Obtaining orbital state vectors given two position vectors and an eccentricity

Question

I was looking into the dynamics of basic 2-system orbits, and I came across a PDF (attached below) showing how to calculate orbital elements given a position and velocity vector. This led me to the question: can I derive the velocity vector V of a point, given two position vectors and the orbital eccentricity e=||e||? Note, this is not the e vector, but instead the planar eccentricity of the ellipse

Assuming that these two positions actually fit into the given orbit, I feel like this should be possible, yet I cant seem to get it to work. Since the two vectors can be used to form the orbital plane, and the shape is a simple ellipse with a given eccentricity, there should be a way to find this right? If so, does anyone know the solution?

For reference, the only assumptions I am making are

1. This is a simple 2-body system (ie planet and satellite, with the satellite having no effect on the planet)
2. The orbit is a simple ellipse (no hyperbola or parabola nonsense, just a simple ellipse with 0<e<1)
3. I am only given two position vectors and the planar eccentricity of the orbit, and the points represent a VALID orbit

EDIT: Based on the comments, I realize that this is not enough to define even a 2-D orbit, so I am adding another given. What if we are now given the semi-major axis of the orbit in addition to everything else? is that enough to get the velocity vector at one position, or do we still need more?

Thank you in advance!

https://downloads.rene-schwarz.com/download/M002-Cartesian_State_Vectors_to_Keplerian_Orbit_Elements.pdf

Answer 1

No, one focus, two points on the orbit, and the orbital eccentricity is not enough information to determine the velocity vector.

Attached below are images of two different ellipses that share the same two points on the orbit ($P_1,P_2$), both share the focus $F_0$, and both have eccentricity $e=0.8$

e=0.8, a=45	e=0.8,a=173

The Geogebra Graph I used to create the two images can be found here: Set of Orbits defined by Two Points - v2 In it, $P_1$ and $P_2$ are freely draggable, $F_1$ can be slid anywhere along the focal hyperbola.

$P_1$ and $P_2$ are the only points that are on the boundaries of both ellipses. Two different coplanar Keplerian ellipses around the same body meet at two points, or not at all, and if two different Keplerian orbits meet at a point, their orbital velocities at that point must differ, in magnitude, direction, or both.

Since the ellipses have different tangents and different semi-major axes at both points, the velocities at both points must differ in both magnitude and direction.

As for how I found these ellipses:

Given any two points on a conic section $(P_1,P_2)$ and one of the foci $(F_0)$, the second focus $F_1$ must lie on a "focal hyperbola" (the dotted gray shape in the images) that uses $P_1$ and $P_2$ as foci, and passes through $F_0$.

For an answer on why this works, see Ellipses given focus and two points on Mathematics Stack Exchange.

If the two foci are on the same lobe of the "focal hyperbola", the conic section is itself a hyperbola. If they're on different lobes, the conic section is an ellipse.

Sliding the second focus on the opposite lobe of the focal hyperbola results in a parabola $(e=1)$ when the the second focus is at infinite distance, down to some minimum eccentricity that I believe happens when the line between the foci is parallel to the line between the points, and back out to a parabola when $F_1$ reaches infinite distance again.

Sliding the second focus down the same lobe makes the resulting conic section go from parabola to an infinite-eccentricity hyperbola (when the line between the foci is perpendicular to the line between the points) and back to parabola again.

As a result, for almost every set of focus, two points, and an eccentricity, there are either two valid ellipses or two valid hyperbolas that meet the solution.

And even if you define whether the orbit is prograde or retrograde, and know the mass of the central body, that's not enough to uniquely determine the velocity vector.

If you had three points, however, you wouldn't even need eccentricity, just the mass of the central body and the direction of travel of the orbit.

Answer 2

Not a complete answer, but: Let $A$ and $B$ be the two given positions, $O$ the position of the primary body, and $F$ the second focus of the orbit. Then we have $|OA|+|AF| = |OB|+|BF|$ and $|OF| = e(|OA|+|AF|)$, which gives us a system of equations on the coordinates of $F$. I don't know if it can be solved analytically, but you can solve it numerically, and from there, you will have enough information to construct the ellipse.

Answer 3

Another partial answer:

Assuming the two points $\mathbf{x_1}, \mathbf{x_1}$ are not identical nor coincident with the origin, then the three points exactly define a plane, so let's solve the 2D orbit problem within this plane first.

In polar coordinates, with the focus at the origin:

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

1. transform your $\mathbf{x_1}, \mathbf{x_1}$ in 3D space to $(x_1, y_1), (x_2, y_2)$ positions in the plane,
2. convert those to two $(r_1, \theta_1), (r_2, \theta_2)$ positions in the plane
3. write the problem as two equations and two unknowns $a$ and $\theta_0$:

$$r_1 = \frac{a(1-e^2)}{1-e \cos(\theta_1-\theta_0)}$$ $$r_2 = \frac{a(1-e^2)}{1-e \cos(\theta_2-\theta_0)}$$

Solve for $a$ and $\theta_0$.

$\theta_0$ will be your periapsis point, defining your line of nodes.

Transform your ellipse in the plane back to your 3D space.

To get your six-dimensional state vector, choose any point on the ellipse. That gives you $x, y, z$.

Get the radius $r = \sqrt{x^2 + y^2 + z^2}$

Get the speed from the vis-viva equation:

$$v^2 = GM \left(\frac{2}{r} - \frac{1}{a} \right)$$

Calculate the normal tangent to the ellipse $\hat{n}_T$

$$v_x, v_y, v_z = v \hat{n}_T$$