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Inspired by this question:


Consider a trinary star system, in which the three stars are arranged in an xy plane and all revolving the same direction, equidistant from each other. The inward pull of gravity is balanced by the centrifugal force, establishing an orbit about a common barycentre.

Now consider adding a fourth star, displaced in the z direction directly above the barycentre of the system (along the axis of revolution). This fourth star will, as far as I can tell, be pulled downward along the axis, past the barycentre, reach a distance away from the plane of revolution equal to the distance from which it started, and begin oscillating indefinitely along the axis of revolution.

The other three stars will gain degrees of freedom as a result of this fourth star. First, they'll move up and down along the z axis as their barycentre is displaced by the gravity of the fourth. Second, they'll begin to move in and out so that their circular orbit becomes more of a sinusoidal ring, as the gravitational attraction of the fourth star will increase until it passes between the other three, then decrease as it moves away from the plane of revolution.

This star system might be best described as a modified Klemperer rosette, which can be in an ideal sense dynamically unstable - one perturbation and the system collapses. However, I’m wondering whether it’s possible for this system to be dynamically unstable in the same way that a traditional rosette is.

Can such a modified Klemperer rosette ever be dynamically unstable?

How this configuration came to be is out of the scope of the question - blame it on an incredibly powerful toddler alien who is learning to play with galaxies much as human children learn to build towers out of blocks.

Although I describe the system as "adding" a fourth star, I'm only interested in the stability of the end result (the simple harmonic oscillation of the system) and am aware that creating an additional sun in this way would destabilize the initially stable trinary star system.

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    $\begingroup$ I wanna place a bounty here as soon as it's available. $\endgroup$ – Renan Dec 13 '18 at 6:20
  • $\begingroup$ most likely yes, although there would be differences is some way, due to the additional gravity of a fourth star. At that point, though, the system may collapse if it too unstable. $\endgroup$ – ThatCamal Dec 13 '18 at 18:02
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A set of N isolated bodies will always orbit around the center of mass of the system, and the orbits will end up being a conic curve.

What you describe as "adding a 4th star out of the plane containing the 3 stars" is already offsetting the center of mass, thus it won't go as you picture.

Either some of the stars will be kicked out of the system along an hyperbolic trajectory (which is a conic curve) or the orbits will be flattened out to roughly be contained in a plane. That's why galaxies are disc shaped, even though they collide with each other at random angles.

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  • $\begingroup$ In more detail: initially there are the material points revolving around their common barycenter. Then a prankster adds a fourth material point somewhere outside the plane of the initial three objects. The barycenter is of course moved in the direction of the fourth object, outside the initial orbital plane. Each of the three initial objects will attempt to orbit the new barycenter, thus placing the three orbits in three different planes, breaking the symmetry. Meanwhile, the fourth object, which was initially at rest, will begin accelerating towards the now moving barycenter. Chaos ensues. $\endgroup$ – AlexP Dec 13 '18 at 13:25
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    $\begingroup$ This answer is incorrect. Masses only orbit along conic curves in the N=2 case. For the three body and up problem acceleration will not be towards the barycenter in the generic case. $\endgroup$ – Anders Sandberg Dec 14 '18 at 9:49
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Yes, you can arrange four stars in tetrahedron, provided the following conditions are met:

  • the stars all of equal mass;
  • the period of revolution of each star around the common barycenter is exactly the same;
  • each star orbits in a perfect circle around the common barycenter;
  • the plane of the orbit of each star is exactly perpendicular to the ground plane formed by the other three stars at any given point in time;
  • the star's initial position is that of a perfect tetrahedron.

With these conditions, the tetrahedron maintains its shape and symmetry and there are no intrinsic perturbations in the system. In fact, any platonic solid can be formed this way. A configuration of five stars with one star in the barycenter which may then differ in mass from the other four will also work.

Note that these systems, like Klemperer Rosettes are unstable under any perturbation, because any deviation from the perpendicular brings a star closer to a number of neighbors and further from a different number of other neighbors; the gravitational imbalance becomes greater towards the closer neighbors and less for the farther neighbors, pulling the perturbed object further towards its closer neighbors, amplifying the perturbation rather than damping it. An inward radial perturbation or deviation from the perfect circle causes the perturbed star to get closer to all other stars, increasing the force on the star and increasing its orbital velocity—which leads indirectly to a perpendicular perturbation and the argument above. Differences in mass work the same way as deviations from the perfect circle.

You cannot create this system by just adding a star to an already existing formation of stars in the manner you described because the polyhedral configuration is a repeller in the systems phase space, in other words, if the initial conditions are not just right (which holds true for the construction method you envisaged), then the system will evolve away from the tetrahedral formation to more stable configurations such as two binary systems or a binary system and two solitaries or four solitaries.

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  • $\begingroup$ Wait, any Platonic solid?? I don’t see how an icosahedron could be constructed in a dynamically unstable way. $\endgroup$ – Dubukay Dec 13 '18 at 16:27
  • $\begingroup$ Is it even possible to have all the conditions met at the same time? $\endgroup$ – Mołot Dec 17 '18 at 12:51

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