<p>Generally, using declination and right ascension to calculate ecliptic latitude and longitude is a good idea, but if you want to see a potential way to do it without using them, consider this:</p>
<p>Consider a geocentric frame where the observer is in the center of the Earth, the moon is orbiting around the Earth in a circular orbit with a radius of <span class="math-container">$3.84\times10^8 m$</span>. (this is a highly simplified model and does not really match the reality)</p>
<p>The moon revolves around the Earth in a orbit with inclination, <span class="math-container">$i = 5.14^{\circ}$</span>, the orbit of Moon will trace a great circle on the celestial sphere (now call it <em>lunar great circle</em>), with a angle of <span class="math-container">$i' = 5.14^{\circ}$</span> to the orbit of Sun (ecliptic). Thus the two great circle have two point where they meet.</p>
<p>Now we establish a coordinate system with <em>lunar great circle</em> as its great circle, it is just like any other spherical coordinate system. For simplicity, define lunar longitude as the angular separation between the ascending node and the projection of the object on the <em>lunar great circle</em>. (basically equatorial system but changing the great circle and ascending node)</p>
<p>Since the Moon is always on <em>lunar great circle</em>, its <em>lunar longitude</em> is <span class="math-container">$0^{\circ}$</span>, the Moon orbit around Earth in a circular orbit, the motion of the Moon on the <em>lunar great circle</em> is a uniform motion. Thus at any given moment, the <em>lunar longitude</em> of the moon is:</p>
<p><span class="math-container">$\lambda' = \frac{t}{T_0} \times 360^{\circ}$</span></p>
<p>Where <span class="math-container">$t$</span> is the time after the last time where Moon pass through the ascending node, <span class="math-container">$T_0$</span> is the orbital period of the Moon.</p>
<p>Now you can use spherical trigonometry to convert coordinate system, it is the same method you use to obtain an equation of Moon's ecliptic latitude and longitude to get declination and right ascension.</p>
<p>You can make it more accurate by adding <span class="math-container">$\omega(t)$</span>, this is a function of angular velocity on celestial sphere, and integrate it over time to get the distance Moon travels on the celestial sphere.</p>

CC BY-SA 4.0
2024-07-02T16:32:53.990Z
How does this simple model help the OP to calculate the Moon's actual position?
PM 2Ring
CC BY-SA 4.0
2024-07-02T17:40:44.547Z
@PM 2Ring, Spherical trigonometry, i define a new coordinate system, give the way to compute moon's coordinate, all that is left is to use spherical trigonometry
Polaris5744
CC BY-SA 4.0
2024-07-02T15:56:16.310Z
2024-07-02T15:56:16.310Z
Polaris5744
<p>Is there a way to compute the Moon's ecliptic latitude and longitude without declination or right ascension?</p>
<p>Note: I have a formula to compute the Moon's position that uses an ecliptic latitude and longitude to compute declination or right ascension so that's why I can't use them to compute ecliptic latitude and longitude.</p>

CC BY-SA 4.0
2024-07-02T15:14:42.443Z
As mentioned in the comments on [your earlier question](https://astronomy.stackexchange.com/q/57873/16685), there isn't a simple way to calculate the Moon's position with high accuracy. The Moon's orbit is [complicated](https://astronomy.stackexchange.com/q/55048/16685)! See [here](https://astronomy.stackexchange.com/a/50974/16685) for other complicated lunar patterns.
PM 2Ring
CC BY-SA 4.0
2024-07-03T04:10:53.090Z
Please don't use so much bolding. It makes the text hard to read.
Nilay Ghosh
CC BY-SA 4.0
2024-07-04T02:38:24.283Z
We need to know this information.  A) How accurate do you want the position?  B) what time frame is acceptable?  For example,  the Astronomical Almanac gives equations for each year, to within X fraction of a degree.
JohnHoltz
<the-moon><mathematics><ecliptic>
Moon's ecliptic latitude and longitude
Ahmed Dyaa
# Moon's ecliptic latitude and longitude

(Asked by: username_0 on 2024-07-02)

Is there a way to compute the Moon's ecliptic latitude and longitude without declination or right ascension?

Note: I have a formula to compute the Moon's position that uses an ecliptic latitude and longitude to compute declination or right ascension so that's why I can't use them to compute ecliptic latitude and longitude.

**Tags:** the-moon, mathematics, ecliptic
> **Comments:**
> **username_1**: As mentioned in the comments on [your earlier question](https://astronomy.stackexchange.com/q/57873/16685), there isn't a simple way to calculate the Moon's position with high accuracy. The Moon's orbit is [complicated](https://astronomy.stackexchange.com/q/55048/16685)! See [here](https://astronomy.stackexchange.com/a/50974/16685) for other complicated lunar patterns.  
> **username_2**: Please don't use so much bolding. It makes the text hard to read.  
> **username_3**: We need to know this information.  A) How accurate do you want the position?  B) what time frame is acceptable?  For example,  the Astronomical Almanac gives equations for each year, to within X fraction of a degree.  

---

## 1 answers

### Answer by username_4 on 2024-07-02. Score: 0

Generally, using declination and right ascension to calculate ecliptic latitude and longitude is a good idea, but if you want to see a potential way to do it without using them, consider this:

Consider a geocentric frame where the observer is in the center of the Earth, the moon is orbiting around the Earth in a circular orbit with a radius of $3.84\times10^8 m$. (this is a highly simplified model and does not really match the reality)

The moon revolves around the Earth in a orbit with inclination, $i = 5.14^{\circ}$, the orbit of Moon will trace a great circle on the celestial sphere (now call it *lunar great circle*), with a angle of $i' = 5.14^{\circ}$ to the orbit of Sun (ecliptic). Thus the two great circle have two point where they meet.

Now we establish a coordinate system with *lunar great circle* as its great circle, it is just like any other spherical coordinate system. For simplicity, define lunar longitude as the angular separation between the ascending node and the projection of the object on the *lunar great circle*. (basically equatorial system but changing the great circle and ascending node)

Since the Moon is always on *lunar great circle*, its *lunar longitude* is $0^{\circ}$, the Moon orbit around Earth in a circular orbit, the motion of the Moon on the *lunar great circle* is a uniform motion. Thus at any given moment, the *lunar longitude* of the moon is:

$\lambda' = \frac{t}{T\_0} \times 360^{\circ}$

Where $t$ is the time after the last time where Moon pass through the ascending node, $T\_0$ is the orbital period of the Moon.

Now you can use spherical trigonometry to convert coordinate system, it is the same method you use to obtain an equation of Moon's ecliptic latitude and longitude to get declination and right ascension.

You can make it more accurate by adding $\omega(t)$, this is a function of angular velocity on celestial sphere, and integrate it over time to get the distance Moon travels on the celestial sphere.

> **Comments:**
> **username_1**: How does this simple model help the OP to calculate the Moon's actual position?  
> **username_4**: @PM 2Ring, Spherical trigonometry, i define a new coordinate system, give the way to compute moon's coordinate, all that is left is to use spherical trigonometry  

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