4
$\begingroup$

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 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 - 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 at $\alpha = -11.7819$, which in turn produced $x=12.6312$ and $z=22.97$.

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.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – PM 2Ring
    Nov 14, 2021 at 23:14
  • $\begingroup$ @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 $\endgroup$
    – d_e
    Nov 14, 2021 at 23:22
  • $\begingroup$ The ecliptic longitude of the lunar ascending node goes (mostly) retrograde from 360° to 0°, so its RA must range from 24h to 0h. $\endgroup$
    – PM 2Ring
    Nov 14, 2021 at 23:25
  • $\begingroup$ @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. $\endgroup$
    – d_e
    Nov 14, 2021 at 23:31
  • 1
    $\begingroup$ You might find it helpful to explore the interactive anim I linked in my answer to the previous question. $\endgroup$
    – PM 2Ring
    Nov 14, 2021 at 23:54

3 Answers 3

4
+50
$\begingroup$

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 of the normals of the planes. We can find those normals using rotation matrices. 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 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

Here's the single plot for $i=5.14°$ RA vs omega for i=5.14°

And here's the Sage / Python script that made that plot.

Lastly, here's an interactive anim 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.

$\endgroup$
6
  • $\begingroup$ 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. > $\endgroup$
    – d_e
    Nov 20, 2021 at 15:51
  • $\begingroup$ > 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). $\endgroup$
    – d_e
    Nov 20, 2021 at 15:58
  • $\begingroup$ 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 and here the result: 20.33696 where we expect 20.322037. This large gap of 0.01 deg. $\endgroup$
    – d_e
    Nov 20, 2021 at 16:45
  • $\begingroup$ @d_e It's probably just approximation errors. The equation $$\sin(RA) = \pm\sin(inc) / \sin(OE)$$ is correct. $\endgroup$
    – PM 2Ring
    Nov 20, 2021 at 21:56
  • $\begingroup$ 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. ;) $\endgroup$
    – PM 2Ring
    Nov 20, 2021 at 22:00
4
$\begingroup$

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

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


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

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

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

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 occur when the graph above crosses $0^\text{m}$, that is also when the poles align, and so they occur $9.3$ years apart.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you very much for this elaborate answer. $\endgroup$
    – d_e
    Dec 2, 2021 at 22:57
1
$\begingroup$

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. 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, via an osculating elements query.

The JPL Planetary and Lunar Ephemerides DE440 and DE441 offers this formula for the obliquity of the ecliptic, which can also be found on Wikipedia:

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

where $\varepsilon$ is in arc-seconds, and $T$ is the time in Julian centuries since the J2000.0 epoch, JD=2451545.0, 2000-Jan-01 12:00:00 TT.

This Sage / Python script 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

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


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 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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – JohnHoltz
    Nov 15, 2021 at 4:39
  • $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Nov 15, 2021 at 4:49
  • $\begingroup$ 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... $\endgroup$
    – PM 2Ring
    Nov 15, 2021 at 4:53
  • 1
    $\begingroup$ @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. $\endgroup$
    – PM 2Ring
    Nov 15, 2021 at 15:59
  • 1
    $\begingroup$ @PM2Ring, wonderful. I've also just edited my question added further thoughts. $\endgroup$
    – d_e
    Nov 15, 2021 at 16:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .