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Apr 25, 2019 at 8:35 history edited Martin Sleziak CC BY-SA 4.0
removed the deprecated (geometry) tag - see the tag info: https://mathoverflow.net/tags/geometry/info; if there are some other geometry-related tags which are suitable, please use some of them instead
S Apr 25, 2019 at 8:18 history suggested Glorfindel CC BY-SA 4.0
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference); for more info, see https://meta.mathoverflow.net/a/4058/70594
Apr 25, 2019 at 7:40 review Suggested edits
S Apr 25, 2019 at 8:18
Feb 13, 2013 at 10:25 answer added jbc timeline score: 2
Feb 10, 2013 at 3:24 comment added user21349 Two quibbles with the 9Feb13 restatement: "Accurate" should be "infinitely accurate," and "n stationary/fixed point masses" should be "n stationary/fixed point masses in general position." Without "infinitely," we can make differing and arbitrarily good explanations of the same trajectory by forming a regular-polytope shell centered on the test particle and expanding the shell to an arbitrary size. Without "in general position," we already have several counterexamples.
Feb 10, 2013 at 2:30 history edited Joseph O'Rourke CC BY-SA 3.0
Clarified as per Douglas Zare's answer.
Feb 10, 2013 at 2:14 answer added Douglas Zare timeline score: 6
Feb 10, 2013 at 1:07 history edited Joseph O'Rourke CC BY-SA 3.0
Dated update.
Feb 10, 2013 at 1:02 history edited Joseph O'Rourke CC BY-SA 3.0
Repose question in light of answers & comments.
Feb 9, 2013 at 13:56 comment added user21349 @Brendan McKay: For the reasons given in my answer, I think your question needs to be refined by saying, basically, "relative to what?" If you have a finite number $j$ of masses, there's a well-defined center of mass. Let's say the trajectory is given relative to that center of mass. The masses have to interact, because otherwise you're not talking about a well-defined physical theory. So in an $n$-dimensional space, there are $2nj$ degrees of freedom, and I think infinitely precise knowledge of $2j+2$ points on the trajectory of a test particle should suffice. Why do you want $n=j$?
Feb 9, 2013 at 5:49 comment added Brendan McKay Although the original question has been answered, I wonder: is it true that in $n$-dimensional space $n$ unit masses in general position (no $k$ of them lying on a common $k-2$-dimensional subspace, for each $k$) have a unique effect on any trajectory?
Feb 9, 2013 at 4:21 answer added user21349 timeline score: 7
Feb 9, 2013 at 2:25 answer added Abhinav Kumar timeline score: 3
Feb 9, 2013 at 2:13 comment added Joel David Hamkins If the trajectory lies in a plane, then couldn't one distribute masses outside of that plane, but symmetrically, in a variety of ways so as to realize the same trajectory?
Feb 9, 2013 at 1:50 answer added Joel David Hamkins timeline score: 5
Feb 9, 2013 at 1:11 history asked Joseph O'Rourke CC BY-SA 3.0