I want to find the true anomaly of satellites in elliptic orbits. How can I do that with following information about elliptic orbit:

Altitude, km 543.8832

Period, sec 6114

Eccentricity 0.0727

Semi-major axis, km 7227.5

Inclination, deg 81.0192


2 Answers 2


Given the information you've provided, you can't fully determine the orbital elements but you can have a bit of a guess. For a Kepler orbit:

$$r(\theta) = {a(1-e^2) \over 1 + e \cos \theta}$$

You've given us $a$ and $e$, but converting the altitude you've supplied to an orbital radius is potentially a bit more of a faff. With a suitable model of Earth's shape and the inclination of the orbit (which you do know, at least) you might be able to manage it, but I'm not going to try.

Obviously you can rearrange the above equation to give you $\cos \theta = {a(1-e^2) \over er} - \frac{1}{e}$, and if Earth was perfectly spherical and had a nice tractable radius like 6378.1km (which it wouldn't) you could get out a true anomaly of ±57.9° but you couldn't tell which was the true true anomaly, nor which direction the orbit was in. There are some chunky error bars on that number due to the shape of the Earth that I'm not going to try and quantify.

To fully constrain an orbit you do need 6 elements, and you've supplied 5-ish (more like 4-and-a-bit, given Earth's frustrating wrinkliness and non-sphericity). Without additional information, you can't do much better.

  • $\begingroup$ Altitude is typically expressed with respect to a fictitious spherical Earth with a radius of 6378.137 km (or sometimes 6378.136 km). $\endgroup$ Feb 28 at 5:05
  • $\begingroup$ That said, two angles that are solutions. $\endgroup$ Feb 28 at 5:23
  • $\begingroup$ @DavidHammen that would certainly make this easier than I'd originally thought. I was imagining at the very least height above local sea level. $\endgroup$ Feb 28 at 9:16
  • $\begingroup$ @StarfishPrime there are some geophysical research satellites which use radar altimetry to measure things like sea surface temperature and ocean roughness, and thus must worry about such details, but they are very unusual. Normally "altitude" signals that we are about to talk only approximately, like "assume a spherical cow." 6378 is a good reference value for orbits with near zero inclination, but the average value for the whole earth is more like 6371. moral of the story, when people talk about satellite altitude, mentally add at least +/- 10 km uncertainty. $\endgroup$
    – Ryan C
    Feb 28 at 23:27

There's not enough information in your list to compute the true anomaly. The true anomaly reflects an angular position of an object along its orbit. The parameters in the question define the shape of the orbit, but not a location along the orbit.

  • $\begingroup$ can you give an numerical example? $\endgroup$
    – u1997
    Feb 27 at 17:33
  • $\begingroup$ @u1997 It's not a matter of analytical or numerical. There's just not enough information to say where along the orbit. $\endgroup$ Feb 27 at 17:36
  • $\begingroup$ Which other information needs so I can calculate true anomaly. $\endgroup$
    – u1997
    Feb 27 at 17:45
  • 2
    $\begingroup$ @u1997, for example: state vectors, eccentric anomaly, or mean anomaly would work. $\endgroup$ Feb 27 at 18:05
  • 1
    $\begingroup$ @u1997 An orbit is like a bus route. five of the six classical Keplerian elements (semi-major axis, eccentricity, inclination, right ascension of the ascending node, and argument of perigee) describe the size and shape of the orbit, but the object may be at any point along it. To use it, you have to know both when the bus starts its route (often perigee), and what time it is now. at a different time, it will be at a different place. true anomaly says how far along the route it is, and you need Kepler's equation to convert between that angle and time. $\endgroup$
    – Ryan C
    Feb 28 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.