Planar motion in central forces.

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SUMMARY

The discussion centers on the behavior of bodies under central forces, specifically in a three-body problem where one body is significantly more massive. It concludes that while angular momentum is conserved, the motions of the two lighter bodies can indeed be non-planar, potentially existing in orthogonal planes. However, the solar system's flatness is attributed to the conservation of angular momentum and the historical dynamics of pre-planetary collisions, which tend to align orbits in a common plane.

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  • Understanding of angular momentum conservation in physics
  • Familiarity with central force dynamics
  • Knowledge of the two-body problem in classical mechanics
  • Basic concepts of orbital mechanics and planetary formation
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precise
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I understand that in a two body problem under central force, corresponding to a potential V(r)(assume one body is massive compared to the other so that its motion is negligible), conservation of angular momentum implies the motion of the body to be in a plane spanned by position r and momentum p vectors.

But if we have three bodies, one of them massive, are the motions of other two bodies still restricted to a plane? Now the total angular momentum is L = L1 + L2 = r1 x p1 + r2 x p2, which is conserved. Mathematically, L could be kept constant while L1 and L2 are changing. Which means we could have motions of the two bodies in two planes orthogonal to each other, a non-planar motion. Is this allowed? If not, why? Then, what is reason for the planar motion?

In specific why is the solar system flat?
 
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precise said:
why is the solar system flat?

https://www.youtube.com/watch?v=tmNXKqeUtJM
 
If the forces on the planets are directed towards a fixed point O, (a good approximation if the Sun is much more massive than any planet), then each planet's angular momentum about O is separately conserved, since if r is the planet's displacement from O, and F is the force on the planet (towards O)…
\mathbf r \times \mathbf F = \mathbf r \times (-F\ \mathbf {\widehat{r}}) = 0
But…
\mathbf r \times \mathbf F = \mathbf r \times \frac{d \mathbf p}{dt} = \frac{d}{dt} (\mathbf r \times \mathbf p) - \frac{d \mathbf r}{dt} \times \mathbf p
The last term is zero because
\frac{d \mathbf r}{dt} \times \mathbf p = \frac{d \mathbf r}{dt} \times m \frac {d \mathbf r}{dt} = 0
So
\frac{d}{dt} (\mathbf r \times \mathbf p ) = 0\ \ \ \ \ \ \ \ \text {so} \ \ \ \ \ \ \ \ \mathbf r \times \mathbf p = \mathbf{constant\ vector}
Non-mathematically, the argument is simply that for any planet a force towards O can't give rise to a torque about O, so angular momentum about O is conserved.

So orbital planes can be at angles to each other, but in general won't be, because of collisions in the pre-planetary swirl, as A.T.'s excellent clip explains.
 
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precise: are you clearer now?
 
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