Cometary orbit

Since $\vec r_1 = -\vec r_2$ , both vectors are perpendicular to the large half-axis of the ellipse The comet has different radial distances $\begin{cases} r_1 = r_2 = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \approx 1.414 & \\ r_3 = \sqrt{0^2 + 3^2 + \left(\frac{1}{2}\right)^2} = \sqrt{37}/2 \approx 3.041 & \end{cases}$ from the sun. The angle $\phi = \angle(\vec r_2 , \vec r_3)$ between the vectors is calculated with the dot product $\begin{aligned} \vec r_2 \cdot \vec r_3 &= r_2 r_3 \cos \phi = 1 \cdot 0 + 1 \cdot 3 + 0 \cdot \frac{1}{2} = 3\\ \Rightarrow \quad \phi &= \arccos \frac{\vec r_2 \cdot \vec r_3}{r_2 r_3} \approx 45.77^\circ \end{aligned}$ The distances $r_2'$ , $r_3'$ to the other focus read $\begin{cases} r_2' = \sqrt{r_2^2 + d^2} & \\ r_3' = \sqrt{r_3^2 + d^2 - 2 r_3 d \sin \phi} & \end{cases}$ with the distance $d$ between the two foci. The sum $r + r' = 2a$ is a constant ( $a$ : large half-axis). Therefore, we get a system of equations for the unknown $a$ and $d$ : $\begin{aligned} & \begin{cases} 2a = r_2 + \sqrt{r_2^2 + d^2} & \\ 2a = r_3 + \sqrt{r_3^2 + d^2 - 2 r_3 d \sin \phi} & \end{cases} \\ \Rightarrow & \begin{cases} (2a - r_2)^2 = r_2^2 + d^2 & \\ (2a - r_3)^2 = r_3^2 + d^2 - 2 r_3 d \sin \phi & \end{cases} \\ \Rightarrow & \begin{cases} 4a^2 - 4 a r_2 = d^2 & \\ 4a^2 - 4 a r_3 = d^2 - 2 r_3 d \sin \phi & \end{cases} \end{aligned}$ If one subtracts both equations from each other, then one gets $\begin{aligned} 4 a (r_3 - r_2) &= 2 r_3 d \sin \phi \\ \Rightarrow \quad d &= 2 a\frac{r_3-r_2}{r_3 \sin \phi} \end{aligned}$ Insertion into the first line of the equation system results $\begin{aligned} 4a^2 - 4 a r_2 &= 4 a^2\left(\frac{r_3-r_2}{r_3 \sin \phi}\right)^2\\ \Rightarrow \quad a &= \frac{r_2}{1 - \left(\frac{r_3-r_2}{r_3 \sin \phi}\right)^2} \approx 3.195 \\ \Rightarrow \quad d &= 2 a\frac{r_3-r_2}{r_3 \sin \phi} \approx 4.771 \end{aligned}$ The maximum distance of the comet from the sun is thus $r_\text{max} = a + \frac{d}{2} \approx 5.581$

Now the tilt angle $\theta$ of the ellipse in relation to the ecliptic has to be determined. A vector orthogonal to the plane results from the cross product $\begin{aligned} & & \vec r_2 \times \vec r_3 &= \left( \begin{array}{c} 1 \cdot \frac{1}{2} - 0 \cdot 3 \\ 0 \cdot 0 - 1 \cdot \frac{1}{2} \\ 1 \cdot 3 - 1 \cdot 0 \end{array} \right) = \left( \begin{array}{c} \frac{1}{2} \\ -\frac{1}{2} \\ 3 \end{array} \right) \\ & & (\vec r_2 \times \vec r_3) \cdot \vec e_z &= |\vec r_2 \times \vec r_3| \cos \theta \\ \Rightarrow & & \theta &= \arccos \frac{(\vec r_2 \times \vec r_3) \cdot \vec e_z}{|\vec r_2 \times \vec r_3|} \approx 13.26^\circ \\ \Rightarrow & & z_\text{max} &= r_\text{max} \sin \theta \approx 1.28 \end{aligned}$

The ellipse of the comet lies in a plane that passes through the origin and the three given vectors

The normal to the plane can be found by cross multiplying $r_1$ with $r_3$ .

$n = r_1 \times r_3$

After normalizing $n$ , we can find two mutually perpendicular unit vectors $u_1 , u_2$ that are orthogonal to $n$ , and thus span the plane.

Now define the reference frame using the three vectors

$R = [ u_1 , u_2 , n ]$

Then if $p_0$ and $p_1$ are the coordinate vectors in the absolute frame and in the frame we just defined, we have the following relation

$p_0 = R p_1$

Note that the choice of $u_1$ and $u_2$ is not unique.

Now we can find the coordinate vectors of $r_1$ , $r_2$ and $r_3$ with respect to the new frame,

$[r'_1, r'_2, r'_3 ] = R^T [r_1, r_2, r_3 ]$

Note that each of the three vectors is of the form

$r'_i = \begin{bmatrix} x'_i \\ y'_i \\ 0 \end{bmatrix}$

Now we use the polar equation of an ellipse with one of its foci at the origin

It is given by

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

where $\phi$ is the angle that the major axis makes with the x'-axis.

To fully specify the ellipse, we need to find the parameters $a$ (the major semi-axis) , $e$ (the eccentricity), and $\phi$ (the tilt angle). To that end, by reciprocating the above equation we get

$\dfrac{1}{r} = a' + b' \cos \theta + c' \sin \theta$

where

$a' = \dfrac{1}{a(1 - e^2) }$

$b' = - \dfrac{ e \cos \phi } {a (1 - e^2) }$

$c' = - \dfrac{ e \sin \phi }{a (1 - e^2) }$

Given the three sets of coordinates of the known vectors in the plane, we can write a linear system of three equations

in the three unknowns $a'$ , $b'$ and $c'$ . Solving the linear system, we can determine that

$\phi = \tan^{-1} (\dfrac{c'}{b'} ) + \pi$

Next $e$ can be determined as

$e = - \dfrac{b'}{a' \cos \phi }$

and

$a = \dfrac{1}{a'(1 - e^2) }$

In the final step, we determine the position vector of the aphelion as that corresponding to $\theta = \phi$ , hence

$r^* = \dfrac{ a (1 - e^2)} {1 - e} = a (1 + e)$

From which, the aphelion has coordinates

$x' = a(1 + e) \cos(\phi)$

$y' = a(1 + e) \sin(\phi)$

$z' = 0$

Using these coordinates we find the coordinates of aphelion with respect to the absolute frame as

$r = R r'$

where R is defined above. And the answer will be the z-coordinate obtained from the last equation.