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Let's assume a simplified 2-body problem, with negligibly small satellite mass.

I know that the full state vector $\{x,\dot{x},t\}$ (position, velocity, time) at any point of the orbit uniquely determines the orbit. $\{x,t\}$ (Position+time), given 2 points allows to determine the former. But how many points of Position $x$ alone are needed to determine the orbit?

extra-hard mode: position of the central body is not known. We know the satellite is in orbit, but not around what exactly. How many position readouts would give us the orbit? (and will the central body's position uniquely determinable)?

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  • $\begingroup$ I don't think there is a solution - without any time information and not knowing the mass of the central body you can eventually get the full trajectory but have no measure of orbital period. $\endgroup$ – asdfex May 24 '16 at 11:38
  • $\begingroup$ I'm pretty sure you mean spherically symmetric body - no equatorial bulge or other weirdness. Basically a particle of negligible mass around a central attractive force $-\frac{\mathbf{\hat{r}}}{r^2}$ - I think there's a name for that but I can't remember. $\endgroup$ – uhoh May 24 '16 at 12:45
  • $\begingroup$ If it is indeed a 1/r^2 central force problem (spherically symmetric body), then the orbit is a conic section. If you know the mass, then two points, if you don't, then three (as long as you don't pick silly points). Doesn't give you epoch, but I don't think you're asking for that. However I can't prove this, needs an answer from someone who's handy with orbital math. $\endgroup$ – uhoh May 24 '16 at 14:49
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It takes five points to define an arbitrary ellipse in a plane, not knowing a priori where the center is. The five points, or really any three of them, also define the plane in three dimensions. (You never said the number of dimensions in your problem, so I am assuming three.) You will then know where the foci of the orbit are, but you won't know which one is where the body is, with no information about time.

It takes three points to define an ellipse if you know where a focus is.

In both cases, it is possible to pick points that give you no new information, so not all sets of five or three points on an ellipse will give you a unique solution. However those points are of measure zero out of all possible points, so if you pick points randomly as opposed to maliciously, five or three will be enough.

In neither case will you really define the "orbit", since you don't know anything about time. E.g. what the orbit period is, unless you know the mass of the central body, which you made no mention of, and even if you know that, you don't know when the object is at any given position in the orbit. So you are left with one degree of freedom that cannot be determined, no matter how many positions you have.

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  • $\begingroup$ "In both cases, it is possible to pick points that give you no new information" - really? The only exception I see is if two of the points are identical. $\endgroup$ – asdfex May 25 '16 at 13:43
  • $\begingroup$ @asdfex: You may be right. It comes down to the inversion of a 5x5 matrix, and I thought for sure that I could find sets of points that make the matrix singular. However trying some simple symmetric examples didn't make it singular. I will look a little harder. $\endgroup$ – Mark Adler May 25 '16 at 16:44

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