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I want to compute the derivatives of argument of periapsis and longitude of the ascending node of the orbit of a GPS satellite from the following formula.

$$\frac{d\Omega}{dt} = -K \cos{i} \\ \frac{d\omega}{dt} = K ( 2 - 2.5 \sin^2{i}) \\ K = \frac{nK_1}{a^2(1-e^2)^2}$$

But what is $K_1$?

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  • $\begingroup$ Okay, I found the solution by myself. $K_1$ is a constant describing the flattening of the earth $ K_1 = 66063.1704*10^6 [m^2] $ $\endgroup$
    – Roman Podolski
    Jan 31, 2014 at 15:18
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    $\begingroup$ The OP has answered the question themselves $\endgroup$
    – John Rennie
    Jan 31, 2014 at 15:28
  • $\begingroup$ @RomanPodolski could you post your answer using the text box below, under "Your Answer"? It's bad etiquette here to edit the answer into the question, so I've removed it, but you're welcome and encouraged to post it as an actual answer. (P.S. apparently you have to wait a few hours until the system will let you do so) $\endgroup$
    – David Z
    Jan 31, 2014 at 18:13
  • $\begingroup$ @JohnRennie: true, but I don't believe that itself is a valid reason to close a question. $\endgroup$
    – David Z
    Jan 31, 2014 at 18:14
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    $\begingroup$ @RomanPodolski Indeed, soon enough you should be able to self-answer, and this is encouraged :) $\endgroup$
    – user10851
    Jan 31, 2014 at 18:21

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I found the solution by myself. $K_1$ is a constant describing the flattening of the earth $$K_1 = 66063.1704 ∗ 10^6 \space[m^2]$$

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