(I apologize if this question is too theoretical for this site.)

This is related to the answer here, although I came up with it independently of that. $\:$ Suppose we
have a unit mass planet at each integer point in 1-d space. $\:$ As described in that answer, the sum
of the forces acting on any particular planet is absolutely convergent. $\;\;$ Suppose we move planet_0
to point $\epsilon$, where $\: 0< \epsilon< \frac12 \:$. $\;\;$ For similar reasons, those sums will still be absolutely convergent.
Now we let Newtonian gravity apply. $\:$ What will happen?

If it's unclear what an answer might look like, you could consider the following more specific questions:

planet_0 will start out moving right, and all of the other planets will start out moving to the left.
Will there be a positive amount of time before any of them turn around?
(As opposed to, for example, each planet_n for $\: n\neq 0 \:$ turning around at time 1/|n|.)

Will there be a positive amount of time before any collisions occur?

"Obviously" (at least, I hope I'm right), planet_0 will collide with planet_1. $\:$ Will that be the first collision?

How long will it be before there are any collisions? $\:\:$ (perhaps just an approximation for small $\:\epsilon\:$)


The acceleration of planet number $n$ except for the planet $0$ will go like $-1/n^3$ because the shift of planet $0$ from zero to $\epsilon$ is equivalent to adding a "dipole" (a pair of positive and negative mass, relatively shifted) at the location $0$ relatively to the balanced (but unstable) uniform chain and this dipole acts with inverse cube, instead of the inverse law.

We see that indeed the planets $+1$ and $-1$ are most affected and fastest to get some acceleration. However, planet $-1$ will move to the left, away from a potential collision. Nevertheless, planet $-2$ is trying to escape from planet $-1$, although by a smaller speed, but that will be enough to guarantee that the $0-1$ collision will be the first one. Other collisions will follow. You may numerically simulate it – the problem isn't integrable even for small $\epsilon$, I think, simply because you're interested in the moments when the distance $\epsilon$ grew to a large number $O(1)$, anyway.


Lumo's answer is good. I would accept it. Only because I'm bored and it's easy here is a numerical simulation of 31 planets (a poor man's infinity), placed initially at rest at the integers, except the centre body is displaced upward by 0.1:

enter image description here

Note that, of course, the outer bodies fall toward the middle since there are only finitely many planets here. Planets 0 and 1 fall towards each other much faster than any of the other bodies (apart from the outer two), and the distance between 0 and -1 is increasing. This agrees with Lumo's prediction. The other bodies are nearly stationary, which gives confidence that the boundary conditions aren't numerically important for understanding what is happening in the middle. Shortly after this plot ends bodies start colliding and this simple simulation breaks down. So in units where everything is 1 it takes about 1 unit of time for bad things to happen. :)

  • $\begingroup$ Good job. But you could easily treat the planets at the end of your truncated sequence much more accurately, so that the collisions near the end of the world - which are artifacts of your truncation, I hope you agree - wouldn't be there etc. $\endgroup$ – Luboš Motl Mar 15 '13 at 7:04
  • $\begingroup$ @LubošMotl Thanks. :) Yes, of course. I gave half a thought to doing periodic boundary conditions or artificially pinning the end ones in place (please don't egg me), but I had already uploaded the picture. :) $\endgroup$ – Michael Brown Mar 15 '13 at 7:08

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