This should be fairly straightforward, but, how do I get the semi-major axis (a) from at TLE.

For example if I have the TLE:

1 25544U 98067A 01260.91843750 .00059354 00000-0 74277-3 0 4795
2 25544 51.6396 342.1053 0008148 106.9025 231.8021 15.5918272116154

I know that 15.5918272 is the mean motion of the body ($n$). I also know that $n = \sqrt{\frac{\mu}{a^3}}$. If I use the given $n$ value I get a semi-major axis $a=11.79$ which is obviously incorrect. What am I missing here?

  • 1
    $\begingroup$ According to CSpOC, their TLE must always be used with the SGP4/SDP4 propagator; see this link (login required): "SGP4 is an analytic method based on a general perturbation theory for generating ephemerides for satellites in earth-centered orbits. It is the proper means for correctly propagating a USSPACECOM 18th Space Control Squadron (18 SPCS) Two Line Element (TLE).". See satprobe.altervista.org/calc_sat.html $\endgroup$ Jun 23, 2020 at 17:17

1 Answer 1


The TLE gives mean motion ($n$) in $\frac{rev}{day}$. This needs to be converted to $\frac{rad}{s}$ which can be accomplished by multiplying the $n$ TLE value by $\frac{2\pi}{86400}$.

Therefore, to go directly from $n$ in TLE to the semi-major axis $a$. We can use the following formula: $a=\frac{\mu^{1/3}}{\frac{2n\pi}{86400}^{2/3}}$.

For $n=15.5918272 \,\, \frac{rev}{day}$, we get $a=6768.16 \,\, km$.

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    $\begingroup$ Bear in mind that the TLE is given in "mean elements", not osculating elements. So the resulting semimajor axis may not exactly be physically correct, depending on what you're trying to do with it. It will be within a couple of percent though. $\endgroup$ Sep 27, 2016 at 16:16
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    $\begingroup$ In this answer, $u$ refers here to $\mu$, the Standard gravitational parameter ($3.986004418 ×10^{14} m^3 s^{−2}$ for the Earth) $\endgroup$
    – Covich
    Oct 10, 2017 at 15:16

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