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.


1 Answer 1


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.


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