# Can an orbit be calculated using two points and transit time?

Working in only two dimensions and assuming that the central body is at the origin of the coordinate system, given two points in space and knowing the transit time between those points, as well as the direction of motion, is it possible to calculate a body's orbit?

It seems to me that there should be enough information - at $T=0$, the object was at $(θ_0, r_0)$ and at $T = T_1$, $(θ_1, r_1)$.

Two points + knowing one focus gives a set of ellipses - but the additional information of transit time seems to me like it should be enough to narrow it down to 1-2 orbits in the general case, travelling in opposite directions if there are two. Is it possible? If so, how would I go about doing so (numeric approximations are fine)?

• Counterexample: you can come up with an ellipse or a parabola that passes through two points in the same times. – Meow Mar 10 '13 at 18:39

Have you looked at Lambert's Problem?

It can be used to solve for a conic orbit that goes from pointA at timeA to pointB at timeB around the same centre of attraction. It only works for the two body problem though, so it wouldn't take into account the gravity of any other bodies.

It is used in preliminary calculations for interplanetary missions.

• Thank you! Lambert's Problem is exactly what I didn't know I was looking for. I'd long ago given up on this problem and moved on to developing different computer games which didn't require an answer, but perhaps I'll revisit this one sometime. – Peter Frauenglass Dec 21 '13 at 2:39

In general, two points and transit time do not define a single orbit. Let us imagine an (non-circular) ellipse with a focus in the origin and the major axis on the abscissa. If we build a second ellipse by reflecting the first ellipse symmetrically with respect to the ordinate, these two ellipses will intersect in two points (A and B) on the ordinate, and the origin will be a focus of each of the two ellipses. So a point can travel from A to B either along one of the ellipses or along the other one. The transit time will be the same for both ellipses (if we choose the right arcs). EDIT(12/22/2012): I don't think I'll have time to give a complete answer to the edited question, but I would suggest the following approach. The equation of an ellipse with a focus in origin O has three free parameters in polar coordinates: major semi-axis, eccentricity, and the direction of the major semi-axis. If the points in question are A and B, distances OA and OB and angle AOB should fix these three parameters, yielding (in a general case) a discrete set of solutions. The equation for the transit time can be obtained by finding the area of ellipse sector AOB (integration in polar coordinates seems to yield an elementary function) and using the second and the third Kepler laws.

• But does two points and a transit time determine a small set of orbits related by symmetry? If so this is probably good enough for many circumstances. – John Rennie Dec 21 '12 at 12:00
• This is certainly possible, but I did not look into that. – akhmeteli Dec 21 '12 at 13:12
• I'm sorry, I'm not understanding the "reflecting symmetrically" bit. If I reflect an ellipse horizontally ("symmetrically with respect to the ordinate"), it shares only the two points on the X-axis with the old ellipse, and likely includes neither point A nor B. threewestwinds.com/Example.jpeg – Peter Frauenglass Dec 21 '12 at 22:15
• @Peter Frauenglass: I call those two intersection points on the ordinate of your figure: "point A" and "point B". – akhmeteli Dec 22 '12 at 1:30
• That's a special case, where the foci and the two points in question are co-linear. But even in that situation, the reflection only produces a second ellipse that matches the criteria, not an arbitrary number of them. – Peter Frauenglass Dec 22 '12 at 5:24

In general the answer is no.

In addition to akhmeteli's counterexample above, there is a more dramatic case where an observation fails to narrow down the number of possible orbits even to a finite set: if the two observations correspond to the planet's positions at periapsis and apoapsis, we will have gained information about the semimajor axis and eccentricity, but cannot fix the orientation of the orbit; this is essentially because we don't get any information about the longitude of the ascending node.

You can see this by rotating the orbit around the axis that runs through apoapsis, periapsis, and the parent star. There are continuously many possible orbits.

• However, if you move one of those two points by even the tiniest distance, the symmetry is broken and it narrows it down to exactly two orbits again, as in akhmeteli's answer. – Nathaniel Mar 21 '13 at 15:16
• This will only be very approximate, though. In the nearly-collinear case the uncertainty in the orbit's orientation will be very large. – Emilio Pisanty Mar 21 '13 at 15:20
• Ah, very good point, thank you. Would it be proper to accept this answer and create a new question, since it does answer the question I asked, but not the one I meant to ask (I'm interested in the 2-D case since this is for a computer game, not actual astronomy), or should I edit the original question? – Peter Frauenglass Mar 21 '13 at 19:01
• @PeterFrauenglass: In the 2D case knowing the positions at apoapsis and periapsis still only narrows it down to two orbits, since this carries no information about the direction; in the 3D case this uncertainty is contained in the angular degeneracy. – Chay Paterson Mar 25 '13 at 17:31
• @PeterFrauenglass: In general there will be an additional factor of two from not knowing the order in which the particle passes through the two points, so you can only ever narrow it down to four orbits: akhmeteli's plus their time reversals, but this is only one bit of information so it might not answer your question. A sensible way to fix the orbit in two dimensions is to get the position and velocity of the particle at one point in time, which is always sufficient if the gravitational sources are known. This is essentially your original idea but with an infinitesimal transit time. – Chay Paterson Mar 25 '13 at 17:40