<p>You want to know the range of RA where the Moon's orbital plane crosses the celestial equator from south to north, so we need to find the line of intersection of those two planes.</p>
<p>The vector corresponding to the line of intersection of two planes equals the <a href="https://en.wikipedia.org/wiki/Cross_product" rel="nofollow noreferrer">cross product</a> of the normals of the planes. We can find those normals using <a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">rotation matrices</a>. The relevant parameters are given relative to the ecliptic plane, so it's convenient to work in that coordinate system. I'll use a right-handed coordinate system, treating the celestial sphere as a unit sphere, so our planes of interest are represented as circles of radius 1.</p>
<p>The <a href="https://en.wikipedia.org/wiki/First_Point_of_Aries" rel="nofollow noreferrer">First Point of Aries</a> is the point on the celestial sphere where the Sun is found at the March equinox, when it crosses the celestial equator from south to north, so it's a point where the celestial equator &amp; ecliptic intersect. In our system it's on the X axis with Cartesian coordinates <span class="math-container">$(1,0,0)$</span>. Both ecliptic longitude and RA equal zero at this point.</p>
<p>We have the Moon's inclination <span class="math-container">$i$</span>, which is approximately 5.14°, but it ranges from roughly 4.98° to 5.30°. The obliquity of the ecliptic <span class="math-container">$\varepsilon$</span> is currently a little over 23.436°, but we need to negate that to get the inclination of the equator relative to the ecliptic. The final lunar parameter we need is <span class="math-container">$\omega$</span>, the longitude of the Moon's ascending node, which ranges from 360° to 0° over the course of the lunar precession cycle.</p>
<p>To find our plane normals, we start with a circle in the XY plane, so its normal is the <span class="math-container">$Z$</span> vector <span class="math-container">$(0, 0, 1)$</span> then we rotate it around the X axis by its inclination, using matrix multiplication. To precess the lunar plane we then rotate it around the Z axis by <span class="math-container">$\omega$</span>. This leads us to the following equation for the vector of the line of intersection:</p>
<p><span class="math-container">$$P = (Rx(\varepsilon)Z) \times (Rz(\omega)Rx(i)Z)$$</span>
where <span class="math-container">$Rx(.)$</span> &amp; <span class="math-container">$Rz(.)$</span> are rotation matrices.</p>
<p>Generally, <span class="math-container">$P$</span> is <em>not</em> a unit vector, but we can easily normalize it so that does have unit length, if necessary. Bear in mind that our XY plane is the ecliptic. We can rotate <span class="math-container">$P$</span> around the X axis by <span class="math-container">$-\varepsilon$</span> to determine its coordinates in the equatorial plane.</p>
<p>I won't do all the algebra here, but the equatorial coordinates of <span class="math-container">$P$</span> are</p>
<p><span class="math-container">$$\begin{align}
x &amp; = \cos(\varepsilon) \cos(\omega) \sin(i) - \sin(\varepsilon)\cos(i)\\
y &amp; = -\sin(\omega)\sin(i)\\
z &amp; = 0
\end{align}$$</span></p>
<p>The RA <span class="math-container">$\alpha$</span> corresponding to P is just
<span class="math-container">$$\alpha=\tan^{-1}\left(\frac yx\right)$$</span></p>
<p>Finding the <span class="math-container">$\omega$</span> corresponding to the maximum &amp; minimum of <span class="math-container">$\alpha$</span> involves some tedious trigonometric calculus and algebra. Fortunately, the results are fairly simple. :)</p>
<p><span class="math-container">$$\begin{align}
\omega &amp; = \cos^{-1}(\tan(i) / \tan(\varepsilon))\\
\alpha &amp; = \sin^{-1}(\sin(i) / \sin(\varepsilon))
\end{align}$$</span></p>
<p>Hopefully, I haven't messed up a minus sign somewhere while transcribing all those equations. ;)</p>
<p>Finally, here are some actual results. Using the value for <span class="math-container">$\varepsilon$</span> mentioned earlier,</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">inc</th>
<th style="text-align: center;">RA</th>
<th style="text-align: center;">omega</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">4.98</td>
<td style="text-align: center;">-12.606952</td>
<td style="text-align: center;">101.596318</td>
</tr>
<tr>
<td style="text-align: center;">5.14</td>
<td style="text-align: center;">-13.017906</td>
<td style="text-align: center;">101.976326</td>
</tr>
<tr>
<td style="text-align: center;">5.30</td>
<td style="text-align: center;">-13.429439</td>
<td style="text-align: center;">102.357061</td>
</tr>
</tbody>
</table>
</div>
<p><a href="https://i.sstatic.net/Gnpmd.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Gnpmd.png" alt="RA vs omega, for 3 inclinations" /></a></p>
<p>Here's the single plot for <span class="math-container">$i=5.14°$</span>
<a href="https://i.sstatic.net/YUOlC.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YUOlC.png" alt="RA vs omega for i=5.14°" /></a></p>
<p>And here's the <a href="https://sagecell.sagemath.org/?z=eJyNU21v2jAQ_p5fcco-4HRJFqCtpkqWth8wCVX7RBUhNzmCp8RmtgNpf_3OTkBANakIwp3v8T3PveQLrFrt4Pkn6C3UWPECDhZ0h42IDsKwmZslUY0NcNhL-Abz70W0JueAldOGsSIF-s6TBCKCbeF5YG6XPEVAH4OuNwo64YwcWDh6CU_6n4d7RZmeDsirtPWXU8isVN66ik5nZ1gZYuWZ-P1zxB9IrlTcktxK9LVeEP-QyqERlQsSNkyqij_k8_sUNPJssczvl4-TJgrBHacWN8HVOHmUxft7Q6lYFMfxOAq3Q9hrOoPjDg0G_5fWakbDMa_SQWW0tWhDoMIWrZOiBfzbC5pLGgkLAra9qpzUyicMI00DHKtW7p2soNWqka6v8cQ4MQhboaqlakDpGnOvKpl0GkHDV0z4RvkflZXQWnhTY-JNqikJ0OOIDJ0UZ6Q3b5Fj8b7yVnbS2TglnquQcGMBQKHjSUvYT93NRuSQwlsKGyJd56E5m73RdV85xmgpM-KEO78lme6CNQRF3lwnudKmE618x5qN2XzrKJUguQtGeYdJ8ES98u8DvTjM45juuKM8vp4UmAtrsnwskmTCfuXjLC17YcdQG8EIABmQm5FfXkBbqXBRE7S4gAaz9HvZasNjgzV1wiOte2uRx0_xfzJkFymyz-WwO31kqxS2RnTIf5seU2iMrAOUx534ow3dFAPaTStesbUbS63j83z-cHXMX-IwNQLTcOMy-QfoyTkV&amp;lang=sage" rel="nofollow noreferrer">Sage / Python script</a> that made that plot.</p>
<p>Lastly, here's an <a href="https://sagecell.sagemath.org/?z=eJx9VeuKIkcU_u9THHogVpvWsR1mGBYashACgSyYZWFlhtmhbEutTFtVW1Xt6Ia8U_YV9sly6ta2JhtRuy7n-p3vnM6yDH5rBdUg5IqB0qxmxnApYMV2EgYDwM-HLYOGCwZyDVxYpg2rrZPBvXV33sAPwD631ErNaQOqoYIZr82NF6q1NAYdyFVb26jJnVu9o42Jnj5qbi0TsDzC_B3M3nOxgdl0Vk7KcjKbDjKMdjBYsQ1UoDhcQ3k_HUjttnsMSWpCpgW4b54PFr3Tsjt9-Jdsmedo9ArmVNMds5rXcPMztIJbqLmuGwYU49V8wwW6XoMKp2Q0enmlemPyNz70PdVkaIe532hmWy1AdSafVSPtzYqQWhpi8wIMF_45xZ8LZDZSHJdO7FlJRNlU5RTPOy8-xrdgeWPZKkX2yu0WY7NbuZECYacHBN0F6e5JXXjxAmijtjTGOXfYxRT0ZlnLRuqqDmHzdRANkkH6xyo6Iwg0olXASStarvx_AWveNNUH3bJ80Ok63mDej-OHAh6e-sr_JbQoYPE9IY8KSi3OIzD8C6vuz2CfT7S01LIFGbv8HXQ_ed7S2np0ngkXNeJbgGTV-AafxjJlKrcKGZXX9wWwuuHK8trnVCR6x53CNlCuD_YsntDWyudWrdBz9QtymkXE0ReMKmyojd9KFneR81exAaVecptax9-4cwjwk23LyHRSIkXQ3Bnu-YWdy0YO1Qu92BawdxaRhGgm0tCt_O3v0NW6hRFgxfYFBKCzZUPrlyzJnUgx_p5kjAoplWA8scobSCSMale3t7dZR8OI9YVKD4m73BXvf4F4_xaJtePWwOuWuYGGcQjqB1dVwdTLHjHlhAHOE7eUkb4Hd_VZW1LCGI6fZr3cExUfyaGAY2xiXI79-qmDQrNVFhl6l59ImTwojWbIOsNANRUb9ga-ff1Ty1asCF0aQl00RxcWkgU93OV_ffs766tmcof2BaF-qtAuDbdEJ0k1AbN208hgWo9PYS-RdTskFBjvP82h8H_teyLvz4LKpe_Rjck8ELlL1gP07-gL8_M-VhC2ODq_SIHMPjN06lEM9MzCrxdvmPB28V1xesd08odIZ2dlFFY7t-qKOk5Fjdcd21P5x_7-TKm714kDh08zzLxjgS-BxUtyQIw1kiA8pnkvyd5g86NT2RMzNpoxkV1IJ1opeyEXOVTOEt-nk9uTbijrhCrFkDnzhCYVfIcAk3CdT9IExIFF_KhCSI9hXE_MVr4GwTC5Cvei_iPUoBr2Rt3QNWdvDwylYRheQZqqLa-H-T80tG78&amp;lang=sage" rel="nofollow noreferrer">interactive anim</a> that demonstrates that the above algebra actually works. ;)</p>
<p>The large black circle is the ecliptic. The blue plane is the celestial equator, the orange plane is the lunar orbital plane. The <code>alpha</code> parameter controls the transparency of the lunar &amp; ecliptic planes; set it to zero if you want unfilled circles for those planes. The blue ball on the equator is the First Point of Aries. The orange ball on the lunar plane is the Moon's north node. Both of those balls sit on the ecliptic. The green line and ball show the intersection of the equator &amp; lunar plane, and the small red balls show the maximum travel of the green ball.</p>
<p>By default, the anim uses -30° for <code>oe</code> = <span class="math-container">$\varepsilon$</span> and 10° for <code>inc</code> = <span class="math-container">$i$</span> to make it a little easier to see what's happening. You can change those values to whatever angles you want, but <code>inc</code> should be positive and <code>oe</code> should be negative, otherwise north nodes &amp; south nodes get swapped. Also, the absolute value of <code>inc</code> <em>must</em> be less than the absolute value of <code>oe</code> or the program will crash.</p>

CC BY-SA 4.0
2021-11-20T15:51:26.863Z
Thanks for this. Looks like this rotations Matrix does this trick. I should look more closely into this somewhat later on when I have more time. interestingly, my original estimate ($sin(x) = sin(5) / sin(23.5$)) turned up - if I get your answer correctly - to be correct (thought the estimate erred in 2 things that somewhat magically canceled each other) . My final edit however produced somewhat different result. I took  0.087266 for "inc", and 0.41 radians for "oe"; it resulted in little higher  ~0.0018 deg than expected. It might be approx stuff, or maybe I have an error in my last edit. >
d_e
CC BY-SA 4.0
2021-11-20T15:58:41.783Z
> I took more than 5 digits after the decimal point when I did the calculation in WA. So it should not amount to 0.002 degrees diff (even its little more than this). Moreover my expression looks like much complex than the answer. maybe they are indeed equivalent. Will look at this later (which unfortunately might be some days from now).
d_e
CC BY-SA 4.0
2021-11-20T16:45:52.313Z
Well, I'm now convinced the expressions are not equivalent.  I took "oe"=30 (hence its sin is 0.5) and "inc"=10. [here calculation for alpha](https://www.wolframalpha.com/input/?i=Derivative+arctan%28sin%28pi%2F18+%2B+arcsin%28sin%28a%29*0.5%29%29+%2F+%28cos%28a%29*0.5%29%29+-+a+%3D+0) and [here the result](https://www.wolframalpha.com/input/?i=arctan%28sin%28pi%2F18+%2B+arcsin%28sin%28a%29*sin%280.523598775%29%29%29+%2F+%28cos%28a%29*sin%280.523598775%29%29%29+-+a+where+a+%3D-0.3207669328): 20.33696  where we expect 20.322037. This large gap of 0.01 deg.
d_e
CC BY-SA 4.0
2021-11-20T21:56:01.903Z
@d_e It's probably just approximation errors. The equation $$\sin(RA) = \pm\sin(inc) / \sin(OE)$$ is correct.
PM 2Ring
CC BY-SA 4.0
2021-11-20T22:00:23.100Z
FWIW, I used a combination of Sage & hand calculation to find the maximum of RA. Sage can do algebra, including symbolic differentiation, but it's not very good at simplifying trig expressions. ;)
PM 2Ring
CC BY-SA 4.0
2021-11-20T22:06:49.773Z
Annoyingly, the Sage 3D graphics rotation methods `.rotateX`, `.rotateY`, and `.rotateZ` rotate clockwise around the axis, which is the opposite convention to how the standard right-handed rotation matrices work. That made things a little confusing while I was working on my animation code.
PM 2Ring
CC BY-SA 4.0
2021-11-20T11:37:21.397Z
2021-11-20T21:46:29.753Z
2021-11-20T21:46:29.753Z
PM 2Ring
<p><strong>Interaction of the Planes</strong></p>
<p>The orbital plane of the Moon maintains a relatively constant inclination to the ecliptic, which currently is about <span class="math-container">$5^{\large\circ}09'$</span>, and the nodes precess along the ecliptic at a fairly constant rate. The equator of the Earth is inclined about <span class="math-container">$23^{\large\circ}27'$</span> to the ecliptic.</p>
<p>In the following animation, the various planes are intersected with a sphere: the ecliptic is in red, the Earth's equator is in green, and the Lunar orbital plane is in blue:</p>
<p><a href="https://i.sstatic.net/dx7WX.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dx7WX.gif" alt="enter image description here" /></a></p>
<p>The following animation shows the locations of the north ecliptic pole in red, the north celestial pole in green, and the north pole of the Lunar orbit in blue, and how they affect the location of the lunar ascending node along the celestial equator.</p>
<p><a href="https://i.sstatic.net/ErkPx.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ErkPx.gif" alt="enter image description here" /></a></p>
<hr />
<p><strong>Spherical Trigonometry</strong></p>
<p>Let's look at the geometry of the triangles on the sphere. Let <span class="math-container">$E$</span> be the north ecliptic pole, <span class="math-container">$C$</span> the north celestial pole, <span class="math-container">$L$</span> the north pole of the Lunar orbit. Furthermore, let <span class="math-container">$F$</span> be the intersection of the equator and the ecliptic, that is, the First Point of Aries, and <span class="math-container">$A$</span> be the ascending node of the Lunar orbit on the equator.</p>
<p><a href="https://i.sstatic.net/sWEcj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Nq5Um.png" alt="enter image description here" /></a></p>
<p>Since an equator is <span class="math-container">$90^{\large\circ}$</span> from its pole, each of the arcs <span class="math-container">$LA$</span>, <span class="math-container">$CA$</span>, <span class="math-container">$EF$</span>, and <span class="math-container">$CF$</span> is <span class="math-container">$90^{\large\circ}$</span>. The Law of Cosines then gives that
<span class="math-container">$$\require{cancel}
\begin{align}
CL&amp;=CAL\tag{1a}\\
CE&amp;=CFE\tag{1b}\\
FA&amp;=FCA\tag{1c}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(1a)}$</span>: <span class="math-container">$\cos(CL)=\overset{0}{\cos(LA)}\,\overset{0}{\cos(CA)}+\overset{1}{\sin(LA)}\,\overset{1}{\sin(CA)}\cos(CAL)$</span><br />
<span class="math-container">$\text{(1b)}$</span>: <span class="math-container">$\cos(CE)=\cos(EF)\cos(CF)+\sin(EF)\sin(CF)\cos(CFE)$</span><br />
<span class="math-container">$\text{(1c)}$</span>: <span class="math-container">$\cos(FA)=\cos(CF)\cos(CA)+\sin(CF)\sin(CA)\cos(FCA)$</span></p>
<p>The Law of Sines then says that
<span class="math-container">$$
\begin{align}
LCA&amp;=90^{\large\circ}\tag{2a}\\
ECF&amp;=90^{\large\circ}\tag{2b}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(2a)}$</span>: <span class="math-container">$\frac{\sin(LCA)}{\sin(LA)}=\frac{\sin(CAL)}{\sin(CL)}=1$</span><br />
<span class="math-container">$\text{(2b)}$</span>: <span class="math-container">$\frac{\sin(ECF)}{\sin(EF)}=\frac{\sin(CFE)}{\sin(CE)}=1$</span></p>
<p>Equations <span class="math-container">$\text{(2a)}$</span> and <span class="math-container">$\text{(2b)}$</span> show that
<span class="math-container">$$
\begin{align}
LCA&amp;=ECF\tag{3a}\\
LCF+FCA&amp;=ECL+LCF\tag{3b}\\
FCA&amp;=ECL\tag{3c}
\end{align}
$$</span>
<span class="math-container">$\text{(1c)}$</span> and <span class="math-container">$\text{(3c)}$</span> then give
<span class="math-container">$$
FA=ECL\tag4
$$</span>
<a href="https://i.sstatic.net/ZJHU5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u9sOa.png" alt="enter image description here" /></a></p>
<p>For a general spherical triangle, <span class="math-container">$ECL$</span>,
<span class="math-container">$$
\begin{align}
\tan(ECL)
&amp;=\frac{\sin(EL)\sin(LEC)}{\cos(EL)\sin(EC)-\sin(EL)\cos(EC)\cos(LEC)}\tag{5a}\\
&amp;=-\frac{\sin(EL)\sin(CEL)}{\cos(EL)\sin(EC)-\sin(EL)\cos(EC)\cos(CEL)}\tag{5b}
\end{align}
$$</span>
Plugging in <span class="math-container">$EL=5^{\large\circ}09'$</span> and <span class="math-container">$EC=23^{\large\circ}27'$</span> gives the following graph of the right ascension of the lunar ascending node in minutes, where <span class="math-container">$1^\text{m}=\frac14{}^{\large\circ}$</span>:</p>
<p><a href="https://i.sstatic.net/YR27x.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3mISU.png" alt="enter image description here" /></a></p>
<p>The extremes on the graph occur when <span class="math-container">$CL$</span> is tangent to the circle traversed by the pole of the Lunar orbit. At that point,
<span class="math-container">$$
\begin{align}
\sin(ECL)&amp;=\frac{\sin(EL)}{\sin(EC)}\tag{6a}\\
ECL&amp;=52^\text{m}09^\text{s}\tag{6b}\\[4pt]
&amp;=13^{\large\circ}02'\tag{6c}
\end{align}
$$</span>
and
<span class="math-container">$$
\begin{align}
\cos(CEL)&amp;=\frac{\tan(EL)}{\tan(EC)}\tag{7a}\\
CEL&amp;=78^{\large\circ}00'\tag{7b}
\end{align}
$$</span>
Since <span class="math-container">$78^{\large\circ}$</span> is less than <span class="math-container">$90^{\large\circ}$</span>, it takes a bit less than <span class="math-container">$8.1$</span> years for the ascending node to go from its furthest west to its furthest east point, whereas it takes a bit more than <span class="math-container">$10.5$</span> years to return to its furthest west point.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Lunar_standstill" rel="nofollow noreferrer">Lunar Standstills</a> occur when the graph above crosses <span class="math-container">$0^\text{m}$</span>, that is also when the poles align, and so they occur <span class="math-container">$9.3$</span> years apart.</p>

CC BY-SA 4.0
2021-12-02T22:57:43.463Z
Thank you very much for this elaborate answer.
d_e
CC BY-SA 4.0
2021-12-02T19:43:07.280Z
2021-12-03T14:43:59.823Z
2021-12-03T14:43:59.823Z
robjohn
<p>I'm leaving this old answer here because the OP found it useful. But please see my new answer. ;)</p>
<hr />
<p>The Moon's ascending node progresses with (mostly) retrograde motion around the ecliptic. It's customary to specify it in terms of its ecliptic longitude, but we can easily convert it to equatorial coordinates using the <a href="https://en.wikipedia.org/wiki/Astronomical_coordinate_systems#Equatorial_%E2%86%94_horizontal" rel="nofollow noreferrer">standard spherical trigonometry equations</a>. Since the ecliptic latitude of a point on the ecliptic is zero, those equations simplify to:</p>
<p><span class="math-container">$$\tan\alpha = \frac{\sin\lambda\cos\varepsilon}{\cos\lambda}$$</span>
<span class="math-container">$$\sin\delta = \sin\lambda\sin\varepsilon$$</span>
where<br />
<span class="math-container">$\alpha$</span> is right ascension<br />
<span class="math-container">$\delta$</span> is declination<br />
<span class="math-container">$\lambda$</span> is ecliptic longitude<br />
<span class="math-container">$\varepsilon$</span> is the obliquity of the ecliptic</p>
<p>We can obtain the ecliptic longitude of the Moon's ascending node from <a href="https://ssd.jpl.nasa.gov/horizons/" rel="nofollow noreferrer">JPL Horizons</a>, via an osculating elements query.</p>
<p><a href="https://iopscience.iop.org/article/10.3847/1538-3881/abd414" rel="nofollow noreferrer">The JPL Planetary and Lunar Ephemerides DE440 and DE441</a> offers this formula for the obliquity of the ecliptic, which can also be found on <a href="https://en.wikipedia.org/wiki/Axial_tilt#Short_term" rel="nofollow noreferrer">Wikipedia</a>:</p>
<p><span class="math-container">$$\varepsilon = 84381.448 - 46.815T -0.00059T^2 + 0.001813T^3$$</span></p>
<p>where <span class="math-container">$\varepsilon$</span> is in arc-seconds, and <span class="math-container">$T$</span> is the time in Julian centuries since the <a href="https://en.wikipedia.org/wiki/Epoch_(astronomy)#Julian_dates_and_J2000" rel="nofollow noreferrer">J2000.0 epoch</a>, JD=2451545.0, 2000-Jan-01 12:00:00 TT.</p>
<p>This <a href="https://sagecell.sagemath.org/?z=eJx1V3tzm0gS_59PMUuyC9gSBj2yiuqoW9mLb521LZWl3FXOUVFjMZJwgCHMYEtJ5btf9wzoYedUtsT09PRrfv3ANE1yx2SZsCdGuFhUKZVJviIsZRnLJYmppGRZ8oz8xcvkG88FWfKSUCJpuWKSPPB4axD4XFc5BbpYsDxGATmPGXlipUh4rhj-UyZSspw8bMnkhnTukKnjdXzXh7--YYIhhpFkBS8lKVkL_r9WTEhhKO3LKl9IzlNBapa0rKIFXawZnIrZigSkSMgZ8QeeYbz50PE8z_UMVvDFGrY6vb7f7_Vd3CPjhzT5WiVyS_iSSBDAFmlSyGTRIklOShonNBfAd8nLDMKhvV9LWYjh2VnME5eXqzPfc7uD3u9nfr87aHcHA_-MPsQ9vwe2LAlvNNiPsTNU3s_ACliRNlE2OWBq912_01ebbwgtF4IthFo9IavtuwO_y9rdkxmc6bvvWbvnqOfeO9jp4_MpGfS6A9_t9Qa1mMZ4XJVMVmUO0k50ZN71BhAUQxnYeByV1E553iKc1XaWFLTDnecdWyQ5bjogYMGFDSwt9YA051AFnDklnRPQkixx8Q_iAX4Eg2fDqMoUJJpN_ISI3ccidXMqqLviT2e0SM7WNbSiZZIyFygIhQcqWLTIYjxtmsbN6O8wCid_hTfBp3BqqKdo9mkSBuF1eBPezqbGBXyHd4HV97w_uu_fW8ZdeBlNP01ncMa6uri71JTJ9eg2DKzw4vpqMru6sIzxx1n08fZqNg2sv2_aU8sIr2-i69F5eA0UUGYZs4lWZY3Op-Prj7PQMi6m_44ux3c3o1nNMz7_EP05mo0C63Zs1XD-Y4dSO6MbkXxjQZJLuwvxw2tYMrlYR5hitoB8ki0iJC_wmxX1fegALDECv7x9Ox1fGt-bwPwwLsY3N6PbP4HB6nq-ZUxno7tZNLu6CZH0Xcn8geTx5JDKC0UMJ9H06r8NkSERVISgQhuvb_gr7Dep6BZcSBtutKXqQvDdglqQUWkNiSXZRlo_WgSvUMBOkhcVbtgWmGq10BHnh4ZNpkS6ggHq13Zp2v8Uzue3n8G70HZP1GM4Dk1VAlyUq48BuDKSCHLLc6aDg5-ixIiat2OiIBHeXU1N58XusZwD5KIonTo_Y6yJFvnVghzoepqaJjkTOiiK1xVFmkhFtTVHbdLn3HQfOSYR7t33h_67-XHiZO6q5FVh-869P5wTnZsi5QWzC69FCr8GgTmDKqXoTclCkbowFR6RHFhNxbmBY1sPi6Gn1z6sfVz7h4ptoLWBEcuQvcHnDTxDcVyztLD_VdJinSyEAwAGT1hJF1KZllHwRqEqMKHCvmt_qPJ2p2Nq3CKt029_oHnbe29qFAfm7yQ2W6AbAMMUZQDg4VIEs7KCKr-oyifWPPOUl4G5KhnL1RmVMANw6TmJ5Trw4Uk8rYJLCsWlRWgleVQVKFiT6mgBTnIuiY1aCPQqpcJ5hZgptLiFhFpHUkaFJAAFjO7RKfMVYnRaqK4YvMpfV0AjLWxHx-NwxXarHZaVjP8LZw59Yg0tuAQO3YN5lce_vDLo0Bz8eY3GevNe6C3balnO_bDtz1UjF9jzkGWuIY-0kj831GGTQfz53pu3cOu-N3fqMLwhE9wlynnym_KafKhS6EFweiuApK79MCsuAbRDQMcJigeZ98PO_ChtZnwIiG8YwE7FUWuUEZQg9Gbe-MYOlmh80mrsZ3kF8QMOGyUdAOANCTcSQU0WNIVxBeYWFLTbByquMcfBXX9e6z4-uncTvISWuEpkFe_y83gO2h1_jKHVZmylLmSZcqojW4F_aHuFZtuA8r4zP1B5wXMYpOSuax-og9S_G-1YOdp8NHzs0YIqD9u-NuOEwOyk2j9UAnjc8atAQysuwA3bRrtL6jgHkQAEJ-TXXVqTICDePsT42arWVQezBh8UuvkxU4tkUBCAEdhrJrN9APPdNR8ZE59mp9v7znC-sylmqYZ6A81RUZR8k2R4kzDPrGCa1UCUK6nyQdXZk0KHvsDQf4MEVY63tP9Kwfwnh_Qu4nIOow_unb7Ya6ujO2POqySNyTn7lrC6tNS2FDjtod4T9aUOQ8UAkToEUhX0FnnCop5Du7HhiN7SYvboR9jDwtZHarvf0JLh1Xu7BIEpA4L-hP6CqP2VxeighFYgvT1tg2VF4rC6p6E58RY2wCbAz6alZkz4PcoSzJHaXzRhv6cMOoURF_rUKdnCvw0iQW-8RRB2cN5UKjuHWFPOHuML_HUhIIgJdBoEoS1KKgiEINrojaJp6UiBhTM_RpeOYwMvkOq8VLOP8asraaLZXPRUlUIsASvsokhUOxPgbfqqve8B0Gz2Tk0wLAVMDFI0MFQdsN-0RvW9O_wiJOr0g4p4VFC5trVfx2drO788w9sbQidOFtImtYxVmcSqcwT39xWU5n1NUgkI5d_M6CMvTXhaljSr-3Z9mm6YiFL6AJN_cG_uyyOUchMqCx46YImUZ_jmhyPjSq3wyyH7kLkRw2IbfXmOhasbva0t1zFoEj2OD9rMNkvgfWYL4za4N3FxiS0Yn-gGnjSTqjfAAuBTv211DqEHr439Xa5gqcEy1oTgRbRx_LNNYp50EckZMKvihEhbcrhFdMoUGU1Ts3WAqXrOeUjp4gtEB2t7AmUyommyyvGtOzChe8DcZe4b7Sh-rGBIoaLAoQW6WsL1gKfc3YAPrZeub3aub7TrPwmHklELhZTc6FhsVCzOjmNT34lgMtIHImWF7bkDCFst40yPao3dgFIY2A6i5gr6xGyreeFzYdfap1uMbwYrUBAn0AnoNspoTlestJ2DuAfwLp6ldozDs86nCAfhSEnWClrkO7yJWEr6YeDhjNwWkOvjSsLbyVUGwqdPKwcZXdBrO27MFtCw7V1bEWv-bE-c_wGx6dzS&amp;lang=sage" rel="nofollow noreferrer">Sage / Python script</a> fetches the necessary data from Horizons, converts to RA (in degrees), and plots the results as a graph. I chose the time interval</p>
<p>From: 2453908.5 A.D. 2006-Jun-22 00:00:00<br />
To: 2460684.5 A.D. 2025-Jan-09 00:00:00<br />
with time steps of 7 days
because it's an almost complete precession cycle.</p>
<p><a href="https://i.sstatic.net/NqGRj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NqGRj.png" alt="RA of Moon's node" /></a></p>
<p>As you can see, the right ascension of the node covers the full range from 0° to 360°.</p>
<p>Here's the corresponding graph of the declination.</p>
<p><a href="https://i.sstatic.net/tBDUK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tBDUK.png" alt="Declination of Moon's node" /></a></p>
<hr />
<p>You can easily use the script to create plots covering other intervals: Horizons can supply the necessary data for any time interval from 9999 BC to 9999 AD. However, the formula for the obliquity is probably not valid at the extremes of that range.</p>
<hr />
<p>Here's a new version of my <a href="https://sagecell.sagemath.org/?z=eJx1VMtu2zAQvOsrFskhpCsrkl3XaQEB9aE9-ZBDDkGLIKBl2iLKkCpFOfbfd_mQbQWqIAjmcnZ2dzj0gRlyd0xPdzTZ8j2U0Ai4h-IhT7RxS5Kn4F6aJLewAiuk5VuohKkkh3dha9DG1nqvFZPAjrxFmp3fJ3Xq4SnshJT0WwL4VAVy1h13m3n24D807MwuOwW-IfqIwVCMYD9-x-w3lZbalBUumGxqVubZ11Cl9KVi5qcSpFB8viW_4xTTgqb9RAV9ueaajWVNizg8ZvU_x7IMt51R8JgZbZnlz8TNjYp9F8pywyrrRXn14XKRFZ9TaC1v2nKOlLySorGiKn8y2XJc_-2YRfIn011WgskY8GNGKOusfu2aLdYMoaiyqwOTEvBEk7OMjcZucCyvY-j_ZiNZ9ecGSQ176yl8wi38iG1BI5niPih252ZDnbNiH45owE6H0EZq1wUOTo7uRLIFpmQLJ_FpuI4st8vlElvUDauEPeFZL8b7ZejDK7mu-tYcx5_NswVMvCT9LAE7MgrJsy9YMapNk1hi3Slm0O8bYQf8azinFQPPD_KU3roswyvetkKrKFrrQV0KB8ei2-AdNIhQ0UYfVe7N3CHqMCr2CPr_YPTsM5aeokoXUZyAHNWF3GNWCGBKvKHR8FogPayj13-R2QT_MiYg7r2lKT1fAkdMYacNCBAKWsPUnpOAegm1V1l_Q8SBE5q1tX4n9B9DQEvZ&amp;lang=sage" rel="nofollow noreferrer">interactive orbit plane anim script</a> which makes it easier to see that the Moon's RA is indeed close to 0° when its declination is 0° (and going from south to north of the celestial equator). Setting the <code>equatorial</code> checkbox rotates the system so that the equatorial plane is horizontal.</p>

CC BY-SA 4.0
2021-11-15T04:39:55.830Z
I believe the original question is this: Where does the Moon's orbit intersect the celestial equator? If that is the question, your answer does not provide that answer. The "ascending node" of the plane of the Moon's orbit and the celestial equator is always near 0 hour RA.
JohnHoltz
CC BY-SA 4.0
2021-11-15T04:49:07.320Z
@JohnHoltz Ah. That would explain the OP's last comment. :) But I'll leave this answer here for now, in case it has some useful info for the OP or other readers.
PM 2Ring
CC BY-SA 4.0
2021-11-15T04:53:39.203Z
I guess I could do a plot of the Moon's RA when its declination is 0°. But that will have to wait for another day...
PM 2Ring
CC BY-SA 4.0
2021-11-15T10:24:45.660Z
Thanks for that demonstration. It shows how this precession of the node versus the ecliptic looks like in the equatorial coordinate system.  My question, however, as JohnHoltz says is about other nodes, namely the nodes of intersection with the celestial equator. Sorry for not explaining this well enough previously.
d_e
CC BY-SA 4.0
2021-11-15T15:59:11.670Z
@d_e Sorry for misunderstanding, but I'm glad that this answer isn't totally useless in its current state. :) 
 I've just added a new version of the anim script to this answer. I'll have to think about the answer to your real question.
PM 2Ring
CC BY-SA 4.0
2021-11-15T16:09:58.123Z
@PM2Ring, wonderful. I've also just edited my question added further thoughts.
d_e
CC BY-SA 4.0
2021-11-16T03:55:43.803Z
@d_e Here's a [script](https://sagecell.sagemath.org/?z=eJyNT8FuwyAMvfMVVi-FysmSdJ2mSnxCL7tO0YRS1jIlkBDaZPv6GdROa3cZMhi_Z_weZ-X50uCEA4alYGyv3-Fl5uEotgxoeR1O3kKngjczT9BrOimXCAVFjVeAqsaN8THCaGy83JDZBfxpqxNZX3W__qd7r3Hj4Y_GvUOK8rfuAST0Bh6gfC7Yjgr6_iBgFe1MKc_cCDYjfCK8Eb_LG9eeOssrwVTbHxVhKqrGrYKyFc9itxCCppIAY05TT7XONzQuAsY2BGzy8vEC9K0LPA3jRhKLMMmwIgZhkJnT9Asekvf1U0HFwZt9a6we5aJTH84vxDfXbHXa&lang=sage) which plots a graph of the variation in RA.
PM 2Ring
CC BY-SA 4.0
2021-11-16T07:49:36.760Z
@PM2Ring, Cool! We can see the max value is at around $\alpha=$78deg (as I found above, just need to add +90 to my $\alpha$ because the way I defined it). Thanks for your plots.
d_e
CC BY-SA 4.0
2021-11-15T03:44:57.433Z
2021-11-20T09:58:31.043Z
2021-11-20T09:58:31.043Z
PM 2Ring
<p>In the span of 18.6 years there is nodal precession of the Moon with respect to the ecliptic of 360 deg. But if we consider this precession with respect to the equator, we will find (unless I'm hugely mistaken) that the nodes hardly move, certainly not 360 deg. <strong>I ask in what range does it move, and how to calculate this range</strong>. <a href="https://astronomy.stackexchange.com/a/47495/25913">In a recent answer</a> about node precession of the Moon, I wrote on that in point 4:</p>
<blockquote>
<p>[T]he node precession of Moon-orbit with respect to the equator is not complete. in the sense the nodes (again with respect to the equator) are not running over 360 deg in 18 years. But they run over much smaller span of maybe around 25 deg, ~12.5 deg from each side of the solar equinox. The precession works as regular (360 deg in 18 years) when we consider the nodes of the intersection not with Earth equator but with the ecliptic.</p>
</blockquote>
<p>I have provided there a rough estimate of the range. The calculation of this was as follows: If at the Solar-node we make the declination of the Moon +5 deg. (due to his tilt 5deg to the ecliptic), and if we assume (this is the estimate) that in this setting the max declination of the Moon is 23.5, this Solar node is 12.62 deg from the the Lunar node. (<span class="math-container">$\sin(x) = \sin(5)/\sin(23.5)$</span>); hence my estimate of ~12.5 on each side of the Solar node.</p>
<p>But I believe the answer should come a little lower, for at this setting the max declination of the Moon is higher than 23.5 and also it seems not to be 5deg at the Solar node. I tried to calculate this but got somewhat confused, and I'm not sure how to calculate this at the end. Also, I didn't find the correct figure online so I was missing something to validate whatever answer I will come up with.</p>
<hr />
<p>I hope my question was clear, but here is another way to formulate this question: for a period of 18.6 years to take all the RAs of the ascending (or descending) node [dec=0] of the Moon. What is the maximum/minimum RA we will find?</p>
<ul>
<li>For calculation lets assume no other perturbation exists.</li>
</ul>
<hr />
<p><strong>Edit:</strong> It struck me that the inclination that was previously estimated to be 23.5 (and really thought it should be little higher) can be calculated relatively easily. this is because we have 2 points on the arc (RA, dec): (x, 5) and (90+x, 23.5) (because we know that if we tilt the 5 deg at a given location, the locations 90 deg from there not going to change their original declination.) Hence we know where <span class="math-container">$-x$</span> is the node location and <span class="math-container">$z$</span> the inclination:</p>
<p><span class="math-container">$ \sin(x)*\sin(z) = \sin(5) $</span>. and also:<br />
<span class="math-container">$ \sin(90+x)*\sin(z) = \sin(23.5)$</span>. So we get by division: <span class="math-container">$$\tan(x) = \dfrac{\sin(5)}{\sin(23.5)}$$</span> =&gt; <span class="math-container">$x=12.32$</span>  and <span class="math-container">$z=24.089$</span>.</p>
<p>I now suspect this is the correct answer but not quite convinced yet. Any kind of validation from this forum would be appreciated. I'm afraid that actual observational data - though will be very close for sure - cannot give us the accuracy I seek, because the 5 deg tilt I used is really ranging between 5 and 5.3 and moves quite fast as @PM 2Ring showed in the last graph of <a href="https://astronomy.stackexchange.com/a/47503/25913">this answer</a> - hence the data should range (if we use the more accurate 23.44 ) between 12.35 and 13.07 and perhaps little lower because of the resolution of observation once a month in which the moon actually passes through the node.</p>
<p><strong>Second Edit (Hopefully the last one):</strong></p>
<p>In the method employed in the previous edit we selected <span class="math-container">$\alpha = 0$</span> as the angle (or call it RA=0) where we perform the 5 deg tilt. But the same method can be applied to every <span class="math-container">$\alpha$</span>, so we can actually see at what <span class="math-container">$\alpha$</span> we get the highest <span class="math-container">$x$</span>. <a href="https://www.wolframalpha.com/input/?i=Derivative%20arctan%28sin%280.087266%20%2B%20arcsin%28sin%28a%29*sin%280.41%29%29%29%20%2F%20%28cos%28a%29*sin%280.41%29%29%29%20-%20a%20%3D%200" rel="nofollow noreferrer">Wolfram yielded maximal value</a> at <span class="math-container">$\alpha = -11.7819$</span>, which <a href="https://www.wolframalpha.com/input/?i=arctan%28sin%280.087266+%2B+arcsin%28sin%28a%29*sin%280.41%29%29%29+%2F+%28cos%28a%29*sin%280.41%29%29%29+-+a+where+a+%3D-0.2056334257" rel="nofollow noreferrer">in turn produced</a> <span class="math-container">$x=12.6312$</span> <a href="https://www.wolframalpha.com/input/?i=arcsin%28%28cos%28a%29*sin%280.41%29%29+%2F+cos%28x%2Ba%29%29+where+a%3D-0.205633+and+x%3D0.220457" rel="nofollow noreferrer">and</a> <span class="math-container">$z=22.97$</span>.</p>
<p>Unless I have an error it is what it is.</p>
<p>For the record, the general equations with <span class="math-container">$\alpha$</span> as parameter are (reminded in those equations the node is at <span class="math-container">$-x$</span> not <span class="math-container">$+x$</span>):</p>
<p><span class="math-container">$ \sin(x+\alpha)*\sin(z) = \sin(5+\arcsin(\sin(\alpha)*\sin(23.5))) $</span>
<span class="math-container">$ \sin(90+x+\alpha)*\sin(z) = \sin(90+\alpha)*\sin(23.5) $</span></p>
<p><strong>Third edit:</strong></p>
<p>When comparing second edit to PM 2Ring's answer, there is a discrepancy of less than 0.002 deg in Moon numbers. (For more info read my comment to that answer), nevertheless in other hypothetical numbers - especially if we increase the 5 deg inclination to the ecliptic - the difference becomes bigger: up to more than a full degree even. <strong>So the method in the second edit is wrong</strong>. I think the mistake was as follows: When I performed the 5 deg tilt, I made this tilt in the equatorial coordinate system by adding +5 deg to the declination in a selected point (RA or <span class="math-container">$\alpha$</span>) on the ecliptic. But this operation results in a plane which is not 5 deg inclined to the ecliptic. (only 5 deg inclined to the &quot;ecliptic&quot; after some another rotation operation). In other words, when we do the right thing: and perform the tilt in the ecliptic coordinate system by adding 5 to latitude at a given ecliptic longitude, and only then convert the orbit to the equatorial system, we will get somewhat different orbit that cannot be obtained by adding 5 to the declination anywhere. The sole exception being when <span class="math-container">$\alpha=90$</span> then this method works.</p>
<p>Another final note: My initial estimate calculation <span class="math-container">$\sin(x) = \sin(5)/\sin(23.5)$</span> turned out to be the correct answer. I guess this is not a mere coincidence. It means that the intersection of the Moon orbit with the equator after this 5 deg tilt, is equal to the intersection of another imaginary plane path that keeps the 23.5 inclination, but increase the declination in 5 deg at the node.</p>

CC BY-SA 4.0
2021-11-14T23:14:22.707Z
What do you mean by Solar node? The ecliptic & equatorial planes meet at the line passing through the equinox points. The lunar ascending & descending nodes move (retrograde) around the ecliptic, with a period of ~18.6 years to complete a full cycle.
PM 2Ring
CC BY-SA 4.0
2021-11-14T23:22:36.453Z
@PM2Ring, I mean the RA of the Solar equinox. I agree with what you wrote. My question is exactly how "The lunar ascending & descending nodes move (retrograde) around the **celestial equator**
d_e
CC BY-SA 4.0
2021-11-14T23:25:31.633Z
The ecliptic longitude of the lunar ascending node goes (mostly) retrograde from 360° to 0°, so its RA must range from 24h to 0h.
PM 2Ring
CC BY-SA 4.0
2021-11-14T23:31:59.853Z
@PM2Ring, how can be it be then, when the inclination of the Moon about the equator is about 23 deg, it means that we might end up somewhere in the 18 year period with 23 dec and RA=0 (when the ascending node will fall on RA=90) ; which is impossible. since it would make the tilt between the Moon Orbit and the ecliptic 23 degrees apart which is really 5. As I said this is subtle issue - hope I'm not falling with words to explain.
d_e
CC BY-SA 4.0
2021-11-14T23:48:51.197Z
Sorry, I'm getting dizzy trying to understand your last comment. ;) By definition, the Moon's nodes are always on the ecliptic. So if a lunar node has RA=6h (90°) it has dec~=23°. When the Moon itself is at RA=6h, its declination is in the range from 18° to 28°. When it's at RA=0h or 12h, its declination is in the range from 5° south to 5° north.
PM 2Ring
CC BY-SA 4.0
2021-11-14T23:54:56.470Z
You might find it helpful to explore the interactive anim I linked in [my answer](https://astronomy.stackexchange.com/a/47503/16685) to the previous question.
PM 2Ring
CC BY-SA 4.0
2021-11-14T23:55:54.140Z
@PM2Ring, Let's remove the Sun and the ecliptic. The orbit of the Moon is a plane around the Earth. It has nodes with respect to the equator. where the celestial equator and the Moon Plane intersects. That's the nodes I'm talking about.  I also become dizzy now - need a break. But I have empirical evidence: Select a random month, spot the date of the ascending node, then check on that date what is the RA. you will see it is always close to 0. https://in-the-sky.org/graphs.php?gtype=2&startday=27&startmonth=11&startyear=2005&duration=2&obj1type=0&obj1txt=the+Moon&obj2type=0&obj2txt=the+Sun
d_e
<the-moon><declination><precession><space-geometry>
How to calculate the nodal precession range of the Moon with respect to the celestial equator?
d_e
# How to calculate the nodal precession range of the Moon with respect to the celestial equator?

(Asked by: username_0 on 2021-11-14)

In the span of 18.6 years there is nodal precession of the Moon with respect to the ecliptic of 360 deg. But if we consider this precession with respect to the equator, we will find (unless I'm hugely mistaken) that the nodes hardly move, certainly not 360 deg. **I ask in what range does it move, and how to calculate this range**. [In a recent answer](https://astronomy.stackexchange.com/a/47495/25913) about node precession of the Moon, I wrote on that in point 4:

> [T]he node precession of Moon-orbit with respect to the equator is not complete. in the sense the nodes (again with respect to the equator) are not running over 360 deg in 18 years. But they run over much smaller span of maybe around 25 deg, ~12.5 deg from each side of the solar equinox. The precession works as regular (360 deg in 18 years) when we consider the nodes of the intersection not with Earth equator but with the ecliptic.

I have provided there a rough estimate of the range. The calculation of this was as follows: If at the Solar-node we make the declination of the Moon +5 deg. (due to his tilt 5deg to the ecliptic), and if we assume (this is the estimate) that in this setting the max declination of the Moon is 23.5, this Solar node is 12.62 deg from the the Lunar node. ($\sin(x) = \sin(5)/\sin(23.5)$); hence my estimate of ~12.5 on each side of the Solar node.

But I believe the answer should come a little lower, for at this setting the max declination of the Moon is higher than 23.5 and also it seems not to be 5deg at the Solar node. I tried to calculate this but got somewhat confused, and I'm not sure how to calculate this at the end. Also, I didn't find the correct figure online so I was missing something to validate whatever answer I will come up with.

---

I hope my question was clear, but here is another way to formulate this question: for a period of 18.6 years to take all the RAs of the ascending (or descending) node [dec=0] of the Moon. What is the maximum/minimum RA we will find?

* For calculation lets assume no other perturbation exists.

---

**Edit:** It struck me that the inclination that was previously estimated to be 23.5 (and really thought it should be little higher) can be calculated relatively easily. this is because we have 2 points on the arc (RA, dec): (x, 5) and (90+x, 23.5) (because we know that if we tilt the 5 deg at a given location, the locations 90 deg from there not going to change their original declination.) Hence we know where $-x$ is the node location and $z$ the inclination:

$ \sin(x)\*\sin(z) = \sin(5) $. and also:  
$ \sin(90+x)\*\sin(z) = \sin(23.5)$. So we get by division: $$\tan(x) = \dfrac{\sin(5)}{\sin(23.5)}$$ => $x=12.32$ and $z=24.089$.

I now suspect this is the correct answer but not quite convinced yet. Any kind of validation from this forum would be appreciated. I'm afraid that actual observational data - though will be very close for sure - cannot give us the accuracy I seek, because the 5 deg tilt I used is really ranging between 5 and 5.3 and moves quite fast as @PM 2Ring showed in the last graph of [this answer](https://astronomy.stackexchange.com/a/47503/25913) - hence the data should range (if we use the more accurate 23.44 ) between 12.35 and 13.07 and perhaps little lower because of the resolution of observation once a month in which the moon actually passes through the node.

**Second Edit (Hopefully the last one):**

In the method employed in the previous edit we selected $\alpha = 0$ as the angle (or call it RA=0) where we perform the 5 deg tilt. But the same method can be applied to every $\alpha$, so we can actually see at what $\alpha$ we get the highest $x$. [Wolfram yielded maximal value](https://www.wolframalpha.com/input/?i=Derivative%20arctan%28sin%280.087266%20%2B%20arcsin%28sin%28a%29*sin%280.41%29%29%29%20%2F%20%28cos%28a%29*sin%280.41%29%29%29%20-%20a%20%3D%200) at $\alpha = -11.7819$, which [in turn produced](https://www.wolframalpha.com/input/?i=arctan%28sin%280.087266+%2B+arcsin%28sin%28a%29*sin%280.41%29%29%29+%2F+%28cos%28a%29*sin%280.41%29%29%29+-+a+where+a+%3D-0.2056334257) $x=12.6312$ [and](https://www.wolframalpha.com/input/?i=arcsin%28%28cos%28a%29*sin%280.41%29%29+%2F+cos%28x%2Ba%29%29+where+a%3D-0.205633+and+x%3D0.220457) $z=22.97$.

Unless I have an error it is what it is.

For the record, the general equations with $\alpha$ as parameter are (reminded in those equations the node is at $-x$ not $+x$):

$ \sin(x+\alpha)\*\sin(z) = \sin(5+\arcsin(\sin(\alpha)\*\sin(23.5))) $
$ \sin(90+x+\alpha)\*\sin(z) = \sin(90+\alpha)\*\sin(23.5) $

**Third edit:**

When comparing second edit to PM 2Ring's answer, there is a discrepancy of less than 0.002 deg in Moon numbers. (For more info read my comment to that answer), nevertheless in other hypothetical numbers - especially if we increase the 5 deg inclination to the ecliptic - the difference becomes bigger: up to more than a full degree even. **So the method in the second edit is wrong**. I think the mistake was as follows: When I performed the 5 deg tilt, I made this tilt in the equatorial coordinate system by adding +5 deg to the declination in a selected point (RA or $\alpha$) on the ecliptic. But this operation results in a plane which is not 5 deg inclined to the ecliptic. (only 5 deg inclined to the "ecliptic" after some another rotation operation). In other words, when we do the right thing: and perform the tilt in the ecliptic coordinate system by adding 5 to latitude at a given ecliptic longitude, and only then convert the orbit to the equatorial system, we will get somewhat different orbit that cannot be obtained by adding 5 to the declination anywhere. The sole exception being when $\alpha=90$ then this method works.

Another final note: My initial estimate calculation $\sin(x) = \sin(5)/\sin(23.5)$ turned out to be the correct answer. I guess this is not a mere coincidence. It means that the intersection of the Moon orbit with the equator after this 5 deg tilt, is equal to the intersection of another imaginary plane path that keeps the 23.5 inclination, but increase the declination in 5 deg at the node.

**Tags:** the-moon, declination, precession, space-geometry
> **Comments:**
> **username_1**: What do you mean by Solar node? The ecliptic & equatorial planes meet at the line passing through the equinox points. The lunar ascending & descending nodes move (retrograde) around the ecliptic, with a period of ~18.6 years to complete a full cycle.  
> **username_0**: @PM2Ring, I mean the RA of the Solar equinox. I agree with what you wrote. My question is exactly how "The lunar ascending & descending nodes move (retrograde) around the **celestial equator**  
> **username_1**: The ecliptic longitude of the lunar ascending node goes (mostly) retrograde from 360° to 0°, so its RA must range from 24h to 0h.  
> **username_0**: @PM2Ring, how can be it be then, when the inclination of the Moon about the equator is about 23 deg, it means that we might end up somewhere in the 18 year period with 23 dec and RA=0 (when the ascending node will fall on RA=90) ; which is impossible. since it would make the tilt between the Moon Orbit and the ecliptic 23 degrees apart which is really 5. As I said this is subtle issue - hope I'm not falling with words to explain.  
> **username_1**: Sorry, I'm getting dizzy trying to understand your last comment. ;) By definition, the Moon's nodes are always on the ecliptic. So if a lunar node has RA=6h (90°) it has dec~=23°. When the Moon itself is at RA=6h, its declination is in the range from 18° to 28°. When it's at RA=0h or 12h, its declination is in the range from 5° south to 5° north.  
> **username_1**: You might find it helpful to explore the interactive anim I linked in [my answer](https://astronomy.stackexchange.com/a/47503/16685) to the previous question.  
> **username_0**: @PM2Ring, Let's remove the Sun and the ecliptic. The orbit of the Moon is a plane around the Earth. It has nodes with respect to the equator. where the celestial equator and the Moon Plane intersects. That's the nodes I'm talking about.  I also become dizzy now - need a break. But I have empirical evidence: Select a random month, spot the date of the ascending node, then check on that date what is the RA. you will see it is always close to 0. https://in-the-sky.org/graphs.php?gtype=2&startday=27&startmonth=11&startyear=2005&duration=2&obj1type=0&obj1txt=the+Moon&obj2type=0&obj2txt=the+Sun  

---

## 3 answers

### Answer by username_1 on 2021-11-20. Score: 4 | Accepted answer

You want to know the range of RA where the Moon's orbital plane crosses the celestial equator from south to north, so we need to find the line of intersection of those two planes.

The vector corresponding to the line of intersection of two planes equals the [cross product](https://en.wikipedia.org/wiki/Cross_product) of the normals of the planes. We can find those normals using [rotation matrices](https://en.wikipedia.org/wiki/Rotation_matrix). The relevant parameters are given relative to the ecliptic plane, so it's convenient to work in that coordinate system. I'll use a right-handed coordinate system, treating the celestial sphere as a unit sphere, so our planes of interest are represented as circles of radius 1.

The [First Point of Aries](https://en.wikipedia.org/wiki/First_Point_of_Aries) is the point on the celestial sphere where the Sun is found at the March equinox, when it crosses the celestial equator from south to north, so it's a point where the celestial equator & ecliptic intersect. In our system it's on the X axis with Cartesian coordinates $(1,0,0)$. Both ecliptic longitude and RA equal zero at this point.

We have the Moon's inclination $i$, which is approximately 5.14°, but it ranges from roughly 4.98° to 5.30°. The obliquity of the ecliptic $\varepsilon$ is currently a little over 23.436°, but we need to negate that to get the inclination of the equator relative to the ecliptic. The final lunar parameter we need is $\omega$, the longitude of the Moon's ascending node, which ranges from 360° to 0° over the course of the lunar precession cycle.

To find our plane normals, we start with a circle in the XY plane, so its normal is the $Z$ vector $(0, 0, 1)$ then we rotate it around the X axis by its inclination, using matrix multiplication. To precess the lunar plane we then rotate it around the Z axis by $\omega$. This leads us to the following equation for the vector of the line of intersection:

$$P = (Rx(\varepsilon)Z) \times (Rz(\omega)Rx(i)Z)$$
where $Rx(.)$ & $Rz(.)$ are rotation matrices.

Generally, $P$ is *not* a unit vector, but we can easily normalize it so that does have unit length, if necessary. Bear in mind that our XY plane is the ecliptic. We can rotate $P$ around the X axis by $-\varepsilon$ to determine its coordinates in the equatorial plane.

I won't do all the algebra here, but the equatorial coordinates of $P$ are

$$\begin{align}
x & = \cos(\varepsilon) \cos(\omega) \sin(i) - \sin(\varepsilon)\cos(i)\\
y & = -\sin(\omega)\sin(i)\\
z & = 0
\end{align}$$

The RA $\alpha$ corresponding to P is just
$$\alpha=\tan^{-1}\left(\frac yx\right)$$

Finding the $\omega$ corresponding to the maximum & minimum of $\alpha$ involves some tedious trigonometric calculus and algebra. Fortunately, the results are fairly simple. :)

$$\begin{align}
\omega & = \cos^{-1}(\tan(i) / \tan(\varepsilon))\\
\alpha & = \sin^{-1}(\sin(i) / \sin(\varepsilon))
\end{align}$$

Hopefully, I haven't messed up a minus sign somewhere while transcribing all those equations. ;)

Finally, here are some actual results. Using the value for $\varepsilon$ mentioned earlier,

| inc | RA | omega |
| --- | --- | --- |
| 4.98 | -12.606952 | 101.596318 |
| 5.14 | -13.017906 | 101.976326 |
| 5.30 | -13.429439 | 102.357061 |

[![RA vs omega, for 3 inclinations](https://i.sstatic.net/Gnpmd.png)](https://i.sstatic.net/Gnpmd.png)

Here's the single plot for $i=5.14°$
[![RA vs omega for i=5.14°](https://i.sstatic.net/YUOlC.png)](https://i.sstatic.net/YUOlC.png)

And here's the [Sage / Python script](https://sagecell.sagemath.org/?z=eJyNU21v2jAQ_p5fcco-4HRJFqCtpkqWth8wCVX7RBUhNzmCp8RmtgNpf_3OTkBANakIwp3v8T3PveQLrFrt4Pkn6C3UWPECDhZ0h42IDsKwmZslUY0NcNhL-Abz70W0JueAldOGsSIF-s6TBCKCbeF5YG6XPEVAH4OuNwo64YwcWDh6CU_6n4d7RZmeDsirtPWXU8isVN66ik5nZ1gZYuWZ-P1zxB9IrlTcktxK9LVeEP-QyqERlQsSNkyqij_k8_sUNPJssczvl4-TJgrBHacWN8HVOHmUxft7Q6lYFMfxOAq3Q9hrOoPjDg0G_5fWakbDMa_SQWW0tWhDoMIWrZOiBfzbC5pLGgkLAra9qpzUyicMI00DHKtW7p2soNWqka6v8cQ4MQhboaqlakDpGnOvKpl0GkHDV0z4RvkflZXQWnhTY-JNqikJ0OOIDJ0UZ6Q3b5Fj8b7yVnbS2TglnquQcGMBQKHjSUvYT93NRuSQwlsKGyJd56E5m73RdV85xmgpM-KEO78lme6CNQRF3lwnudKmE618x5qN2XzrKJUguQtGeYdJ8ES98u8DvTjM45juuKM8vp4UmAtrsnwskmTCfuXjLC17YcdQG8EIABmQm5FfXkBbqXBRE7S4gAaz9HvZasNjgzV1wiOte2uRx0_xfzJkFymyz-WwO31kqxS2RnTIf5seU2iMrAOUx534ow3dFAPaTStesbUbS63j83z-cHXMX-IwNQLTcOMy-QfoyTkV&lang=sage) that made that plot.

Lastly, here's an [interactive anim](https://sagecell.sagemath.org/?z=eJx9VeuKIkcU_u9THHogVpvWsR1mGBYashACgSyYZWFlhtmhbEutTFtVW1Xt6Ia8U_YV9sly6ta2JhtRuy7n-p3vnM6yDH5rBdUg5IqB0qxmxnApYMV2EgYDwM-HLYOGCwZyDVxYpg2rrZPBvXV33sAPwD631ErNaQOqoYIZr82NF6q1NAYdyFVb26jJnVu9o42Jnj5qbi0TsDzC_B3M3nOxgdl0Vk7KcjKbDjKMdjBYsQ1UoDhcQ3k_HUjttnsMSWpCpgW4b54PFr3Tsjt9-Jdsmedo9ArmVNMds5rXcPMztIJbqLmuGwYU49V8wwW6XoMKp2Q0enmlemPyNz70PdVkaIe532hmWy1AdSafVSPtzYqQWhpi8wIMF_45xZ8LZDZSHJdO7FlJRNlU5RTPOy8-xrdgeWPZKkX2yu0WY7NbuZECYacHBN0F6e5JXXjxAmijtjTGOXfYxRT0ZlnLRuqqDmHzdRANkkH6xyo6Iwg0olXASStarvx_AWveNNUH3bJ80Ok63mDej-OHAh6e-sr_JbQoYPE9IY8KSi3OIzD8C6vuz2CfT7S01LIFGbv8HXQ_ed7S2np0ngkXNeJbgGTV-AafxjJlKrcKGZXX9wWwuuHK8trnVCR6x53CNlCuD_YsntDWyudWrdBz9QtymkXE0ReMKmyojd9KFneR81exAaVecptax9-4cwjwk23LyHRSIkXQ3Bnu-YWdy0YO1Qu92BawdxaRhGgm0tCt_O3v0NW6hRFgxfYFBKCzZUPrlyzJnUgx_p5kjAoplWA8scobSCSMale3t7dZR8OI9YVKD4m73BXvf4F4_xaJtePWwOuWuYGGcQjqB1dVwdTLHjHlhAHOE7eUkb4Hd_VZW1LCGI6fZr3cExUfyaGAY2xiXI79-qmDQrNVFhl6l59ImTwojWbIOsNANRUb9ga-ff1Ty1asCF0aQl00RxcWkgU93OV_ffs766tmcof2BaF-qtAuDbdEJ0k1AbN208hgWo9PYS-RdTskFBjvP82h8H_teyLvz4LKpe_Rjck8ELlL1gP07-gL8_M-VhC2ODq_SIHMPjN06lEM9MzCrxdvmPB28V1xesd08odIZ2dlFFY7t-qKOk5Fjdcd21P5x_7-TKm714kDh08zzLxjgS-BxUtyQIw1kiA8pnkvyd5g86NT2RMzNpoxkV1IJ1opeyEXOVTOEt-nk9uTbijrhCrFkDnzhCYVfIcAk3CdT9IExIFF_KhCSI9hXE_MVr4GwTC5Cvei_iPUoBr2Rt3QNWdvDwylYRheQZqqLa-H-T80tG78&lang=sage) that demonstrates that the above algebra actually works. ;)

The large black circle is the ecliptic. The blue plane is the celestial equator, the orange plane is the lunar orbital plane. The `alpha` parameter controls the transparency of the lunar & ecliptic planes; set it to zero if you want unfilled circles for those planes. The blue ball on the equator is the First Point of Aries. The orange ball on the lunar plane is the Moon's north node. Both of those balls sit on the ecliptic. The green line and ball show the intersection of the equator & lunar plane, and the small red balls show the maximum travel of the green ball.

By default, the anim uses -30° for `oe` = $\varepsilon$ and 10° for `inc` = $i$ to make it a little easier to see what's happening. You can change those values to whatever angles you want, but `inc` should be positive and `oe` should be negative, otherwise north nodes & south nodes get swapped. Also, the absolute value of `inc` *must* be less than the absolute value of `oe` or the program will crash.

> **Comments:**
> **username_0**: Thanks for this. Looks like this rotations Matrix does this trick. I should look more closely into this somewhat later on when I have more time. interestingly, my original estimate ($sin(x) = sin(5) / sin(23.5$)) turned up - if I get your answer correctly - to be correct (thought the estimate erred in 2 things that somewhat magically canceled each other) . My final edit however produced somewhat different result. I took  0.087266 for "inc", and 0.41 radians for "oe"; it resulted in little higher  ~0.0018 deg than expected. It might be approx stuff, or maybe I have an error in my last edit. >  
> **username_0**: > I took more than 5 digits after the decimal point when I did the calculation in WA. So it should not amount to 0.002 degrees diff (even its little more than this). Moreover my expression looks like much complex than the answer. maybe they are indeed equivalent. Will look at this later (which unfortunately might be some days from now).  
> **username_0**: Well, I'm now convinced the expressions are not equivalent.  I took "oe"=30 (hence its sin is 0.5) and "inc"=10. [here calculation for alpha](https://www.wolframalpha.com/input/?i=Derivative+arctan%28sin%28pi%2F18+%2B+arcsin%28sin%28a%29*0.5%29%29+%2F+%28cos%28a%29*0.5%29%29+-+a+%3D+0) and [here the result](https://www.wolframalpha.com/input/?i=arctan%28sin%28pi%2F18+%2B+arcsin%28sin%28a%29*sin%280.523598775%29%29%29+%2F+%28cos%28a%29*sin%280.523598775%29%29%29+-+a+where+a+%3D-0.3207669328): 20.33696  where we expect 20.322037. This large gap of 0.01 deg.  
> **username_1**: @d_e It's probably just approximation errors. The equation $$\sin(RA) = \pm\sin(inc) / \sin(OE)$$ is correct.  
> **username_1**: FWIW, I used a combination of Sage & hand calculation to find the maximum of RA. Sage can do algebra, including symbolic differentiation, but it's not very good at simplifying trig expressions. ;)  
> **username_1**: Annoyingly, the Sage 3D graphics rotation methods `.rotateX`, `.rotateY`, and `.rotateZ` rotate clockwise around the axis, which is the opposite convention to how the standard right-handed rotation matrices work. That made things a little confusing while I was working on my animation code.  

---

### Answer by username_2 on 2021-12-02. Score: 4

**Interaction of the Planes**

The orbital plane of the Moon maintains a relatively constant inclination to the ecliptic, which currently is about $5^{\large\circ}09'$, and the nodes precess along the ecliptic at a fairly constant rate. The equator of the Earth is inclined about $23^{\large\circ}27'$ to the ecliptic.

In the following animation, the various planes are intersected with a sphere: the ecliptic is in red, the Earth's equator is in green, and the Lunar orbital plane is in blue:

[![enter image description here](https://i.sstatic.net/dx7WX.gif)](https://i.sstatic.net/dx7WX.gif)

The following animation shows the locations of the north ecliptic pole in red, the north celestial pole in green, and the north pole of the Lunar orbit in blue, and how they affect the location of the lunar ascending node along the celestial equator.

[![enter image description here](https://i.sstatic.net/ErkPx.gif)](https://i.sstatic.net/ErkPx.gif)

---

**Spherical Trigonometry**

Let's look at the geometry of the triangles on the sphere. Let $E$ be the north ecliptic pole, $C$ the north celestial pole, $L$ the north pole of the Lunar orbit. Furthermore, let $F$ be the intersection of the equator and the ecliptic, that is, the First Point of Aries, and $A$ be the ascending node of the Lunar orbit on the equator.

[![enter image description here](https://i.sstatic.net/Nq5Um.png)](https://i.sstatic.net/sWEcj.png)

Since an equator is $90^{\large\circ}$ from its pole, each of the arcs $LA$, $CA$, $EF$, and $CF$ is $90^{\large\circ}$. The Law of Cosines then gives that
$$\require{cancel}
\begin{align}
CL&=CAL\tag{1a}\\
CE&=CFE\tag{1b}\\
FA&=FCA\tag{1c}
\end{align}
$$
Explanation:  
$\text{(1a)}$: $\cos(CL)=\overset{0}{\cos(LA)}\,\overset{0}{\cos(CA)}+\overset{1}{\sin(LA)}\,\overset{1}{\sin(CA)}\cos(CAL)$  
$\text{(1b)}$: $\cos(CE)=\cos(EF)\cos(CF)+\sin(EF)\sin(CF)\cos(CFE)$  
$\text{(1c)}$: $\cos(FA)=\cos(CF)\cos(CA)+\sin(CF)\sin(CA)\cos(FCA)$

The Law of Sines then says that
$$
\begin{align}
LCA&=90^{\large\circ}\tag{2a}\\
ECF&=90^{\large\circ}\tag{2b}
\end{align}
$$
Explanation:  
$\text{(2a)}$: $\frac{\sin(LCA)}{\sin(LA)}=\frac{\sin(CAL)}{\sin(CL)}=1$  
$\text{(2b)}$: $\frac{\sin(ECF)}{\sin(EF)}=\frac{\sin(CFE)}{\sin(CE)}=1$

Equations $\text{(2a)}$ and $\text{(2b)}$ show that
$$
\begin{align}
LCA&=ECF\tag{3a}\\
LCF+FCA&=ECL+LCF\tag{3b}\\
FCA&=ECL\tag{3c}
\end{align}
$$
$\text{(1c)}$ and $\text{(3c)}$ then give
$$
FA=ECL\tag4
$$
[![enter image description here](https://i.sstatic.net/u9sOa.png)](https://i.sstatic.net/ZJHU5.png)

For a general spherical triangle, $ECL$,
$$
\begin{align}
\tan(ECL)
&=\frac{\sin(EL)\sin(LEC)}{\cos(EL)\sin(EC)-\sin(EL)\cos(EC)\cos(LEC)}\tag{5a}\\
&=-\frac{\sin(EL)\sin(CEL)}{\cos(EL)\sin(EC)-\sin(EL)\cos(EC)\cos(CEL)}\tag{5b}
\end{align}
$$
Plugging in $EL=5^{\large\circ}09'$ and $EC=23^{\large\circ}27'$ gives the following graph of the right ascension of the lunar ascending node in minutes, where $1^\text{m}=\frac14{}^{\large\circ}$:

[![enter image description here](https://i.sstatic.net/3mISU.png)](https://i.sstatic.net/YR27x.png)

The extremes on the graph occur when $CL$ is tangent to the circle traversed by the pole of the Lunar orbit. At that point,
$$
\begin{align}
\sin(ECL)&=\frac{\sin(EL)}{\sin(EC)}\tag{6a}\\
ECL&=52^\text{m}09^\text{s}\tag{6b}\\[4pt]
&=13^{\large\circ}02'\tag{6c}
\end{align}
$$
and
$$
\begin{align}
\cos(CEL)&=\frac{\tan(EL)}{\tan(EC)}\tag{7a}\\
CEL&=78^{\large\circ}00'\tag{7b}
\end{align}
$$
Since $78^{\large\circ}$ is less than $90^{\large\circ}$, it takes a bit less than $8.1$ years for the ascending node to go from its furthest west to its furthest east point, whereas it takes a bit more than $10.5$ years to return to its furthest west point.

The [Lunar Standstills](https://en.wikipedia.org/wiki/Lunar_standstill) occur when the graph above crosses $0^\text{m}$, that is also when the poles align, and so they occur $9.3$ years apart.

> **Comments:**
> **username_0**: Thank you very much for this elaborate answer.  

---

### Answer by username_1 on 2021-11-15. Score: 1

I'm leaving this old answer here because the OP found it useful. But please see my new answer. ;)

---

The Moon's ascending node progresses with (mostly) retrograde motion around the ecliptic. It's customary to specify it in terms of its ecliptic longitude, but we can easily convert it to equatorial coordinates using the [standard spherical trigonometry equations](https://en.wikipedia.org/wiki/Astronomical_coordinate_systems#Equatorial_%E2%86%94_horizontal). Since the ecliptic latitude of a point on the ecliptic is zero, those equations simplify to:

$$\tan\alpha = \frac{\sin\lambda\cos\varepsilon}{\cos\lambda}$$
$$\sin\delta = \sin\lambda\sin\varepsilon$$
where  
$\alpha$ is right ascension  
$\delta$ is declination  
$\lambda$ is ecliptic longitude  
$\varepsilon$ is the obliquity of the ecliptic

We can obtain the ecliptic longitude of the Moon's ascending node from [JPL Horizons](https://ssd.jpl.nasa.gov/horizons/), via an osculating elements query.

[The JPL Planetary and Lunar Ephemerides DE440 and DE441](https://iopscience.iop.org/article/10.3847/1538-3881/abd414) offers this formula for the obliquity of the ecliptic, which can also be found on [Wikipedia](https://en.wikipedia.org/wiki/Axial_tilt#Short_term):

$$\varepsilon = 84381.448 - 46.815T -0.00059T^2 + 0.001813T^3$$

where $\varepsilon$ is in arc-seconds, and $T$ is the time in Julian centuries since the [J2000.0 epoch](https://en.wikipedia.org/wiki/Epoch_(astronomy)#Julian_dates_and_J2000), JD=2451545.0, 2000-Jan-01 12:00:00 TT.

This [Sage / Python script](https://sagecell.sagemath.org/?z=eJx1V3tzm0gS_59PMUuyC9gSBj2yiuqoW9mLb521LZWl3FXOUVFjMZJwgCHMYEtJ5btf9wzoYedUtsT09PRrfv3ANE1yx2SZsCdGuFhUKZVJviIsZRnLJYmppGRZ8oz8xcvkG88FWfKSUCJpuWKSPPB4axD4XFc5BbpYsDxGATmPGXlipUh4rhj-UyZSspw8bMnkhnTukKnjdXzXh7--YYIhhpFkBS8lKVkL_r9WTEhhKO3LKl9IzlNBapa0rKIFXawZnIrZigSkSMgZ8QeeYbz50PE8z_UMVvDFGrY6vb7f7_Vd3CPjhzT5WiVyS_iSSBDAFmlSyGTRIklOShonNBfAd8nLDMKhvV9LWYjh2VnME5eXqzPfc7uD3u9nfr87aHcHA_-MPsQ9vwe2LAlvNNiPsTNU3s_ACliRNlE2OWBq912_01ebbwgtF4IthFo9IavtuwO_y9rdkxmc6bvvWbvnqOfeO9jp4_MpGfS6A9_t9Qa1mMZ4XJVMVmUO0k50ZN71BhAUQxnYeByV1E553iKc1XaWFLTDnecdWyQ5bjogYMGFDSwt9YA051AFnDklnRPQkixx8Q_iAX4Eg2fDqMoUJJpN_ISI3ccidXMqqLviT2e0SM7WNbSiZZIyFygIhQcqWLTIYjxtmsbN6O8wCid_hTfBp3BqqKdo9mkSBuF1eBPezqbGBXyHd4HV97w_uu_fW8ZdeBlNP01ncMa6uri71JTJ9eg2DKzw4vpqMru6sIzxx1n08fZqNg2sv2_aU8sIr2-i69F5eA0UUGYZs4lWZY3Op-Prj7PQMi6m_44ux3c3o1nNMz7_EP05mo0C63Zs1XD-Y4dSO6MbkXxjQZJLuwvxw2tYMrlYR5hitoB8ki0iJC_wmxX1fegALDECv7x9Ox1fGt-bwPwwLsY3N6PbP4HB6nq-ZUxno7tZNLu6CZH0Xcn8geTx5JDKC0UMJ9H06r8NkSERVISgQhuvb_gr7Dep6BZcSBtutKXqQvDdglqQUWkNiSXZRlo_WgSvUMBOkhcVbtgWmGq10BHnh4ZNpkS6ggHq13Zp2v8Uzue3n8G70HZP1GM4Dk1VAlyUq48BuDKSCHLLc6aDg5-ixIiat2OiIBHeXU1N58XusZwD5KIonTo_Y6yJFvnVghzoepqaJjkTOiiK1xVFmkhFtTVHbdLn3HQfOSYR7t33h_67-XHiZO6q5FVh-869P5wTnZsi5QWzC69FCr8GgTmDKqXoTclCkbowFR6RHFhNxbmBY1sPi6Gn1z6sfVz7h4ptoLWBEcuQvcHnDTxDcVyztLD_VdJinSyEAwAGT1hJF1KZllHwRqEqMKHCvmt_qPJ2p2Nq3CKt029_oHnbe29qFAfm7yQ2W6AbAMMUZQDg4VIEs7KCKr-oyifWPPOUl4G5KhnL1RmVMANw6TmJ5Trw4Uk8rYJLCsWlRWgleVQVKFiT6mgBTnIuiY1aCPQqpcJ5hZgptLiFhFpHUkaFJAAFjO7RKfMVYnRaqK4YvMpfV0AjLWxHx-NwxXarHZaVjP8LZw59Yg0tuAQO3YN5lce_vDLo0Bz8eY3GevNe6C3balnO_bDtz1UjF9jzkGWuIY-0kj831GGTQfz53pu3cOu-N3fqMLwhE9wlynnym_KafKhS6EFweiuApK79MCsuAbRDQMcJigeZ98PO_ChtZnwIiG8YwE7FUWuUEZQg9Gbe-MYOlmh80mrsZ3kF8QMOGyUdAOANCTcSQU0WNIVxBeYWFLTbByquMcfBXX9e6z4-uncTvISWuEpkFe_y83gO2h1_jKHVZmylLmSZcqojW4F_aHuFZtuA8r4zP1B5wXMYpOSuax-og9S_G-1YOdp8NHzs0YIqD9u-NuOEwOyk2j9UAnjc8atAQysuwA3bRrtL6jgHkQAEJ-TXXVqTICDePsT42arWVQezBh8UuvkxU4tkUBCAEdhrJrN9APPdNR8ZE59mp9v7znC-sylmqYZ6A81RUZR8k2R4kzDPrGCa1UCUK6nyQdXZk0KHvsDQf4MEVY63tP9Kwfwnh_Qu4nIOow_unb7Ya6ujO2POqySNyTn7lrC6tNS2FDjtod4T9aUOQ8UAkToEUhX0FnnCop5Du7HhiN7SYvboR9jDwtZHarvf0JLh1Xu7BIEpA4L-hP6CqP2VxeighFYgvT1tg2VF4rC6p6E58RY2wCbAz6alZkz4PcoSzJHaXzRhv6cMOoURF_rUKdnCvw0iQW-8RRB2cN5UKjuHWFPOHuML_HUhIIgJdBoEoS1KKgiEINrojaJp6UiBhTM_RpeOYwMvkOq8VLOP8asraaLZXPRUlUIsASvsokhUOxPgbfqqve8B0Gz2Tk0wLAVMDFI0MFQdsN-0RvW9O_wiJOr0g4p4VFC5trVfx2drO788w9sbQidOFtImtYxVmcSqcwT39xWU5n1NUgkI5d_M6CMvTXhaljSr-3Z9mm6YiFL6AJN_cG_uyyOUchMqCx46YImUZ_jmhyPjSq3wyyH7kLkRw2IbfXmOhasbva0t1zFoEj2OD9rMNkvgfWYL4za4N3FxiS0Yn-gGnjSTqjfAAuBTv211DqEHr439Xa5gqcEy1oTgRbRx_LNNYp50EckZMKvihEhbcrhFdMoUGU1Ts3WAqXrOeUjp4gtEB2t7AmUyommyyvGtOzChe8DcZe4b7Sh-rGBIoaLAoQW6WsL1gKfc3YAPrZeub3aub7TrPwmHklELhZTc6FhsVCzOjmNT34lgMtIHImWF7bkDCFst40yPao3dgFIY2A6i5gr6xGyreeFzYdfap1uMbwYrUBAn0AnoNspoTlestJ2DuAfwLp6ldozDs86nCAfhSEnWClrkO7yJWEr6YeDhjNwWkOvjSsLbyVUGwqdPKwcZXdBrO27MFtCw7V1bEWv-bE-c_wGx6dzS&lang=sage) fetches the necessary data from Horizons, converts to RA (in degrees), and plots the results as a graph. I chose the time interval

From: 2453908.5 A.D. 2006-Jun-22 00:00:00  
To: 2460684.5 A.D. 2025-Jan-09 00:00:00  
with time steps of 7 days
because it's an almost complete precession cycle.

[![RA of Moon's node](https://i.sstatic.net/NqGRj.png)](https://i.sstatic.net/NqGRj.png)

As you can see, the right ascension of the node covers the full range from 0° to 360°.

Here's the corresponding graph of the declination.

[![Declination of Moon's node](https://i.sstatic.net/tBDUK.png)](https://i.sstatic.net/tBDUK.png)

---

You can easily use the script to create plots covering other intervals: Horizons can supply the necessary data for any time interval from 9999 BC to 9999 AD. However, the formula for the obliquity is probably not valid at the extremes of that range.

---

Here's a new version of my [interactive orbit plane anim script](https://sagecell.sagemath.org/?z=eJx1VMtu2zAQvOsrFskhpCsrkl3XaQEB9aE9-ZBDDkGLIKBl2iLKkCpFOfbfd_mQbQWqIAjmcnZ2dzj0gRlyd0xPdzTZ8j2U0Ai4h-IhT7RxS5Kn4F6aJLewAiuk5VuohKkkh3dha9DG1nqvFZPAjrxFmp3fJ3Xq4SnshJT0WwL4VAVy1h13m3n24D807MwuOwW-IfqIwVCMYD9-x-w3lZbalBUumGxqVubZ11Cl9KVi5qcSpFB8viW_4xTTgqb9RAV9ueaajWVNizg8ZvU_x7IMt51R8JgZbZnlz8TNjYp9F8pywyrrRXn14XKRFZ9TaC1v2nKOlLySorGiKn8y2XJc_-2YRfIn011WgskY8GNGKOusfu2aLdYMoaiyqwOTEvBEk7OMjcZucCyvY-j_ZiNZ9ecGSQ176yl8wi38iG1BI5niPih252ZDnbNiH45owE6H0EZq1wUOTo7uRLIFpmQLJ_FpuI4st8vlElvUDauEPeFZL8b7ZejDK7mu-tYcx5_NswVMvCT9LAE7MgrJsy9YMapNk1hi3Slm0O8bYQf8azinFQPPD_KU3roswyvetkKrKFrrQV0KB8ei2-AdNIhQ0UYfVe7N3CHqMCr2CPr_YPTsM5aeokoXUZyAHNWF3GNWCGBKvKHR8FogPayj13-R2QT_MiYg7r2lKT1fAkdMYacNCBAKWsPUnpOAegm1V1l_Q8SBE5q1tX4n9B9DQEvZ&lang=sage) which makes it easier to see that the Moon's RA is indeed close to 0° when its declination is 0° (and going from south to north of the celestial equator). Setting the `equatorial` checkbox rotates the system so that the equatorial plane is horizontal.

> **Comments:**
> **username_3**: I believe the original question is this: Where does the Moon's orbit intersect the celestial equator? If that is the question, your answer does not provide that answer. The "ascending node" of the plane of the Moon's orbit and the celestial equator is always near 0 hour RA.  
> **username_1**: @JohnHoltz Ah. That would explain the OP's last comment. :) But I'll leave this answer here for now, in case it has some useful info for the OP or other readers.  
> **username_1**: I guess I could do a plot of the Moon's RA when its declination is 0°. But that will have to wait for another day...  
> **username_0**: Thanks for that demonstration. It shows how this precession of the node versus the ecliptic looks like in the equatorial coordinate system.  My question, however, as JohnHoltz says is about other nodes, namely the nodes of intersection with the celestial equator. Sorry for not explaining this well enough previously.  
> **username_1**: @d_e Sorry for misunderstanding, but I'm glad that this answer isn't totally useless in its current state. :)   I've just added a new version of the anim script to this answer. I'll have to think about the answer to your real question.  
> **username_0**: @PM2Ring, wonderful. I've also just edited my question added further thoughts.  
> **username_1**: @d_e Here's a [script](https://sagecell.sagemath.org/?z=eJyNT8FuwyAMvfMVVi-FysmSdJ2mSnxCL7tO0YRS1jIlkBDaZPv6GdROa3cZMhi_Z_weZ-X50uCEA4alYGyv3-Fl5uEotgxoeR1O3kKngjczT9BrOimXCAVFjVeAqsaN8THCaGy83JDZBfxpqxNZX3W__qd7r3Hj4Y_GvUOK8rfuAST0Bh6gfC7Yjgr6_iBgFe1MKc_cCDYjfCK8Eb_LG9eeOssrwVTbHxVhKqrGrYKyFc9itxCCppIAY05TT7XONzQuAsY2BGzy8vEC9K0LPA3jRhKLMMmwIgZhkJnT9Asekvf1U0HFwZt9a6we5aJTH84vxDfXbHXa&lang=sage) which plots a graph of the variation in RA.  
> **username_0**: @PM2Ring, Cool! We can see the max value is at around $\alpha=$78deg (as I found above, just need to add +90 to my $\alpha$ because the way I defined it). Thanks for your plots.  

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