Hi,
thanks for the interesting problem to think about
I can get the semimajor axis (a) and the standard gravitational parameter (mu) from the vis-viva equation;
v^2 = mu*(2/r - 1/a)
But then I get stuck!
I'm a bit suspicious of the orbital speeds given in the opening post, but here is the general procedure for solving this problem for Keplerian orbits where the points are separated by less than half an orbit:
Let's suppose, as above, that you are given two (3-vector) points on an Keplerian orbit [MATH]\mathbf{R}_i[/MATH] and [MATH]\mathbf{R}_f[/MATH] and that you know the (scalar) orbital speeds at those points, [MATH]V_i[/MATH] and [MATH]V_f[/MATH] respectively.
Now calculate:
[MATH]
\begin{aligned}
r_i &= \sqrt{\mathbf{R}_i.\mathbf{R}_i} \\
r_f &= \sqrt{\mathbf{R}_f.\mathbf{R}_f} \\
\theta &=\cos^{-1}\left(\frac{\mathbf{R}_i.\mathbf{R}_f}{r_i\,r_f}\right) \\
A &= r_i + r_f \\
B &= 2\,\sqrt{r_i\,r_f}\,\cos\left(\frac{\theta}{2}\right)
\end{aligned}
[/MATH]
In addition, calculate the semi-major axis at either the first or last point using:
[MATH]V_i^2 = \mu\,\left(\frac{2}{r_i}-\frac{1}{a}\right)[/MATH]
or
[MATH]V_f^2 = \mu\,\left(\frac{2}{r_f}-\frac{1}{a}\right)[/MATH]
Then (and here's the magic!), solve for the two roots of the quadratic equation:
[MATH]A-B\,X = 2\,a\,(1 - X^2)[/MATH]
Call these roots [MATH]X_1[/MATH] and [MATH]X_2[/MATH]. These two numbers encode information about the two Keplerian arcs that join the points [MATH]\mathbf{R}_i[/MATH] and [MATH]\mathbf{R}_f[/MATH] with semi-major axis [MATH]a[/MATH].
Then, the radial and transverse components of the velocity vector at the [MATH]\mathbf{R}_i[/MATH] are given by:
[MATH]
\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right)
[/MATH]
and the radial and transverse components of the velocity vector at the [MATH]\mathbf{R}_f[/MATH] are given by:
[MATH]
\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right)
[/MATH]
As mentioned, this gives two possible solutions - corresponding to the two possible Keplerian arcs that pass through the initial and final points with set-major-axis [MATH]a[/MATH]. For each solution, we calculate the eccentricity in the usual way. In practice, for problems involving nearly circular orbits, one value of [MATH]X[/MATH] will give a high eccentricity; and the other will give one close to zero. Obviously, we choose the solutions corresponding to the low orbital eccentricity - thereby removing the degeneracy.
(N.B., I have a proof of the above but it's a little long-winded - too long for this post. But I can share if anyone is sufficiently interested)
---------- Post added at 02:35 PM ---------- Previous post was at 01:23 PM ----------
Let's take a simple worked example:
Suppose that we choose units such that [MATH]\mu=1[/MATH] with:
[MATH]
\begin{aligned}
\mathbf{R}_i &= \left(-0.9803802, 0.241571, 0\right) \\
\mathbf{R}_f &= \left(-0.9972285, -0.159303, 0\right) \\
\\
V_i &= 0.9903428 \\
V_f &= 0.9901760
\end{aligned}
[/MATH]
Find the velocity vectors at [MATH]\mathbf{R}_i[/MATH] and [MATH]\mathbf{R}_f[/MATH].
Using the above algorithm we calculate:
[math]
\begin{aligned}
r_i &= \sqrt{\mathbf{R}_i.\mathbf{R}_i} = 1.009704\\
r_f &= \sqrt{\mathbf{R}_f.\mathbf{R}_f} = 1.009872 \\
\theta &=\cos^{-1}\left(\frac{\mathbf{R}_i.\mathbf{R}_f}{r_i\,r_f}\right) = 0.4000000\\
A &= r_i + r_f = 2.019576\\
B &= 2\,\sqrt{r_i\,r_f}\,\cos\left(\frac{\theta}{2}\right) = 1.979319
\end{aligned}
[/math]
We next calculate the semi-major axis using
[math]
V_i^2 = \mu\,\left(\frac{2}{r_i}-\frac{1}{a}\right)
[/math]
or
[math]
V_f^2 = \mu\,\left(\frac{2}{r_f}-\frac{1}{a}\right)
[/math]
and in both cases we calculate the same semi-major axis [MATH]a = 1.00000[/MATH]. (You may find this value a little suspicious - but it's just a result of me setting up a test problem with simple parameters.)
Next, we solve the quadratic:
[math]
A-B\,X = 2\,a\,(1 - X^2)
[/math]
The two roots of this are [MATH]X_1 = 0.00999118[/MATH] and [MATH]X_2 = 0.9796683[/MATH].
With [MATH]X = X_1 = 0.00999118[/MATH], the radial and transverse components at [MATH]\mathbf{R}_i[/MATH] are:
[math]
\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(0.9702056, 0.1986958\right)
[/math]
and the radial and transverse components at [MATH]\mathbf{R}_f[/MATH] are:
[math]
\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(-0.9700421, 0.1986627\right)
[/math]
And with [MATH]X = X_2 = 0.9796683[/MATH], the radial and transverse components at [MATH]\mathbf{R}_i[/MATH] are:
[math]
\left(v_{i,r},v_{i,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(\sqrt{\frac{r_f}{r_i}}\,\cos\left(\frac{\theta}{2}\right)- X,\sqrt{\frac{r_f}{r_i}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(0.002392613, 0.9903399\right)
[/math]
and the radial and transverse components at [MATH]\mathbf{R}_f[/MATH] are:
[math]
\left(v_{f,r},v_{f,\theta}\right) = \sqrt{\frac{2\,\mu}{A-B\,X}}\,\left(X - \sqrt{\frac{r_i}{r_f}}\,\cos\left(\frac{\theta}{2}\right),\sqrt{\frac{r_i}{r_f}}\,\sin\left(\frac{\theta}{2}\right)\right) = \left(-0.001577536, 0.9901747\right)
[/math]
Based on this, if we go through the straightforward exercise of calculating the eccentricity with [math]X=X_1[/math] we find that the orbital eccentricity is 0.9796683; and with [math]X=X_2[/math] we find that the orbital eccentricity is 0.010000. Since I know that the orbit is nearly circular (i.e., so that the eccentricity is close to zero), the appropriate solution to use is the one corresponding to [math]X=X_2[/math].
So, finally, the radial and transverse components of the velocity vector at [MATH]\mathbf{R}_i[/MATH] are [MATH](0.002392613, 0.9903399)[/MATH]; and the radial and transverse components of the velocity vector at [MATH]\mathbf{R}_f[/MATH] are [MATH](-0.001577536, 0.9901747)[/MATH].
