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Having found the right ascension of the node, for example,

$$ \begin{align*} \boldsymbol{\Omega} &= cos^{-1}\frac{N_X}{N} \\ &= \underline{15.60^\circ} \end{align*} $$

How do we know in which quadrant does this lies?

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  • $\begingroup$ Can you add the definition of the terms $N_X$ and $N$ and cite the source of that equation? Thanks! $\endgroup$
    – uhoh
    Jan 24 at 23:22
  • $\begingroup$ @uhoh Those are respectively the X component and the magnitude of the cross product between the unit vector pointing along the positive Z axis and the specific angular momentum vector, which in turn is the cross product between the position and velocity vectors. $\endgroup$ Jan 25 at 6:50
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    $\begingroup$ @uhoh The formula posted in the question is all over the internet, and also in many textbooks. I don't understand why the much cleaner two argument arctangent calculation isn't widely taught. That simple one-liner (see my answer) eliminates calculating an unneeded vector ($\vec N$), eliminates calculating the magnitude of that unneeded vector ($N$), and finally eliminates an if statement to handle cases where the result of the inverse cosine calculation needs to be adjusted. $\endgroup$ Jan 25 at 6:58
  • $\begingroup$ @DavidHammen you may be able to find it all over the internet because you know what you are looking for, but I and likely many other readers won't know what it is or where it comes from. Is there any way to get at least a hint please? Does it have a name? $\endgroup$
    – uhoh
    Jan 25 at 14:21
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    $\begingroup$ @uhoh I don't think it has a name. You can find it by searching for "Cartesian to Keplerian orbital elements." $\endgroup$ Jan 25 at 14:53
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Don't use that formula. Instead use the two argument arctangent function.

The specific orbital angular momentum vector, $\vec h$, is the cross product of the position vector $\vec r$ and velocity vectors $\vec v$. After too much mathematical drudgery to reproduce here, the result of this cross product is $$\vec h = r^2\dot\theta \left( \sin\Omega \sin I \hat x -\cos\Omega \sin I \hat y +\cos I \hat z \right)$$ There is no reason to introduce the vector $\vec N$ defined by $$\vec N \equiv \hat z \times \vec h = r^2\dot\theta\left(\cos\Omega \sin I \hat x + \sin\Omega \sin I \hat y\right)$$ Simply use the specific orbital angular momentum vector $\vec h$ and the two argument arctangent function. Note that $\frac{h_x}{-h_y} = \tan\Omega$. The $r^2\dot\theta\sin I$ term cancels in this division, and each term is non-negative. With most computer languages, the two argument arctangent function is of the form atan2(numerator,denominator). Denoting the $x$ and $y$ components of the specific orbital angular momentum vector as h_x and h_y, one can thus use Omega=atan2(h_x,-h_y).

Note very well: some languages and most spreadsheets reverse the arguments to their implementations of the two argument arctangent function, in which case you'll have to use Omega=atan2(-h_y,h_x). Note also that some languages and most spreadsheets use a name other than atan2. But the function will still be there. That function is far too useful to not be there. If the function doesn't exist in your tool of choice, choose a different tool.

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  • $\begingroup$ If you feel compelled to introduce the vector $N$, even though there is no reason to do so, you can still use the two argument arctangent function with Omega=atan2(N_y,N_x). $\endgroup$ Jan 25 at 6:47
  • $\begingroup$ Thank you for your detailed answer! I accepted it since it provides really valuable information, but it didn't quite answer my question. My question was: when we obtain the degrees (regardless of which function we use), how do we conclude in which quadrant it lies? $\endgroup$ Jan 30 at 20:25
  • $\begingroup$ @lawndownunder The range of the inverse cosine spans 180 degrees, which means one must look at other characteristics to determine the quadrant. On the other hand, the range of the two argument inverse tangent spans 360 degrees. There is no need for extra information because the two argument inverse range explicitly tells you the quadrant. $\endgroup$ Jan 30 at 23:00

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