# How to get semi-major axis from TLE?

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?

• 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 Jun 23, 2020 at 17:17

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 $$a=6768.16 \,\, km$$.
• In this answer, $u$ refers here to $\mu$, the Standard gravitational parameter ($3.986004418 ×10^{14} m^3 s^{−2}$ for the Earth) Oct 10, 2017 at 15:16