4

The two-body problem deals with two planets revolving around a common center relative to one another. Why doesn't the model ever exhibit a jagged orbit and it is always smoothly elliptical? Is there something about the math actively enforcing the smooth orbit?

develarist
  • 1,584
  • There is something about the physical assumptions that leads to mathematical equations that force smooth orbits. – Gerry Myerson Oct 02 '20 at 02:46
  • 1
    what math is feeding in the physical assumptions? – develarist Oct 02 '20 at 02:48
  • So, wait – you don't know what the equations are that are used to model the motions of two bodies under gravity? Maybe you should look that up, first. – Gerry Myerson Oct 02 '20 at 02:56
  • 1
    I know the formula, but can't tell which variables in the two-body formula are feeding in the physical assumptions about orbit – develarist Oct 02 '20 at 02:57
  • "Smooth" is a technical mathematical term and it's not clear to me whether you intend anything close to that technical sense or something totally different. (For instance, maybe what you really mean is something more like that the argument of each body relative to the center is monotone.) – Eric Wofsey Oct 02 '20 at 04:32
  • a triangular orbit, a zig-zag orbit, a wiggly orbit, a rectangular orbit are all counter-examples of what i mean by "smooth" – develarist Oct 02 '20 at 04:38
  • Perhaps "differentiable and convex" would be what OP means by "smooth". – Gerry Myerson Oct 02 '20 at 07:28
  • 1
    You are aware that the equations are the result of a central gravity field, and that a direct consequence of the equations is the Kepler laws? Or is your question about that the two-body problem decouples into two "independent" one-body problems around the center-of-mass? Which will not happen with more bodies in the system. – Lutz Lehmann Oct 02 '20 at 09:17
  • No, just the sun is fixed and only the 2nd lighter mass body is orbiting – develarist Oct 02 '20 at 11:14
  • 2
    Yes, the mathematical model (central force, proportional to $r^{-2}$) enforces elliptic/parabolic/hyperbolic trajectories, in rather good agreement with observations. If you don't like this universe, you'll have to go to another one! :) –  Oct 02 '20 at 19:08
  • 1
    It is a good question. @ProfessorVector The question is not whether the mathematical model enforces smooth trajectories, but what feature(s) of the mathematical model produces this feature. – sammy gerbil Oct 04 '20 at 18:29
  • A jagged 'orbit' is possible eg if motion is along a straight line, the smaller object passing through a tunnel in the larger object. – sammy gerbil Oct 04 '20 at 18:32
  • You can find a more "jagged" motion if you put an electron into a magnetic dipol field, see https://math.stackexchange.com/questions/3087150/ – Lutz Lehmann Oct 05 '20 at 08:04

1 Answers1

5

As you are asking a question about inertial motion and gravitational forces, this question might have been better suited to Physics SE.

It is not only the two-body problem which has smooth trajectories. n-body problems also result in smooth (although far more complex) trajectories.

In special cases gravitational trajectories can change direction abruptly. For example, the 1D motion of a smaller body oscillating in a tunnel through the centre of a larger body. However, this trajectory is the limiting case of an ellipse with eccentricity $e \to 1$, and the velocities change smoothly through zero at the extremes so there is never an infinite acceleration. Another example is the 'sling-shot' motion of a light point-like particle passing very close to a much heavier moving particle. On a 'grand' scale this looks like a discontinuous change of speed and direction but on a sufficiently fine scale the trajectory is smooth and the acceleration remains finite (although possibly very large).

The motion of objects results from the interaction between a Force Law $f(r)$ and Inertial Law $f(r)=m\ddot r$. The gravitational and electrostatic force laws $f(r)=G\frac{Mm}{r^2}$ and $f(r)=k\frac{Qq}{r^2}$ are long-range and vary smoothly and continuously with separation $r$ between objects. The inertial law (Newton's 2nd Law) which links force and motion is linear; finite accelerations are guaranteed because mass $m$ is constant and finite. Because force $f(r)$ changes smoothly then so does acceleration $\ddot r$.

By contrast, contact forces are discontinuous (for practical purposes) which is why the trajectories of billiard balls are not smooth (when viewed macroscopically). And it is conceivable (in a different universe) that mass could vary discontinuously with force, resulting in infinite accelerations and non-smooth trajectories.

sammy gerbil
  • 920