# Optimal change of argument of periapsis?

If I want to rotate an eccentric orbit around the central body - retain orbital plane, retain apoapsis and periapsis altitudes, but have the orbit rotated in its orbital plane - change the argument of periapsis - what is the optimal maneuver to that end?

I know an easy way to achieve this effect is performing a radial burn (towards the center of the central body) at periapsis, at thrust such that the craft retains altitude, against centripetal acceleration; moving in circular path around the body; "dragging the periapsis along" - the moment the engines are cut off, it enters the new trajectory. I'm also aware this method may be awfully costly, especially for highly eccentric orbits and large changes of argument of periapsis.

Another method is circularizing the orbit at apoapsis, and then returning to desired eccentricity back upon achieving the desired argument of periapsis. This one has a fixed cost, which will be excessive in case the orbit is very eccentric and the desired shift in angle is small.

There's also a method involving only tangential burns (pro/retrograde) at various points of the orbit, but I only have a rough hunch of how it works, no good solid recipe.

Is there an universal strategy to optimally perform this change?

This is an expensive operation. Suppose $\Delta \omega$ is the angle by which you wish to change the argument of periapsis. The instantaneous delta V needed to perform that optimal change is $$\Delta v = 2\sqrt{\frac{\mu}{a(1-e^2)}}\,\sin\left(\frac {\Delta \omega} 2\right)$$ Note that this is very similar in form to the $\Delta v$ needed to change the inclination by an angle $\Delta i$.
• Hmm, I think there's also a $e$ factor missing in the formula here. To change the argument of periapsis by angle $\Delta \omega$, one needs to reverse the radial component of the velocity at true anomaly $\Delta \omega/2$ and these equations in Wikipedia (and my calcluations too long to fit here) say that $\dot{r} = \sqrt{\mu/p}e \sin(\theta)$ where $p=a(1-e^2)$ and $\theta$ is the true anomaly. Then $\Delta v$ is $2\dot{r}$ at $\theta=\Delta \omega/2$. – JiK Oct 14 '17 at 19:34