I'm having trouble finding a satisfying answer to the following question:

To describe perfectly a Keplerian orbit, one needs 6 orbital parameters: $a$ the semi-major axis, $e$ the eccentricity, $i$ the inclination, $\omega$ the argument of periapsis, $\Omega$ the longitude of ascending node and $\nu$ the true anomaly.

We sometimes see corner plots in scientific papers discussing orbital mechanics and orbital elements predictions where $\omega$ and $\Omega$ are not represented separately but jointly as $\Omega-\omega$ and $\Omega+\omega$.

Is there a specific reason for such a representation? I can't figure out what $\Omega-\omega$ and $\Omega+\omega$ represent in terms of angular quantities.

If one of you knows, I'd be glad to hear an explanation.


1 Answer 1


A node in orbital elements is the intersection of some reference plane (e.g., the ecliptic plane) with the plane defined by the orbit of said object. The ascending node is where the intersection occurs on the northward-bound part of the orbital object's trajectory. The argument of periapsis, $\omega$, is the angle between the ascending node and the periapsis vector, $\mathbf{r}_{p}$. The longitude of the ascending node, $\Omega$, is the angle between some reference direction (e.g., First Point of Aries) and the ascending node. The true anomaly, $\nu$, is the angular position along the orbital trajectory of the object. The true longitude is then defined as: $$ \lambda = \Omega + \omega + \nu_{o} \tag{0} $$ where $\nu_{o}$ is the true anomaly at some defined reference epoch (e.g., J2000).

The longitude of the periapsis is defined as: $$ \varpi = \Omega + \omega \tag{1} $$ describes the longitude of the periapsis of an orbital body if the inclination of the orbit were zero.

From inspection, I am not sure to what the parameter $\Omega - \omega$ refers, however.

Courtesy of Wikimedia Commons


  • Fränz, M. and D. Harper "Heliospheric coordinate systems," Planet. Space Sci. 50, pp. 217--233, doi:10.1016/S0032-0633(01)00119-2, 2002.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.