebra is correct and my calculations do have similar results.

Perhaps oversimplifying will help me...

If the orbits were perfectly circular and with no troublesome neighbors, would stability be infinite regardless of the lesser mass Trojan?

With some eccentricity, acceleration will take place upon the Trojan but the motion would be the same regardless of mass -- a feather falls at the same speed as a rock, in space. So wouldn't any resonant effect be the same, and independent of mass?
Last edited by George; 2018-Jan-07 at 10:12 PM.

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but I'm not clear why Tony's simulation should show a range of unstable values.
I think I found the answer.

The simulation is very much in agreement with
The payoff for computing these values is that Marchal presents a nice graph (his Figure 13 on page 49) showing zones of stability for this "Trojan" arrangement. For circular orbits, there are two zones in which the orbit can be proved stable:
0 <= R < 0.02860...
and
0.02860... < R <= 0.03852....
In the simulation, Earth and Trojan start at 1 AU and ecc = 0. When I add significant mass to either Earth or the Trojan, their eccentricity rises. To reset it to zero, I would have to add speed to their orbits.
So in this simulation, which simulates the strange gap in stability quite nicely, ecc isn't 0. And you can see that the moment you press play as the objects have radial motion in the rotating frame.
http://orbitsimulator.com/gravitySim...nMassTest.html

I modified this simulation to re-circularized their orbits when mass is added. When you change the masses, it recomputes their circular velocities, v = sqrt(G*(Msun+Mearth+Mtrojan) / 1 AU) to restore their ecc to 0.
http://orbitsimulator.com/gravitySim...sTest0ecc.html
No more gap!

5. Originally Posted by tony873004
I think I found the answer.

The simulation is very much in agreement with

In the simulation, Earth and Trojan start at 1 AU and ecc = 0. When I add significant mass to either Earth or the Trojan, their eccentricity rises. To reset it to zero, I would have to add speed to their orbits.
Hmmm, would the increase in mass thus change the barycenter thus cause additional eccentricity? You showed one plot of stability up to or near Marchal's R value, I think, but no stability thereafter. So my guessing here is still inadequate even if correct.

6. Originally Posted by George
Thanks Grant. I am curious if there is a somewhat simple physical explanation why there is a sweet, or perhaps sour, spot for that R value, where even a circular orbit is unstable?
Marchal's stability diagram goes back to Danby (750KB pdf) in 1964, who did the first stability analysis of the Trojan points. He treated only the case of zero test masses, so uses a different mass parameter, but the shape of the graph is recognizably the same. He says that the line separating the demarcation between stable and unstable is the point at which the period of one Trojan oscillation is equal to twice the period of the planet. If you reflect the right side of the graph in the x-axis, you end up with a straight line passing through zero eccentricity and continuing below the axis, which shows how the apparent singularity at zero eccentricity is just another point on a curve generated from the roots of a quadratic equation that determine stability. I can't think of an intuitive way of translating the maths into a physical argument, except to say that when you smoothly vary the mass function at zero eccentricity, you will hit a single point at which this killer resonance intrudes.

But there's more. Nauenberg (720 KB pdf) wrote a nice paper in 2002, exploring yet another mass parameter versus eccentricity, and flipping the axes, but still with recognizably the same stability graph. What he showed was that if you cross the conventional stability line, you don't necessarily end up with an unstable escape - there are period-doubling chaotic oscillations at the edges of stability, so that the Trojans don't actually wander off until you're some way into the unstable zone. And with some mass ratios, resonances appear within the "stable" zone, causing Trojans to escape.
So it's a complicated situation.

Grant Hutchison

7. Originally Posted by tony873004
I think I found the answer.
...
No more gap!
Excellent!

Grant Hutchison

8. Subscribe.

9. I just stumbled across this animation on imgur.

Fascinating correlation between Jupiter's Trojan and asteroid belt.

https://i.imgur.com/FmfgUl8.gifv

10. Originally Posted by DaveC426913
I just stumbled across this animation on imgur.

Fascinating correlation between Jupiter's Trojan and asteroid belt.
The asteroids in red are the Hilda group, which have a 3:2 resonance with Jupiter. They always pass Jupiter when they're at perihelion.

Grant Hutchison

11. Originally Posted by grant hutchison
If you reflect the right side of the graph in the x-axis, you end up with a straight line passing through zero eccentricity and continuing below the axis, which shows how the apparent singularity at zero eccentricity is just another point on a curve generated from the roots of a quadratic equation that determine stability.
Yes and, for increasing the mass ratio a little, it is interesting that at zero ecc., instability will come, but perhaps they limit the mass range so I may be overstating it. It may be more that zero eccentricity cannot be maintained at the slightly greater masses, thus instability comes more quickly with these greater masses.

Still, valid quadratic or not, it seems odd that there is any positive slope even for that limited range of mass ratio. Reviewing Poincare's work is not high on my list, however.

I can't think of an intuitive way of translating the maths into a physical argument, except to say that when you smoothly vary the mass function at zero eccentricity, you will hit a single point at which this killer resonance intrudes.
Ken has said in the past something to the effect that if we can't explain the physics then we don't fully understand the math. I don't think, however, that CM is one of his great areas of expertise. [Yes, this is a stage que, FWIW. ]

But there's more. Nauenberg (720 KB pdf) wrote a nice paper in 2002, exploring yet another mass parameter versus eccentricity, and flipping the axes, but still with recognizably the same stability graph. What he showed was that if you cross the conventional stability line, you don't necessarily end up with an unstable escape - there are period-doubling chaotic oscillations at the edges of stability, so that the Trojans don't actually wander off until you're some way into the unstable zone. And with some mass ratios, resonances appear within the "stable" zone, causing Trojans to escape.
It looks like the early work (Danby) was limited to first order computations due to the lack of very fast computers at the time. Looks like it was very close to being right on.

So it's a complicated situation.
Yes, and it speaks volumes about the complexity of the 3-body problem.
Last edited by George; 2018-Jan-09 at 05:37 PM. Reason: grammar

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