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?
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?
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.
Omega=atan2(N_y,N_x).
– David Hammen
Jan 25 '21 at 06:47
ifstatement to handle cases where the result of the inverse cosine calculation needs to be adjusted. – David Hammen Jan 25 '21 at 06:58