# How many position data points define an orbit uniquely?

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)?

• 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. – asdfex May 24 '16 at 11:38
• 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. – uhoh May 24 '16 at 12:45
• 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. – uhoh May 24 '16 at 14:49