# 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. 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.

Image courtesy of Wikipedia user Wmc824 under the Creative Commons Attribution 3.0 Unported license.

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.

• Yeah, but I think it's supposed to be in degrees, but I keep getting a number like 6 million. I'm just not entirely sure what units it's in. Feb 3 '17 at 14:39
• Okay, so your saying that an answer like 6 million makes sense because it's just a point on the edge of a map? Feb 3 '17 at 15:00
• @JamesKirk No 6 million should not make sense. Since this is an angle, it should only be defined between 0 and 360 degrees. A value of 6 million just doesn't make sense. Depending on your functions and equations, you may find that you wind up with something between -90 and 90 degrees. We couldn't know how you're getting that answer though without seeing your equations and numbers. Feb 3 '17 at 17:25
• @JamesKirk What zephyr said. That's way too large. Check any conversions you might have made. Feb 3 '17 at 18:47
• Okay, I'll double check it. Feb 3 '17 at 19:11