In a question about determining orbital elements from two points on an orbit, some discussion arouse regarding how many points are actually necessary, since two is not sufficient.

Determining a conic section requires 5 points, though one of them can be the central mass, so only 4 points on the orbit.

But are 4 points actually necessary in practice?

This polar equation seems to suggest 3 points are enough, though with arguably some extra assumptions: $$r = \frac{a(1−e^2)}{1+e\cosθ}$$

The only counterexample with ambiguous orbits I could come up with is "unrealistic", as it involves the phantom half of a hyperbola.

Does disregarding such purely theoretical trajectories reduce the requirement to 3 points on the orbit?


  • $\begingroup$ I wouldn't call this "purely theoretical trajectory" - that would be something possible, but not realised. What you show is commonly called an "unphysical solution" of the equations, because of the additional constraint that an orbit needs to be continuous which is not reflected in the formulae. $\endgroup$
    – asdfex
    Commented Oct 9, 2021 at 16:24
  • 1
    $\begingroup$ When you give a point on the trajectory, that's 2 coordinates, but the information obtained is the same as if you gave any other point in a 1-dimensional line (the trajectory), so the number of parameters is decreased by 2-1=1. When you give the location of central mass, the number of parameters is decreased by 2. (This does not prove that 3 points on the orbit and the central mass uniquely specify the orbit, just explains why one would expect this information to restrict the set of possible orbits to a discrete one, not a continuous infinite one, as when just 4 points on the orbit are given.) $\endgroup$
    – Litho
    Commented Oct 9, 2021 at 18:17
  • $\begingroup$ It is similar to this: you can specify a circle by giving its center and 1 point on the circle; but 2 points on the circle are not enough to specify it, you need 3. $\endgroup$
    – Litho
    Commented Oct 9, 2021 at 18:19
  • $\begingroup$ Thanks for this question, I think I learned quite a bit about conic sections and GeoGebra working this one out. $\endgroup$
    – notovny
    Commented Oct 11, 2021 at 1:31

2 Answers 2


I believe the answer is yes, given one focus, and three coplanar points on the orbit, you can define the orbital shape, and obtain semi-major axis, orbital eccentricity, and the direction of the periapsis.

Since the three points must be coplanar within the orbital plane. I leave the exercise of converting the three coplanar points you have to orbital-plane coordinates as an exercise for the reader.

I wound up not solving the polar equations to do this, but instead, relying on the definitions of ellipses and hyperbolas, and a lot of help from GeoGebra:

Ellipse from 3 points and focus Hyperbola from 3 points and focus
Ellipse from 3 points and focus Hypebola from 3 points anf focus

Given 3 points, $P_1, P_2, P_3$ of format (x,y), and the focus $F_0$ at the origin:

Solving it Geometrically:

Circular case: If all three points are equidistant from the known focus, it's a circular orbit.

Elliptical or Hyperbolic Case: Otherwise, for any pair of the chosen points, $P_1, P_2, P_3$, there exists a hyperbola that passes through the intended focus $F_0$. The ellipse or hyperbola that passes through the pair of chosen points must have both foci on that focal hyperbola; if it's a hyperbolic orbit, the focus will be on the same lobe. If it's an elliptical orbit it will be on the other lobe.

If you construct the three hyperbolas that have each each pair of points as foci ($P_1$,$P_2$) , ($P_1$,$P_3$), ($P_2$,$P_3$) and pass through $F_0$, the other point where those three hyperbolas intersect is where the other focus is.

Three hyperbolas defining the foci of an ellipse.
Three Hyperbolae defining the foci of an ellipse.

You can then examine the orbital ellipse and hyperbolic trajectory defined by the two foci. In most cases, all three points will lie on either the ellipse or the hyperbola, determining which of the two is the valid trajectory.

There are rare cases where the three points can be on both the hyperbola and the ellipse; but if an ellipse and a hyperbola share foci, they will intersect at only four points, requiring at least one of the points be on the non-physical phantom lobe on the hyperbola, disqualifying it.

Parabolic Case: If the secondary focus winds up at infinity, it's the parabolic case. Construct three circles around the points that intersect at the focus. The line that's tangent to all three circles is the directrix, use that and the focus to construct your parabola.

The Geogebra Graph used to create the above images can be found here: Geometric Three-Point and Focus Orbit - Shared All three points are freely draggable, and it only seems to stutter in a few cases.

Solving it analytically:

An ellipse is defined as the set of all points in a plane where the sum of the distance of the point from the two foci is a constant : Twice the semimajor axis. The following equations assume the primary focus, $F_0$ is at the origin, and relate the semi-major axis and the x and y coordinates of the secondary focus.

Solve the following equations: $$ \\(2z - |P_1|)² = (x - P_{1x})² + (y - P_{1y})² \\(2z - |P_2|)² = (x - P_{2x})² + (y - P_{2y})² \\(2z - |P_3|)² = (x - P_{3x})² + (y - P_{3y})² $$

Where $|P_n|$ denotes the distance of the point $P_n$ from the known focus, and $P_{nx}$ and $P_{ny}$ are the x and y coordinates of $P_n$ (I'm open to better notation on these, this is probably a bit confusing)

  • $z$ will be have the same magnitude of your semi-major axis.
  • $F_1=(x,y)$ is the other focus, where $x$ and $y$ are not equal to 0.
  • $2c$ is the distance between the two foci.
  • $e = |c/z|$ gives the orbital eccentricity. If it is less than 1, the orbit is elliptical, if it is greater than 1, the orbit is hyperbolic.
  • $a = \pm|z|$ is the semi-major axis. Use the negative value if hyperbolic, and the positive value if elliptical.

We have a rotated conic section now. The angle of periapsis can be found opposite the direction of the second focus, if elliptical, and in the direction of the second focus, if hyperbolic.

$$ \displaystyle \theta_0 = \begin{cases} \mathrm{atan2}(-F_{1y},-F_{1x}) & \text{if $e < 1$} \\ \mathrm{atan2}(F_{1y},F_{1x}) & \text{if $e>1$} \\ \end{cases} $$

And the polar equation for the conic section is:

$$r(\theta)= \frac{a(1-e^2)}{1+e \cos(\theta - \theta_0)}$$

If the orbit turned out to be hyperbolic, compare the distance from the points to the primary focus and to the secondary focus. If the points are closer to the secondary focus, they're on the phantom lobe, and the solution is nonphysical; the object could not have been at those points in a Keplerian hyperbolic trajectory.

I've mostly been working this out through GeoGebra, and its numerical solver is really finicky on these, so I'm not entirely certain that this is correct, but it seems to work. It also works if the three points are on the hyperbola's phantom lobe.

If you want to fiddle around with it, you can do so here: GeoGebra: Three Point and Focus Fiddler

Dragging around point $P_2$ seems to be generally fine, but the numerical solver for the equations gets a bit stroppy if $P_1$ or $P_3$ are moved around too much. It's still a bit of a work in progress.

And if you know the order the three points are passed in, then, yes, you can figure out whether it's prograde or not, you can go back to 3D and determine the orbital inclination, the longitude of the ascending node, and the argument of periapsis.


If the time between two points is known, only two points are sufficient. The angle between these points should be larger than 0° and less than 180°.

If the position and velocity relative to the observer are available (as is the case with radar observations), these observational data can be adjusted by the known position and velocity of the observer relative to the attracting body at the times of observation. This yields the position and velocity with respect to the attracting body. If two such observations are available, along with the time difference between them, the orbit can be determined using Lambert's method, invented in the 18th century. See Lambert's problem for details.

Even if no distance information is available, an orbit can still be determined if three or more observations of the body's right ascension and declination have been made. Gauss's method, made famous in his 1801 "recovery" of the first lost minor planet, Ceres, has been subsequently polished.

Source: https://en.wikipedia.org/wiki/Orbit_determination

Three points without distance information are sufficient too.

  • 4
    $\begingroup$ This is not correct. Consider Lambert's problem. There are at least two very different orbits that start at one point at a specific time and end at a different point at a later specific time. In the case of points separated by 180°, the number of potential orbits is infinite. (Lambert's problem has a singularity at 180°, and also at n*360°+180°, where n is a positive integer.) If the time difference between the end time and start time is large enough, there can be more than two distinct orbits. $\endgroup$ Commented Oct 9, 2021 at 16:12
  • $\begingroup$ @TooTea The wikipedia article aboutLambert's problem writes only of position and time. $\endgroup$
    – Uwe
    Commented Oct 10, 2021 at 19:21
  • $\begingroup$ @DavidHammen See the edit. I suppose restricting it to 0° ≤ angle < 180° guarantees a unique solution. $\endgroup$
    – TooTea
    Commented Oct 10, 2021 at 19:36
  • $\begingroup$ @TooTea The edit is not sufficient. A transfer orbit might involve a 360° to 540° change in true anomaly and still satisfy the 0° to 180° requirement (which is arbitrary) if the time difference is large enough. $\endgroup$ Commented Oct 10, 2021 at 22:16
  • $\begingroup$ @DavidHammen I understand that restriction as the total change in true anomaly. In other words,saying that the time refers to the first crossing of point 2 after point 1 (thus ruling out once-around trajectories) should be enough to make the solution unique. Perhaps just a bit of rewording would do. $\endgroup$
    – TooTea
    Commented Oct 11, 2021 at 6:24

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