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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?

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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: $a=\frac{u^{1/3}}{\frac{2n\pi}{86400}^{2/3}}$.

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. $\endgroup$ – pericynthion Sep 27 '16 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 '17 at 15:16

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