Is a Methane-Molecule-Like Solar System Stable?

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The discussion explores the stability of a hypothetical solar system modeled after a methane molecule, with a central star representing carbon and planets as hydrogen atoms. For stability, the planets must have equal mass and orbital periods to avoid perturbations, or they must occupy separate orbits. The system's dynamics suggest that if the planets are equidistant from the star, they should not overtake each other, and their angular momentum vectors must differ due to the non-planar nature of hydrogen in methane. Calculations involving Hamiltonian mechanics indicate that a central mass is necessary for stability, while systems without it may exhibit linear instability. Overall, the analysis emphasizes the complexity of gravitational interactions in such a unique configuration.
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I just came across a question in another forum regarding the stability of a rather strange system. A huge methane-molecule-like system with the parent star in the place of carbon, and the planets in the place of H atoms (It's actually for a sci-fi story). If the planets moved in the same direction (that's what the OP said), what are the parameters in terms of mass, radius of orbit etc for this system to be stable. Also the planets should have a g-value so that humans could survive.

I'm guessing that if it started out as a normal system, i.e. with the planets in a plane, then it should have had a violent past for the planets to go elsewhere. I also think that collision(s) should have been fine-tuned to form such a system.
 
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The hydrogen atoms in methane are not planar, so their angular momentum vectors would not be the same, i.e. they cannot be moving in the same direction or planar orbit.

To be 'stable' with four planets which are equidistant from the central star, the planets would have to be the same mass and period in order not to have at least one over take another. Otherwise the planets have to be in separate orbits. The system would be stable if the masses and periods were such that the planets did not perturb each other.
 
I've lost the link, but somebody has solved the problem of 64 planets in mutual attaction orbiting to form a "buckeyball" configuration. He has an animation of it too.
 
Astronuc said:
The hydrogen atoms in methane are not planar, so their angular momentum vectors would not be the same, i.e. they cannot be moving in the same direction or planar orbit.

To be 'stable' with four planets which are equidistant from the central star, the planets would have to be the same mass and period in order not to have at least one over take another. Otherwise the planets have to be in separate orbits. The system would be stable if the masses and periods were such that the planets did not perturb each other.

Yes, the masses must be equal given the symmetry of the system at any instant. If we separate the planets in two pairs, the members of a pair sharing an orbit with the same period, will this make the planets move without perturbing the others? Of course, the main question is how would one determine the numbers.

selfAdjoint said:
I've lost the link, but somebody has solved the problem of 64 planets in mutual attaction orbiting to form a "buckeyball" configuration. He has an animation of it too.
That would be one amazing system to watch! Please do post the link if you come across it.
 
I have no clue how you would go about calculating interactions. Even 3 body solutions are difficult to approximate in gravitationally bound systems.
 
I did a semi-insane amount of work on this and related problems a while back.

If you take the Hamiltonian of this system, H


H = 1/2*px[1]^2+1/2*py[1]^2+w*(px[1]*y[1]-py[1]*x[1])+
1/2*px[2]^2+1/2*py[2]^2+w*(px[2]*y[2]-py[2]*x[2])+
1/2*px[3]^2+1/2*py[3]^2+w*(px[3]*y[3]-py[3]*x[3])+
1/2*px[4]^2+1/2*py[4]^2+w*(px[4]*y[4]-py[4]*x[4])
-1/((x[1]^2-2*x[1]*x[2]+x[2]^2+y[1]^2-2*y[1]*y[2]+y[2]^2)^(1/2))
-1/((x[1]^2-2*x[1]*x[3]+x[3]^2+y[1]^2-2*y[1]*y[3]+y[3]^2)^(1/2))
-1/((x[1]^2-2*x[1]*x[4]+x[4]^2+y[1]^2-2*y[1]*y[4]+y[4]^2)^(1/2))
-1/((x[2]^2-2*x[2]*x[3]+x[3]^2+y[2]^2-2*y[2]*y[3]+y[3]^2)^(1/2))
-1/((x[2]^2-2*x[2]*x[4]+x[4]^2+y[2]^2-2*y[2]*y[4]+y[4]^2)^(1/2))
-1/((x[3]^2-2*x[3]*x[4]+x[4]^2+y[3]^2-2*y[3]*y[4]+y[4]^2)^(1/2))
-M/(x[1]^2+y[1]^2)^(1/2)-M/(x[2]^2+y[2]^2)^(1/2)
-M/(x[3]^2+y[3]^2)^(1/2)-M/(x[4]^2+y[4]^2)^(1/2)where M is the ratio of the central mass to the 4 orbiting masses, you can compute the linear stability by finding the characteristic polynomial of a 16x16 matrix given by Hamilton's equations

<br /> \dot{p_i} = \frac{\partial H}{\partial x_i}<br />
<br /> \dot{x_i} = \frac{\partial H}{\partial px_i}<br />

(The above eq's need to be duplicated for y) - with the appropriate initial conditions for all the x_i,y_i,px_i,py_i which represent the initial state of the system.

This is obviously something that you don't want to try by hand.

Here w^2 = M + 1/4 + sqrt(2)/2

The characteristic polynomial looks like

1/8192*(256*x^8+512*x^6*M+256*x^6+256*x^6*2^(1/2)+
256*x^4*M^2+640*x^4*M+1408*x^4*M*2^(1/2)+736*x^4+
320*x^4*2^(1/2)+384*x^2*M^2+1152*x^2*M^2*2^(1/2)+
2400*x^2*M+960*x^2*M*2^(1/2)+592*x^2+624*x^2*2^(1/2)+
2736*M^2+864*M^2*2^(1/2)+1272*M+1368*M*2^(1/2)+
361+204*2^(1/2))*(4*x^2+4*M+1+
2*2^(1/2))*(8*x^4+4*x^2*2^(1/2)+
2*x^2+9*2^(1/2)+8*x^2*M+36*M*2^(1/2))*x^2

which appears to have no real or imaginary parts for M = 100, suggesting that the system is (marginally) stable with a central mass. Pertubations should not grow or shrink.

For M=100, the characteristic polynomial is

1/8192*

(256*x^8+51456*x^6+256*x^6*2^(1/2)+2624736*x^4+
141120*x^4*2^(1/2)+4080592*x^2+11616624*x^2*2^(1/2)+
27487561+8777004*2^(1/2))*

(8*x^4+802*x^2+4*x^2*2^(1/2)+3609*2^(1/2))*
(4*x^2+401+2*2^(1/2))*x^2

Without a centeral mass, the system is not stable for n=4.

The numerical simulations for very high n's with no central mass mentioned are interesting - I haven't done those, though I've seen the results.

They appear to be a system that is linearly unstable, but has limit cycle oscillations, which implies a non-linear sort of "stability".

[add]
ps - adding the central mass terms to the Hamiltonian was a bit of an afterthought, I may have to review how I did this :-(.
 
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