2
$\begingroup$

The problem

I'm trying to get a formula to calculate the state vectors $\vec{r}$ and $\vec{v} = \dot{\vec{r}}$ on an orbit, given a true anomaly $\nu$. I'm following the process described here : https://downloads.rene-schwarz.com/download/M001-Keplerian_Orbit_Elements_to_Cartesian_State_Vectors.pdf . The first calculation step involves calculating the intermediate simple state vectors $\vec{o}$ and $\dot{\vec{o}}$ laying in the xy plane (which are then rotated in space to get $\vec{r}$ and $\vec{v}$ respectively) :

$$ \vec{o} = r\left(\begin{array}{ c } \cos \nu\\ \sin \nu \\ 0 \end{array}\right), \ \ \dot{\vec{o}} = \frac{\sqrt{\mu a}}{r}\left(\begin{array}{ c } -\sin E\\ \sqrt{1-e^{2}}\cos E\\ 0 \end{array}\right) $$

with $r = \dfrac{p}{1+e\cos\nu}$.

This works well in the case of an elliptic orbit, but it is invalid for a hyperbolic one because of $e > 1$ and $a < 0$ resulting in $\sqrt{1-e^2}$ and $\sqrt{\mu a}$ being undefined.

My current solution

In the case of an hyperbolic orbit, I adapted the second answer from this post : Calculating velocity state vector with orbital elements in 2D to calculate $\dot{\vec{o}}$ with the flight path angle $\phi$, knowing the angular momentum $h = ||\vec{h}||$. We first calculate the radial unit vector $\hat{u_o}$ of the intermediate position vector, and $\hat{u_s}$ the unit vector perpendicular to $\hat{u_o}$ in the xy plane :

$$ \hat{u_o} = \frac{\vec{o}}{r} = \left(\begin{array}{ c } u_{o,\ x}\\ u_{o,\ y}\\ 0 \end{array}\right) , \ \ \hat{u_s} = \left(\begin{array}{ c } -u_{o,\ y}\\ u_{o,\ x}\\ 0 \end{array}\right) $$

We then calculate the sin and cos of the flight path angle : $$ \cos \phi = \frac{h}{rv}, \ \ \sin \phi = \frac{e \sin \nu}{1 + e \cos \nu} \cos \phi $$

with $v$ being the magnitude of the velocity, calculated from the vis-viva equation. And finally, we get the intermediate velocity vector :

$$ \dot{\vec{o}} = v(\sin(\phi)\hat{u_o} + \cos(\phi)\hat{u_s}) $$

A better solution ?

Is there a better, more straightforward way, to compute this intermediate velocity vector in the case of a hyperbolic orbit ? One that doesn't require knowing $h$. For example, is there a formula similar to the one given in the PDF that makes use of the hyperbolic eccentric anomaly $H$ ?

Thanks in advance.

$\endgroup$
3
1
+100
$\begingroup$

Solution for hyperbolic velocity

After some mathematical manipulations I ended up finding an actual solution that makes use of the hyperbolic anomaly $H$.

In the following, $e$ is the eccentricity of the orbit, $\nu$ is the true anomaly and $a$ is the semi major axis.

Prerequisite : proving that $iH = E$ in a hyperbolic orbit

This little proof is here just to show how one can retrieve the equality from the well known formulas of eccentric anomalies.

  • For an elliptical orbit ($e < 1$), the eccentric anomaly $E$ is defined by: $$ \tag{1} \tan\frac{E}{2} =\sqrt{\frac{1-e}{1+e}}\tan\frac{\nu }{2} $$

  • For a hyperbolic orbit ($e > 1$), the hyperbolic anomaly $H$ (also written $F$) is defined by: $$ \tag{2} \tanh\frac{H}{2} =\sqrt{\frac{e-1}{e+1}}\tan\frac{\nu }{2} $$

In the case of a hyperbolic orbit, $1-e < 0$ leads to an undefined definition of $E$ in (1) because of the square root term. Thus the need to use its hyperbolic equivalent (2). However, considering the relation (1) in $\mathbb{C}$ by involving $i = \sqrt{-1}$ allows for a complex definition of $E$ :

$$ \tan\frac{E}{2} = i\sqrt{\frac{e-1}{e+1}}\tan\frac{\nu }{2} $$

in which we notice the right term of (2). This actually directly links $E \in \mathbb{C}$ and $H \in \mathbb{R}$ by:

$$ \tag{3} \tan\frac{E}{2} = i\tanh\frac{H}{2} $$

The relations between hyperbolic and trigonometric functions give : $$ \forall x \in \mathbb{R} \ \ \ i\tanh(x) = \tan(ix) $$

which when applied to (3) leads to:

$$ \tan\frac{E}{2} = \tan\frac{iH}{2} $$

And since $x \mapsto \tan(ix)$ is bijective $\forall x \in \mathbb{R}$ because it is proportional to $x \mapsto \tanh(x)$, we deduce that :

$$ \tag{4} iH = E $$

in the case of a hyperbolic orbit, with $H \in \mathbb{R}$ (and therefore $E \in i\mathbb{R}$).

Adapting the intermediate velocity vector formula to hyperbolic orbits

The equation described by René Schwarz to calculate the intermediate velocity vector (ignoring the z-component with value 0) is:

$$ \tag{5} \dot{\vec{o}} =\frac{\sqrt{\mu a}}{r}\left(\begin{array}{ c } -\sin E\\ \sqrt{1-e^{2}}\cos E \end{array}\right) $$

We suppose now a hyperbolic orbit, therefore $e > 1$ and $a < 0$. Thus (5) cannot be used directly because $\sqrt{1-e^2}$ and $\sqrt{\mu a}$ are undefined in $\mathbb{R}$. By using the fact that $a = -|a|$ and $1 - e^2 = -(e^2 - 1)$, considering the equation in $\mathbb{C}$ gives:

$$ \begin{array}{ c c l } ( 5) \ \ \Leftrightarrow \ \ \dot{\vec{o}} & = & \dfrac{\sqrt{-\mu |a|}}{r}\left(\begin{array}{ c } -\sin E\\ \sqrt{-\left( e^{2} -1\right)}\cos E \end{array}\right)\\ & = & i\dfrac{\sqrt{\mu |a|}}{r}\left(\begin{array}{ c } -\sin E\\ i\sqrt{e^{2} -1}\cos E \end{array}\right)\\ & = & \dfrac{\sqrt{\mu |a|}}{r}\left(\begin{array}{ c } -i\sin E\\ -\sqrt{e^{2} -1}\cos E \end{array}\right)\\ & = & \dfrac{\sqrt{\mu |a|}}{r}\left(\begin{array}{ c } -\sinh iE\\ -\sqrt{e^{2} -1}\cosh iE \end{array}\right) \end{array} $$

because $\forall x \in \mathbb{R} \ \ i\sin x = \sinh ix$ and $\cos x = \cosh ix$.
Finally, involving $iH = E \Leftrightarrow iE = -H$, and the fact that $\cosh$ is even and $\sinh$ is odd, we get:

$$ \tag{6} \dot{\vec{o}} =\frac{\sqrt{-\mu a}}{r}\left(\begin{array}{ c } \sinh H\\ -\sqrt{e^{2} -1}\cosh H \end{array}\right) $$

with $|a|$ written as $-a$ for clarity.

This formula seems to work in practical cases (orbit simulation and determination). Please don't hesitate to comment to correct eventual errors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.