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!

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    $\begingroup$ 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. $\endgroup$
    – Jack
    Commented Aug 19, 2018 at 7:47
  • $\begingroup$ 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! $\endgroup$
    – Jack
    Commented Aug 19, 2018 at 7:52
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    $\begingroup$ 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. $\endgroup$
    – Heopps
    Commented Aug 19, 2018 at 10:36
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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Commented Aug 19, 2018 at 17:31

1 Answer 1


From this Wikipedia page https://en.wikipedia.org/wiki/Low_Earth_orbit LEO is defined at its maximum loosely as being less than 2000km. As other answers have mentioned, you could use that maximum value as how far your apogee may go before the oribt is no longer LEO, and use the minimum value of being outside the Earth's atmosphere as how low your perigee may be and use the semi-major axis you found in combination with these values to work out your eccentricity. This will work out as your maximum eccentricity allowable, so you know that your eccentricity can be between 0 and that. That's the best you can do with that info. Afterward, just assume an eccentricity inside that range, and if you have any reason for being on an ELEO (eg observation) pick a similar currently operational satellite that has the same application with an eccentricity inside your interval, and that can be your justification for choosing that eccentricity.


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