How does one calculate the argument of periapsis of an orbit after an arbitrary maneuver?

Given a satellite in an equatorial orbit, a specific prograde or retrograde burn is executed at an arbitrary point within the orbit, and I need to calculate the resulting orbital ellipse.

The technique I'm using is to first use the position and velocity vectors of the satellite to find the flight path angle, as follows:

$$\varphi = cos^{-1}(\frac{r_pv_p}{r_bv_b})$$

Where $$r_p$$ and $$v_p$$ are the position and velocity vectors at the periapsis of the original orbit, and $$r_b$$ and $$v_b$$ are the position and velocity vectors at the point of the burn, and $$v_b = v_{orig} + \Delta v$$.

Then I calculate the eccentricity of the resulting ellipse as follows:

$$e = \sqrt{(\frac{r_bv^2 _b}{GM}-1)^2 \cos^2(\varphi) + sin^2(\varphi)}$$

From the eccentricity, I can trivially calculate the semi-major axis.

What I do not know how to calculate is the argument of periapsis, $$\omega$$, of the resulting elliptical orbit. I recognize that it is a function of the original orbit's $$\omega$$ and the angular position of the burn, but I'm getting stuck coming up with the right calculation. Does anyone know of a formula to find it?

• One option that should work, but I haven't tried it, is to convert to Cartesian coordinates and back. Aug 19 '19 at 14:22

welcome to SE!

The argument of periapsis is a function of the eccentricity vector and the mean motion vector of an orbit, and is calculated based on the formula:

$$\cos(\omega)=\frac{\boldsymbol{n} \cdot \boldsymbol{e}}{|\boldsymbol{n} ||\boldsymbol{e}|}$$ subject to if $$e_{Z}<1, \implies \omega = 360^{o}-\omega$$

where the mean motion and eccentricity vectors are defined as: $$n=\sqrt\frac{\mu}{a^3}, \boldsymbol{e}=\frac{(v^2-\frac{\mu}{r})\boldsymbol{r}-(\boldsymbol{r} \cdot \boldsymbol{v})\boldsymbol{v}}{\mu}$$

Since our determiner is the cosine of the argument of periapsis, the sign of the Z-vector or third vector of the ECI frame determines where it lies.

So, you take those vectors in the central body's inertial frame, use their dot product and then normalize them by the product of their magnitudes.

There are three special cases, depending on the inclination and eccentricity of the orbit. If the orbit is equatorial but elliptical then, $$\cos(\omega_{true})=\frac{e_X}{|\boldsymbol{e}|}$$

If it is circular but inclined, then $$\cos(\omega)=\frac{\boldsymbol{n} \cdot \boldsymbol{r}}{|\boldsymbol{n} ||\boldsymbol{r}|}$$

And if it's circular and equatorial, then $$\cos(\omega_{true})=\frac{r_X}{|\boldsymbol{r}|}$$

These are standard conversions when you transform radius and velocity states to classical orbital elements and can be found in most astrodynamical books/references.