# Meaning of "equatorial radius in an orbital plane"

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1. Dec 15, 2015

### M.T

I wish to solve the inverse geodesic problem numerically using http://geographiclib.sourceforge.ne...odesic.html#a455300c36e6caa70968115416e1573a4, and to finish off I need to specify the "equatorial radius". I am not too familiar with this, and do not see immediately what I would define as "North" or "South" in the orbital plane.

Does the "equatorial radius" here refer to the semi-major axis, or perhaps the apocenter or pericenter?

2. Dec 16, 2015

### Buzz Bloom

Hi MT:

I noticed that no one has responded to your question for about a day, so I thought I would try to give a useful reply based on my my limited knowledge.

I do not understand the tool you linked to, so this is just a guess. Perhaps the tool takes into account the possibility that the central mass effecting geodesic calculations is not radially symmetric. In that case the equatorial radius may refer to the maximum radius of an ellipsoidal central mass, like the Earth's equatorial radius is larger than its polar radius.

Hope this helps.

Regards,
Buzz

3. Jan 5, 2016

### M.T

Hi Buzz Bloom

The central mass is assumed ellipsoidal, so I agree with you in that it does not have to be radially symmetric. I guess it also makes sense to define the equator along the largest "diameter", that is $2a$ with $a$ being the semi-major axis, but is the radius then (1) $a$ (distance from centre of ellipsoid) or is it (2) $c+a$ with $c$ being the distance from the foci where eg. the Earth is; $c-a$ being the shortest distance between the Earth and the ellipsoid orbit, and $c+a$ the longest distance.

would I be right then in assuming you mean the distance from the centre (1)?

Thanks,
MT

4. Jan 5, 2016

### Buzz Bloom

Hi MT:

Yes, except that no mean is necessary, since for an ellipsoid all points on the equator are equally distant from the center.

Regards,
Buzz