# What is the eccentricity of a near equatorial LEO?

I am trying to do some calculation on the 2-body problem but I am stuck as I need either the eccentricity or the periapsis of the orbit. I already have the orbital period which I assume as ~127 min as it is an LEO but that only gives me the semi-major axis. The rest of my equations (obtained from vis-viva equation) requires the eccentricity which I don't have.

I don't think it is possible to obtained the eccentricity from the period and thus, I think my only solution is to obtain from observational data. However there is not much info on equatorial orbit out there!

• Equatorial just refers to the inclination of the orbit- ie. it’s 0 or 180. A geostationary orbit is equatorial, as is a highly elliptical orbit with a low perigee.
– Jack
Aug 19 '18 at 7:47
• You’ll find that with LEO, the eccentricity must be close to zero otherwise the perigee will be below the Earth’s surface! Beyond that, you can just choose the eccentricity to get the apogee and perigee that you like!
– Jack
Aug 19 '18 at 7:52
• If you set orbital period 127 minutes than you can even calculate maximum possible excentricity (from the demand of periapsis not be lower than 200 km up to Earth surface). If you solve that for minimal eccentricity=0 and for maximum possible eccentricity, you will see if they are much different or not. Aug 19 '18 at 10:36
• I don't understand your question. 127 minutes gives $a$ ~ 8370 km using GMe = 3.986E+14 m^3/s^2. Like the other comment(s) it can be a circle, or any ellipse that stays above 200 km altitude. There is no one, single eccentricity. The eccentricity is pretty much independent of the period or semimajor axis. Your periapsis is any value between 6378 + 200 km and $a$. You can choose both of those limits, then use $e = 1-(r_{peri}/a)$ en.wikipedia.org/wiki/Apsis and get any $e$ between 0 and about 0.214.
– uhoh
Aug 19 '18 at 17:31