1
$\begingroup$

From what I understand, the semi-major axis of a circular orbit should be equal to its radius.

However, checking Wikipedia's info on Hubble, which is in a nearly circular orbit, I find :

sMA $= 6,919$ km

apogee $\approx$ perigee $\approx 540$ km

I'm having trouble understanding that difference.

Assuming I approximate an orbit as being circular, how can I get its radius from the semi-major axis?

Screenshot from https://en.wikipedia.org/wiki/Hubble_Space_Telescope

enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ You are right, there is a problem with the table. $\endgroup$
    – uhoh
    May 3, 2017 at 10:11

1 Answer 1

Reset to default
6
$\begingroup$

The Wikipedia article on Hubble provides the apogee and perigee height with respect to the Earth's surface not its center. As a consequnce the semi-major axis should be equal to the sum of the Earth's radius and the perigee or apogee respectively (of course only for cirular orbits).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Except that Earth isn't a sphere, and only approximately an ellipsoid. Your answer is almost right, but I'd change "...should be equal to..." to "would be equal to... if the Earth were a sphere." Then perhaps point out that it's not obvious from this table what value should be added. 6378 km? 6371 km? Something else? $\endgroup$
    – uhoh
    May 3, 2017 at 10:09
  • $\begingroup$ Note the question asked: "Assuming I approximate an orbit as being circular, how can I get its radius from the semi-major axis?" The semi-major axis is already stated in the Wikipedia article. Instead of "the semi-major axis should be..." perhaps you mean "the radius at perigee and apogee should be..." $\endgroup$
    – uhoh
    May 3, 2017 at 10:16
  • 4
    $\begingroup$ For these types of calculations, you use (by convention -- don't ask me why) earth's equatorial radius of 6,378 km. You'll then see that the semimajor axis is just equal to this number added to the arithmetic mean of the apogee and perigee altitudes, i.e., $6378 + \frac{1}{2}(539 + 543) = 6919$ $\endgroup$
    – Tristan
    May 3, 2017 at 14:45
  • 1
    $\begingroup$ @Tristan Thanks, I was just about to ask a separate q about that. $\endgroup$
    – Russell Borogove
    May 3, 2017 at 15:49
  • $\begingroup$ @RussellBorogove the question/answer structure of SE is the way helpful information is validated and recored so it is easy for people to find in the future, and for people here to link back to in future answers. Linking back to a hard-to-find comment isn't as reliable. I think asking it would be great! In fact, it might be a bit of a challenge to find a reliable reference for the answer, but it being found it would be valuable as well. $\endgroup$
    – uhoh
    May 4, 2017 at 2:39

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.