Where does this formula for the argument of periapsis come from?

Question

I was looking at Poliastro/Hapsira python code determining orbital elements from the position and velocity vector. On this line, argp or $\omega$ is determined as (if my understanding of the @ operator in Python is correct):

$$p_x = \vec{r}\cdot\vec{n}$$

$$p_y = \frac{\vec{r}\cdot(\vec{h}\times\vec{n})}{h}$$

$$\omega=\arctan2(py,px) - \nu \mod (2\pi)$$

where $\vec{r},\vec{v}$ are the position and velocity vector, $\vec{h}$ is the angular momentum vector, and $\vec{n}$ the node vector, with $\nu$ being the true anomaly.

the source mentions Curtis' Orbital Mechanics for Engineering Students, Example 4.3. But in my copy it seems to use this more common formula:

$$ \omega = \left\{ \begin{array}{lcc} cos^{-1}{\left ( \frac{\vec{N}\vec{e}}{Ne} \right )} & if & e_{z} \geq 0 \\ \\ 360^{o} -cos^{-1}{\left ( \frac{\vec{N}\vec{e}}{Ne} \right )} & if & e_{z} < 0 \\ \end{array} \right.$$

After an extensive search I have not been able to find other mentions of this expression. Not being an expert in astrodynamics I was curious to see how it is derived.

Answer 1

The arctan2 expression eliminates the need for having two alternate expressions using the inverse cosine. They are equivalent. But do be careful when using arctan2 because the order of the arguments varies between programming languages.

Answer 2

The argument of periapsis is the angle between the node vector $\vec{n}$ (which points towards the ascending node) and the eccentricity vector $\vec{e}$ which points towards periapsis. By the dot product formula $\vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\|\cos(\theta)$ where $\theta$ is the angle between the two vectors, $\frac{\vec{n} \cdot \vec{e}}{ne} = \cos(\omega)$. The angle is $\omega$ by definition as it is the angle between the ascending node and the periapsis. This is the derivation of the original formula, with the coterminal case required due to the quadrant ambiguity of $\mathrm{cos}^{-1}$.

Remember, $\vec{h}$ points in the direction perpendicular to the plane of the orbit. Thus, $\vec{h} \times \vec{n}$, by the right hand rule, is a vector coplanar to the orbital plane, perpendicular to the ascending node.

Note that $$\|\vec{h} \times \vec{n}\| = \|\vec{h}\|\|\vec{n}\|\sin(\theta)$$ where $\theta$ is the angle between $\vec{h}$ and $\vec{n}$ in the plane containing them. However, remember that $\vec{n} = [0,0,1] \times \vec{h}$ and thus $\vec{h}$ and $\vec{n}$ are orthogonal. Thus, $\theta = \frac{\pi}{2}$ and $\sin(\frac{\pi}{2}) = 1$ and thus the length of this vector is simply $\|\vec{h}\|\|\vec{n}\|$.

Thus, $\vec{r} \cdot (\vec{h} \times \vec{n}) = \|\vec{r}\|\|\vec{h} \times \vec{n}\|\cos(\theta_1) = \|\vec{r}\|\|\vec{h}\|\|\vec{n}\|\cos(\theta_1)$. Which makes $p_y = \|\vec{r}\|\|\vec{n}\|\cos(\theta_1)$. Note that $\theta_1$, the angle between $\vec{r}$ and $\vec{h} \times \vec{n}$, will be $\omega + \nu - \frac{\pi}{2}$. Thus, $$\cos(\theta_1) = \cos(\omega + \nu - \frac{\pi}{2}) = \sin(\omega + \nu)$$ This is because the angle between $\vec{r}$ and $\vec{h} \times \vec{n}$, a vector perpendicular to $\vec{n}$ (in the same plane as $\vec{r}$), is the same as the angle between $\vec{r}$ and $\vec{n}$ (which is $\omega + \nu$) minus the $\frac{\pi}{2}$ angle between $\vec{n}$ and $\vec{h} \times \vec{n}$.

Additionally, $p_x = \vec{r} \cdot \vec{n} = \|\vec{r}\|\|\vec{n}\|\cos(\theta_2)$. And $\theta_2 = \omega + \nu$, as it is the angle between the ascending node and the current radius, which is the argument of periapsis plus the true anomaly.

Dividing these two: $$\frac{p_y}{p_x} = \frac{\|\vec{r}\|\|\vec{n}\|\sin(\omega + \nu)}{\|\vec{r}\|\|\vec{n}\|\cos(\omega + \nu)} = \frac{\sin(\omega + \nu)}{\cos(\omega + \nu)} = \tan(\omega + \nu)$$

Thus we get $\mathrm{arctan2}(p_y, p_x) = \tan^{-1}(\frac{p_y}{p_x}) = \omega + \nu$ and thus, finally, $$\omega = \mathrm{arctan2}(p_y, p_x) - \nu \mod (2\pi)$$

I believe this derivation is correct, but if you spot any mistakes, please don't fret to let me know or correct them yourself!