# What is $\phi '$ in orbital mechanics?

For the last week or so, I have been teaching myself orbital mechanics within the context of Braeunig's Rocket and Space Technology.

I noticed a symbol, $\phi '$, and was wondering what context that was used in? I think it is an angle measurement, but I know that it is almost equal to the semi-major axis on earth.

From the box labeled Geodetic Latitude, Geocentric Latitude, and Declination, found about half way down after selecting Orbital Mechanics here. • No problem. Welcome to Astronomy Stack Exchange! By the way, questions about rockets and space technology are usually the domain of Space Exploration Stack Exchange, but since this one is about generic orbital mechanics it is probably ok. – called2voyage Feb 3 '17 at 14:46
• In mathematical notation in general, a prime could be used to indicate a different reference position, or coordinate system, or a time derivative, or something else. Can you at least do a screen shot of how it's used there, or link to the particular place where you find it - it's hard to answer without more information, and if there were an answer wouldn't be so useful for anyone else reading this later. – uhoh Feb 4 '17 at 10:32
• Ah, it's in the comment. Didn't see it there. I'll move it to the question. I've added the figure, adjusted the linking. This can help make questions easier to answer. In this case you can see it says "The geocentric latitude, $\phi$' is the angle between..." right there on the page you were looking at. – uhoh Feb 4 '17 at 13:37

This appears to be the geocentric latitude ($\psi$), which is the angle between the equatorial plane and the point on the surface of the ellipse. It can be calculated from the geodetic latitude (also known as the geographic latitude) ($\phi$) and the eccentricity of the ellipse ($e$): $$\psi(\phi)=\tan^{-1}\left(\left(1-e^2\right)\tan\phi\right)$$ If $e=0$, then the two lattitudes are the same, because the focus is at the center of the ellipse. It might make more sense to use this latitude to describe the position of an orbiting object, but the vector from the focus to the object's position $(r,\theta)$ is perpendicular to the tangential component of the object's velocity vector, meaning that the geodetic latitude may be better.