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.


Perigee is the Earth-specific name for periapsis. People use longitude (which is a composite angle rather than an angle) because this solves the problems of circular and equatorial orbits.

The reason you need to use an epoch time to specify the Moon's argument of perigee and longitude of ascending node is because the Moon's orbit about the Earth precesses. Neither the argument of perigee nor the longitude of ascending node is constant. Instead, they are functions of time (and hence, so is longitude of perigee).

In the case of an inclined, non-circular orbit, "longitude" is the sum of the right ascension of the ascending node (which is measured on the fundamental plane of the reference system) and some angle (or angle-like measure such as mean anomaly) on the orbital plane. Thus longitude of periapsis (or longitude of perigee in the case of an object orbiting the Earth) is the sum of the object's right ascension of ascending node and it's argument of periapsis.

The use of "longitude" as orbital elements are for two reasons: To be able take advantage of Hamiltonian mechanics (this was done by Delaunay, resulting in the Delaunay orbital elements), and to address problems related to orbits with very small inclinations and/or very small eccentricities (this was done by Poincaré, resulting in the Poincaré orbital elements). Anomaly and longitude are synonymous in the case of an orbit with zero inclination.

While it is invalid to add two angles on different planes and treat the sum as if it were an angle, it is not necessarily invalid to treat that sum as what it is, a composite angle or dogleg angle, which is exactly what the Poincaré orbital elements do. ("Dogleg angles" aren't angles. Analogous terms include "dwarf planets", which aren't planets, and "red herrings", which oftentimes are neither red nor herrings.)

  • $\begingroup$ 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. $\endgroup$ – ubik Aug 5 '15 at 21:16
  • $\begingroup$ @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. $\endgroup$ – David Hammen Aug 6 '15 at 1:02
  • $\begingroup$ Great, thanks! Maybe it's better if you make that clearer in the answer before I accept the question? $\endgroup$ – ubik Aug 6 '15 at 16:45
  • $\begingroup$ @ubik - Answer updated. $\endgroup$ – 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 :

$$ B = \cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) $$

$$ C = \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega \cos i) + \cos ε \sin \omega \sin i $$


$$ C = \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) $$


$$ \alpha = \arctan \left ( \frac B A \right ) = \arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right ) $$

$$ \delta = \arcsin C = \arcsin \left ( \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) \right ) $$


$$ ϖ_{i=0} = $$

$$ = \arctan \left (\frac { \sin (\alpha) \cos (\epsilon) + \tan (\delta) \sin (\epsilon)} {\cos (\alpha)} \right ) $$

$$ = \arctan \left (\frac { \sin (\arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right ) ) \cos (\epsilon) + \tan (\arcsin \left ( \sin ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) \right )) \sin (\epsilon)} {\cos (\arctan \left ( \frac {\cos ε (\cos \omega \sin \Omega + \sin \omega \cos \Omega) } {\cos ω \cos Ω – \sin ω \sin Ω } \right ) )} \right ) $$

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...)

  • 1
    $\begingroup$ when use TeX or MathJaX, if you write \tan instead of tan it looks much much better! $\endgroup$ – James K Aug 24 '20 at 15:48
  • $\begingroup$ @JamesK much better, thanks $\endgroup$ – jumpjack Aug 24 '20 at 15:57
  • $\begingroup$ 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 }) $ $\endgroup$ – jumpjack Aug 24 '20 at 16:03

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