# What is exactly the “longitude of the perigee”

Some moon phase calculation algorithms (apparently derived from Duffet-Smith's book, example here) seem to use a parameter called "longitude of perigee at epoch". What exactly is this?

Can I assume this is the same as the longitude of the periapsis? Can I just calculate it adding the longitude of the ascending node and the argument of periapsis together?

Unfortunately, all the data that I find on this parameter is either related to Duffet-Smith's book or algorithms based on it. So, I have no idea on how this can be calculated for a specific epoch.

• Thanks for the answer. I get that "perigee" and "periasis" mean the same in this context, and I understand the concept of epoch. I don't get however what this "longitude" actually is. I suspect it's $\Omega + \omega$, but would like to confirm. I'm also curious on how you can just add the two angles together even though they may be on different planes. – ubik Aug 5 '15 at 21:16
• @ubik -- That's exactly how it's defined, as $\Omega+\omega$, with a couple exceptions, which are circular orbits (where $\omega$ is ill-defined), and equatorial orbits (where $\Omega is ill-defined). In both of those cases, longitude is still well-defined. The only exception is orbits inclined by 180 degrees, and there are no such beasts in the solar system that we know of. – David Hammen Aug 6 '15 at 1:02 • Great, thanks! Maybe it's better if you make that clearer in the answer before I accept the question? – ubik Aug 6 '15 at 16:45 • @ubik - Answer updated. – David Hammen Aug 6 '15 at 17:07 Assuming that wikipedia formulas are right (didn't check), here it is how to pass from "compound angle" (sum of angles on different planes) to "pure angle" (both angles on ecliptic plane): $$ϖ = \omega + \Omega \text { (compound angle)}$$ $$\tan ϖ = \frac { \sin α \cos ε + \tan δ \sin ε} {\cos α}$$ $$ϖ = \arctan \left (\frac { \sin \alpha \cos \epsilon + \tan \delta \sin \epsilon} {\cos \alpha} \right ) \text { (pure angle)}$$ i, inclination ω, argument of perihelion Ω, longitude of ascending node ε, obliquity of the ecliptic (i.e. rotation axis inclination)  $$A = \cos ω \cos Ω – \sin ω \sin Ω \cos i$$ i=0 : $$A = \cos ω \cos Ω – \sin ω \sin Ω$$ $$B = \cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega \cos i) – \sin ε \sin \omega \sin i$$ i=0 : $$B = \cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega)$$ $$C = \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega \cos i) + \cos ε \sin \omega \sin i$$ i=0: $$C = \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega)$$ i=0 $$\alpha = \arctan \left ( \frac B A \right ) = \arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right )$$ $$\delta = \arcsin C = \arcsin \left ( \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) \right )$$ Hence: $$ϖ_{i=0} =$$ $$= \arctan \left (\frac { \sin (\alpha) \cos (\epsilon) + \tan (\delta) \sin (\epsilon)} {\cos (\alpha)} \right )$$ $$= \arctan \left (\frac { \sin (\arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right ) ) \cos (\epsilon) + \tan (\arcsin \left ( \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) \right )) \sin (\epsilon)} {\cos (\arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right ) )} \right )$$ As it is usually said in such cases, "reader can easily verify" ;-) that $$ϖ_{i=0} \approx \omega + \Omega$$ (Actually a comparison chart would be needed...) • when use TeX or MathJaX, if you write \tan instead of tan it looks much much better! – James K Aug 24 at 15:48 • @JamesK much better, thanks – jumpjack Aug 24 at 15:57 • I wonder if Wikipedia data are actually right: isn't orbit inclination i the same of axis inclination$\epsilon $, at lest for Earth? Hence i=0 would mean$ \epsilon = 0 $and much simpler formulas$ \omega = \alpha = \arctan (\frac {\cos \omega \sin \Omega + \sin \omega \cos \Omega} {\cos \omega \cos \Omega – \sin \omega \sin \Omega }) \$ – jumpjack Aug 24 at 16:03