<p>Below is a derivation of the expected distribution of the Moon's ecliptic latitude over time.  We make the following simplifying assumptions:</p>
<ol>
<li>The Moon's orbit is approximately circular.</li>
<li>The precession of the moon's orbit can be ignored.</li>
<li>The Sun's ecliptic longitude is uncorrelated with the Moon's ecliptic longitude.</li>
</ol>
<p>In particular, assumption #3 effectively says that the New Moon occurs at &quot;random times&quot; during the Moon's orbit.  So if we find the distribution of the Moon's ecliptic latitude over a full orbit, it will follow the same distribution as the Moon's ecliptic latitude at the moment of New Moon.</p>
<p>If it wasn't obvious from the above, we will work in ecliptic coordinates with the origin at the Earth's center, with the primary direction rotating with the line of nodes of the Moon's orbit.  (I don't think this violates assumption #2 above, but I welcome corrections on this matter.)  The <span class="math-container">$x$</span>-direction points along the line of nodes and the <span class="math-container">$z$</span>-direction points towards the ecliptic pole (i.e., the ecliptic plane is the <span class="math-container">$xy$</span>-plane.</p>
<p>Under these assumption, define <span class="math-container">$\hat{n}$</span> to be a unit vector normal to the plane of the Moon's orbit:
<span class="math-container">$$
\hat{n} = (0, \sin \alpha, \cos \alpha)
$$</span>
where <span class="math-container">$\alpha \approx 5.14^\circ$</span> is the inclination of the Moon's orbit to the ecliptic plane.  Let <span class="math-container">$\hat{v}$</span> be a unit vector pointing towards the Moon's position on the sky at a time <span class="math-container">$t$</span>.  This vector must make an angle of 90° with <span class="math-container">$\hat{n}$</span> and an angle of <span class="math-container">$\omega t$</span> with the <span class="math-container">$x$</span>-axis;  in other words, <span class="math-container">$\hat{n}\cdot\hat{v} = 0$</span> and <span class="math-container">$\hat{x}\cdot \hat{v} = \cos \omega t$</span> (assuming that <span class="math-container">$t = 0$</span> corresponds with the moon crossing the line of nodes.  A bit of algebra shows us that we have
<span class="math-container">$$
\hat{v} = (\cos \theta, -\sin \theta \cos \alpha, \sin \theta \sin \alpha),
$$</span>
where <span class="math-container">$\theta \equiv \omega t$</span>.</p>
<p>It can then be shown that the Moon's ecliptic latitude <span class="math-container">$\lambda$</span> satisfies <span class="math-container">$\cos(\pi/2 - \lambda) = \hat{v} \cdot \hat{z} = \sin \theta \sin \alpha$</span>, or
<span class="math-container">$$
\sin \lambda = \sin \theta \sin \alpha.
$$</span></p>
<p>Now, the probability distribution of <span class="math-container">$\theta$</span> over one orbit is
<span class="math-container">$$
\mathcal{P}(\theta) \, d \theta = \frac{d\theta}{2\pi}
$$</span>
and so the probability distribution with respect to <span class="math-container">$\lambda$</span> is given by
<span class="math-container">$$
\mathcal{P}(\lambda) = 2 \mathcal{P}(\theta) \left| \frac{d\theta}{d\lambda} \right| = \frac{1}{\pi} \left| \frac{d\theta}{d\lambda} \right|
$$</span>
where the factor of 2 comes from the fact that there are two possible values of <span class="math-container">$\theta$</span> for every value of <span class="math-container">$\lambda$</span>.  Taking the appropriate derivative, we find that
<span class="math-container">$$
\frac{d \theta}{d \lambda} = \frac{d}{d\lambda} \left[ \sin^{-1} \left( \frac{\sin \lambda}{\sin \alpha} \right) \right] = \frac{\cos \lambda}{\sqrt{ \sin^2 \alpha - \sin^2 \lambda}}
$$</span>
and so
<span class="math-container">$$
\boxed{ \mathcal{P}(\lambda) = \frac{\cos \lambda}{\pi \sqrt{ \sin^2 \alpha - \sin^2 \lambda}}.}
$$</span></p>
<p>This distribution has the shape found in <a href="https://astronomy.stackexchange.com/a/53729/10004">novotny's answer</a>;  it is relatively flat in the region near <span class="math-container">$\lambda = 0$</span> but diverges sharply in the region where <span class="math-container">$\lambda \to \pm \alpha$</span>.  Physically, this is because when the moon is near the extremes of its range of ecliptic latitude, its ecliptic latitude is not changing very fast, and therefore it spends quite a lot of time at such values of <span class="math-container">$\lambda$</span>.  Conversely, when it is near the region where <span class="math-container">$\lambda \approx 0$</span>, its ecliptic latitude is changing relatively quickly, and so it will not spend a lot of time in such regions, leading to a lower probability of being observed at such latitudes.</p>

CC BY-SA 4.0
2023-05-26T12:53:30.293Z
You should be computing $d\lambda/d\theta$ as $\theta$ is the independent variable here. The $\sqrt{\sin^2\alpha - \sin^2\lambda}$ factor in the denominator doesn't quite make sense and leads to an infinity at $\lambda=\alpha$.
David Hammen
CC BY-SA 4.0
2023-05-26T13:46:08.060Z
@DavidHammen:  To go from $\mathcal{P}(\theta)$ to $\mathcal{P}(\lambda)$ you need to take the derivative of the inverse function $\theta(\lambda)$, which is (in this case) $\theta(\lambda) = \sin^{-1}(\sin\lambda/\sin\alpha)$.  The mathematical details can be found [here](https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables), though it's a bit jargony. ...
Michael Seifert
CC BY-SA 4.0
2023-05-26T13:46:40.937Z
... As to the divergence in $\mathcal{P}(\lambda)$, it's a probability density, and has to be integrated over a range of $\lambda$ to get a measurable value.  So for example, the probablity of finding $\lambda$ between $\lambda_0$ and $\alpha$ works out to be$$\int_{\lambda_0}^\alpha\mathcal{P}(\lambda)\, d\lambda = \left[\frac{1}{\pi}\sin^{-1} \left( \frac{\sin\lambda}{\sin \alpha} \right)\right]_{\lambda_0}^\alpha$$ which works out to be finite even though the integrand is divergent.  Effectively, the divergence is sufficiently "weak" that the distribution is still integrable.
Michael Seifert
CC BY-SA 4.0
2023-05-25T15:34:13.063Z
2023-05-25T23:59:19.707Z
2023-05-25T23:59:19.707Z
Michael Seifert
<p>Short Answer: why don’t you ask a Muslim? The Islamic calendar is lunar-centric, and New-moon timing is a big deal. The record for “Newest-moon” Observations is typically re-set by some group of Muslims. And they (usually) aren’t using any super-equipment with mega-funding, either.</p>
<p>Long(ish) Answer: the lunar theory includes the misalignment between the orbit plane and the ecliptic, but… the pull of the Sun and other planets causes that orbit to precess. Lunar theory thus includes the “major” and “minor standstills,” the cycle in which the orbit plane bobs about the ecliptic. In other words, an epicycle or metacycle. Thus, the criterion of ‘closer than Venus’ or not doesn’t have a single answer, but a sliding scale.</p>
<p>If you take an average over ~months, then the precession is a bit low and irrelevant. Take the current nodal state, and average together “X” number of “months” to get the declination. If, instead, you want a long-term or theoretically rigorous answer, then you need to take this (16 year? can’t remember off the top of my head) additional cycle into account as a higher-order term. (Plus the eccentricity, of course.)</p>

CC BY-SA 4.0
2023-05-24T15:43:04.673Z
2023-05-24T15:43:04.673Z
caInstrument
<p><strong>Mean angular separation of about 3.2°, based on <a href="https://rhodesmill.org/skyfield/" rel="nofollow noreferrer"><code>skyfield</code></a> Python library simulation of the 2nd millennium.</strong></p>
<p>This is pretty dependent on your viewing location. The moon's  orbit is inclined roughly 5° to the ecliptic and goes through processive and nutation cycles on a timescale a couple decades.</p>
<p>For the purposes of this answer, I assumed the viewpoint from the center of the Earth, and ignored atmospheric refraction.</p>
<p>I used the following skyfield script to look between 1001-01-01 00:00 UTC and 2001-01-01 UTC for New Moons, and calculate  the angular separation between the Sun and the Moon from the center of the Earth.</p>
<pre><code>import numpy
import pytz
from datetime import datetime
import matplotlib.pyplot as plt
from skyfield import api, almanac


def main():
    time_zone = pytz.timezone(&quot;UTC&quot;)

    ephemeris = api.load(&quot;de422.bsp&quot;)
    timescale = api.load.timescale()

    start_ts, end_ts = timescale.utc(
        [time_zone.localize(datetime(1001, 1, 1)),
         time_zone.localize(datetime(2001, 1, 1))])

    phase_times, phase_types = almanac.find_discrete(
        start_ts, end_ts, almanac.moon_phases(ephemeris))

    new_moon_times = timescale.tt_jd([phase_time.tt for phase_time, phase_type
                                      in zip(phase_times, phase_types)
                                      if phase_type == 0])
    sun, moon, earth = ephemeris['sun'], ephemeris['moon'], ephemeris['earth']

    earth_center = earth.at(new_moon_times)
    separations = earth_center.observe(sun).separation_from(earth_center.observe(moon))

    print(f&quot;{len(separations.degrees)} New Moons found.\n&quot;
          f&quot;Minimum separation {min(separations.degrees):5.3f}°\n&quot;
          f&quot;Average separation {numpy.average(separations.degrees):5.3f}°\n&quot;
          f&quot;Maximum Separation {max(separations.degrees):5.3f}°\n&quot;)

    plt.hist(separations.degrees, edgecolor='black', range=[0, 6], bins=12)
    plt.title(&quot;Angular separation of Sun and Moon at New moon during 2nd millennium&quot;)
    plt.xlabel(&quot;Angular separation (degrees)&quot;)
    plt.ylabel(&quot;Number of New Moons&quot;)
    plt.show()


if __name__ == '__main__':
    main()
</code></pre>
<p>The following results were returned:</p>
<pre><code>12369 New Moons found.
Minimum separation 0.000°
Average separation 3.186°
Maximum Separation 5.011°
</code></pre>
<p>With the following histogram:</p>
<p><a href="https://i.sstatic.net/P6IFq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/P6IFq.png" alt="Angular Separation of Sun and Moon at New Moon during 2nd millennium" /></a></p>
<p>As Michael Seifert mentioned in the comments, the moon in general spends more time in its orbit near the upper range of the angular separations,  resulting in the spike of separations nearing 5° inclination.</p>
<p>That said, 5° is still fairly close visually, about the visual width of the three center fingers on an extended arm, though with angular radii of  about 0.25° each, as David Hammen mentions, probably the first two to four bins would be &quot;skirting the edges&quot; of the solar glare for a human observer.</p>
<p>Venus (maximum elongation 47°) is almost always going to be more angularly distant from the Sun than the New Moon. Lightly modifying the script, I'm finding only 504 of the 12369 New Moons in the 2nd millennium where Venus is angularly closer to the Sun viewed from the center of the Earth, a bit under 4.1% of the total.</p>

CC BY-SA 4.0
2023-05-19T11:16:56.183Z
Rather than comparing to Venus, it would be more useful to compare to the Sun. Given that the angular radii of the Sun and Moon are both about a quarter of a degree, the "skirting around the edges somewhere" cases would be your first two to four bins. The remaining bins are not "skirting around the edges somewhere" -- and that's the vast majority of the cases. Nice answer, BTW.
David Hammen
CC BY-SA 4.0
2023-05-19T14:04:06.093Z
This distribution makes a certain amount of sense.  When the moon is at its farthest above or below the ecliptic plane, it is not moving very fast towards/away from that plane, and so it spends more time at a high angular separation from the ecliptic plane.  Conversely, when the Moon is passing through the ecliptic plane, it is moving relatively fast to that plane and so spends less time at small angular separations from the ecliptic plane.
Michael Seifert
CC BY-SA 4.0
2023-05-19T14:09:51.020Z
If we assume a circularly orbiting moon with an orbital plane inclined at $\alpha$ relative to the ecliptic, and assume that the Moon's orbital period is incommensurate with the Earth's, then (if I've done my math right) the long-term distribution of angular separation $\theta$ at New Moon is $$\mathcal{P}(\theta) \, d \theta = \frac{\cos{\theta}}{\pi\sqrt{\sin^2\alpha - \sin^2\theta}} \, d\theta.$$This distribution has a flat region near small separations and a spike near maximum separation, consistent with your data.
Michael Seifert
CC BY-SA 4.0
2023-05-19T14:11:49.030Z
I should really turn the above into an answer.  Maybe next week after I get my grades in...
Michael Seifert
CC BY-SA 4.0
2023-05-20T17:07:20.910Z
@MichaelSeifert I'm getting something rather different: $P(\theta) = k \sin \alpha \cos \theta$ where $k$ is a normalizing factor, $\alpha$ is the ~5.1 degree lunar orbit inclination with respect to the ecliptic, and $\theta$ is an angle that over the long term is close to being equally distributed over the range 0 to $2\pi$.
David Hammen
CC BY-SA 4.0
2023-05-25T23:56:41.830Z
@DavidHammen:  I agree with you that the ecliptic latitude of the Moon $\lambda$ satisfies $\sin \lambda = \sin \alpha \cos \theta$, and that $\theta$ should be uniformly distributed between 0 and 2π.  If you then calculate what the distribution of $\lambda$ should be given those facts, you get the distribution I gave in my comment.  See [my answer below](https://astronomy.stackexchange.com/a/53793/10004) for a bit more of the process.
Michael Seifert
CC BY-SA 4.0
2023-05-18T22:10:03.557Z
2023-05-20T15:57:01.337Z
2023-05-20T15:57:01.337Z
notovny
<p>In a Solar Eclipse, the Sun, Earth and Moon are exactly colinear (well some area of the Earth where the eclipse is visible), and these are relatively rare.</p>
<p>But in addition, Sun, Earth and Moon are <em>approximately</em> colinear at every New Moon; indeed in a 2D projection it would look like there ought to be a Solar Eclipse every month.</p>
<p>This got me curious: if I could see the New Moon in the sky somehow, and at every New Moon I traced the minimum angular distance between the Sun and Moon, what would this typically be? Presumably this would be in fact 0 degrees at a full Solar Eclipse, but what about an &quot;average&quot; non-eclipsing New Moon?</p>
<p>In other words, how close is a typical New Moon to the Sun? Is it skirting around the edges somewhere, or a fair old distance etc. Would it be closer than Venus, say?</p>
<p>These questions touch on the subject but don't actually answer the angular distance part:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/questions/49480/what-is-the-difference-between-new-moon-day-and-lunar-eclipse">What is the difference between New Moon day and Lunar Eclipse?</a></li>
<li><a href="https://astronomy.stackexchange.com/questions/35698/why-cant-we-see-a-solar-eclipse-every-month">Why can&#39;t we see a solar eclipse every month?</a></li>
</ul>

CC BY-SA 4.0
2023-05-20T22:42:49.293Z
You wrote *indeed in a 2D projection it would look like there ought to be a Solar Eclipse every month*, but that depends on which 2D projection you make. If you projected onto a vertical plane through the axis of the solar system, you would see the angular separation described in the answer.
Lee Mosher
<the-sun><the-moon><solar-eclipse><moon-phases>
How close is a typical New Moon to the Sun in the sky?
Bennet
# How close is a typical New Moon to the Sun in the sky?

(Asked by: username_0 on 2023-05-18)

In a Solar Eclipse, the Sun, Earth and Moon are exactly colinear (well some area of the Earth where the eclipse is visible), and these are relatively rare.

But in addition, Sun, Earth and Moon are *approximately* colinear at every New Moon; indeed in a 2D projection it would look like there ought to be a Solar Eclipse every month.

This got me curious: if I could see the New Moon in the sky somehow, and at every New Moon I traced the minimum angular distance between the Sun and Moon, what would this typically be? Presumably this would be in fact 0 degrees at a full Solar Eclipse, but what about an "average" non-eclipsing New Moon?

In other words, how close is a typical New Moon to the Sun? Is it skirting around the edges somewhere, or a fair old distance etc. Would it be closer than Venus, say?

These questions touch on the subject but don't actually answer the angular distance part:

* [What is the difference between New Moon day and Lunar Eclipse?](https://astronomy.stackexchange.com/questions/49480/what-is-the-difference-between-new-moon-day-and-lunar-eclipse)
* [Why can't we see a solar eclipse every month?](https://astronomy.stackexchange.com/questions/35698/why-cant-we-see-a-solar-eclipse-every-month)

**Tags:** the-sun, the-moon, solar-eclipse, moon-phases

> **Comments:**
> **username_1**: You wrote *indeed in a 2D projection it would look like there ought to be a Solar Eclipse every month*, but that depends on which 2D projection you make. If you projected onto a vertical plane through the axis of the solar system, you would see the angular separation described in the answer.  

---

## 3 answers

### Answer by username_5 on 2023-05-18. Score: 10 | Accepted answer

**Mean angular separation of about 3.2°, based on [`skyfield`](https://rhodesmill.org/skyfield/) Python library simulation of the 2nd millennium.**

This is pretty dependent on your viewing location. The moon's orbit is inclined roughly 5° to the ecliptic and goes through processive and nutation cycles on a timescale a couple decades.

For the purposes of this answer, I assumed the viewpoint from the center of the Earth, and ignored atmospheric refraction.

I used the following skyfield script to look between 1001-01-01 00:00 UTC and 2001-01-01 UTC for New Moons, and calculate the angular separation between the Sun and the Moon from the center of the Earth.

```
import numpy
import pytz
from datetime import datetime
import matplotlib.pyplot as plt
from skyfield import api, almanac

def main():
    time_zone = pytz.timezone("UTC")

    ephemeris = api.load("de422.bsp")
    timescale = api.load.timescale()

    start_ts, end_ts = timescale.utc(
        [time_zone.localize(datetime(1001, 1, 1)),
         time_zone.localize(datetime(2001, 1, 1))])

    phase_times, phase_types = almanac.find_discrete(
        start_ts, end_ts, almanac.moon_phases(ephemeris))

    new_moon_times = timescale.tt_jd([phase_time.tt for phase_time, phase_type
                                      in zip(phase_times, phase_types)
                                      if phase_type == 0])
    sun, moon, earth = ephemeris['sun'], ephemeris['moon'], ephemeris['earth']

    earth_center = earth.at(new_moon_times)
    separations = earth_center.observe(sun).separation_from(earth_center.observe(moon))

    print(f"{len(separations.degrees)} New Moons found.\n"
          f"Minimum separation {min(separations.degrees):5.3f}°\n"
          f"Average separation {numpy.average(separations.degrees):5.3f}°\n"
          f"Maximum Separation {max(separations.degrees):5.3f}°\n")

    plt.hist(separations.degrees, edgecolor='black', range=[0, 6], bins=12)
    plt.title("Angular separation of Sun and Moon at New moon during 2nd millennium")
    plt.xlabel("Angular separation (degrees)")
    plt.ylabel("Number of New Moons")
    plt.show()

if __name__ == '__main__':
    main()

```

The following results were returned:

```
12369 New Moons found.
Minimum separation 0.000°
Average separation 3.186°
Maximum Separation 5.011°

```

With the following histogram:

[![Angular Separation of Sun and Moon at New Moon during 2nd millennium](https://i.sstatic.net/P6IFq.png)](https://i.sstatic.net/P6IFq.png)

As username_2 mentioned in the comments, the moon in general spends more time in its orbit near the upper range of the angular separations, resulting in the spike of separations nearing 5° inclination.

That said, 5° is still fairly close visually, about the visual width of the three center fingers on an extended arm, though with angular radii of about 0.25° each, as username_3 username_3 mentions, probably the first two to four bins would be "skirting the edges" of the solar glare for a human observer.

Venus (maximum elongation 47°) is almost always going to be more angularly distant from the Sun than the New Moon. Lightly modifying the script, I'm finding only 504 of the 12369 New Moons in the 2nd millennium where Venus is angularly closer to the Sun viewed from the center of the Earth, a bit under 4.1% of the total.

> **Comments:**
> **username_3**: Rather than comparing to Venus, it would be more useful to compare to the Sun. Given that the angular radii of the Sun and Moon are both about a quarter of a degree, the "skirting around the edges somewhere" cases would be your first two to four bins. The remaining bins are not "skirting around the edges somewhere" -- and that's the vast majority of the cases. Nice answer, BTW.  
> **username_2**: This distribution makes a certain amount of sense.  When the moon is at its farthest above or below the ecliptic plane, it is not moving very fast towards/away from that plane, and so it spends more time at a high angular separation from the ecliptic plane.  Conversely, when the Moon is passing through the ecliptic plane, it is moving relatively fast to that plane and so spends less time at small angular separations from the ecliptic plane.  
> **username_2**: If we assume a circularly orbiting moon with an orbital plane inclined at $\alpha$ relative to the ecliptic, and assume that the Moon's orbital period is incommensurate with the Earth's, then (if I've done my math right) the long-term distribution of angular separation $\theta$ at New Moon is $$\mathcal{P}(\theta) \, d \theta = \frac{\cos{\theta}}{\pi\sqrt{\sin^2\alpha - \sin^2\theta}} \, d\theta.$$This distribution has a flat region near small separations and a spike near maximum separation, consistent with your data.  
> **username_2**: I should really turn the above into an answer.  Maybe next week after I get my grades in...  
> **username_3**: @username_2 I'm getting something rather different: $P(\theta) = k \sin \alpha \cos \theta$ where $k$ is a normalizing factor, $\alpha$ is the ~5.1 degree lunar orbit inclination with respect to the ecliptic, and $\theta$ is an angle that over the long term is close to being equally distributed over the range 0 to $2\pi$.  
> **username_2**: @username_3:  I agree with you that the ecliptic latitude of the Moon $\lambda$ satisfies $\sin \lambda = \sin \alpha \cos \theta$, and that $\theta$ should be uniformly distributed between 0 and 2π.  If you then calculate what the distribution of $\lambda$ should be given those facts, you get the distribution I gave in my comment.  See [my answer below](https://astronomy.stackexchange.com/a/53793/10004) for a bit more of the process.  

---

### Answer by username_2 on 2023-05-25. Score: 1

Below is a derivation of the expected distribution of the Moon's ecliptic latitude over time. We make the following simplifying assumptions:

1. The Moon's orbit is approximately circular.
2. The precession of the moon's orbit can be ignored.
3. The Sun's ecliptic longitude is uncorrelated with the Moon's ecliptic longitude.

In particular, assumption #3 effectively says that the New Moon occurs at "random times" during the Moon's orbit. So if we find the distribution of the Moon's ecliptic latitude over a full orbit, it will follow the same distribution as the Moon's ecliptic latitude at the moment of New Moon.

If it wasn't obvious from the above, we will work in ecliptic coordinates with the origin at the Earth's center, with the primary direction rotating with the line of nodes of the Moon's orbit. (I don't think this violates assumption #2 above, but I welcome corrections on this matter.) The $x$-direction points along the line of nodes and the $z$-direction points towards the ecliptic pole (i.e., the ecliptic plane is the $xy$-plane.

Under these assumption, define $\hat{n}$ to be a unit vector normal to the plane of the Moon's orbit:
$$
\hat{n} = (0, \sin \alpha, \cos \alpha)
$$
where $\alpha \approx 5.14^\circ$ is the inclination of the Moon's orbit to the ecliptic plane. Let $\hat{v}$ be a unit vector pointing towards the Moon's position on the sky at a time $t$. This vector must make an angle of 90° with $\hat{n}$ and an angle of $\omega t$ with the $x$-axis; in other words, $\hat{n}\cdot\hat{v} = 0$ and $\hat{x}\cdot \hat{v} = \cos \omega t$ (assuming that $t = 0$ corresponds with the moon crossing the line of nodes. A bit of algebra shows us that we have
$$
\hat{v} = (\cos \theta, -\sin \theta \cos \alpha, \sin \theta \sin \alpha),
$$
where $\theta \equiv \omega t$.

It can then be shown that the Moon's ecliptic latitude $\lambda$ satisfies $\cos(\pi/2 - \lambda) = \hat{v} \cdot \hat{z} = \sin \theta \sin \alpha$, or
$$
\sin \lambda = \sin \theta \sin \alpha.
$$

Now, the probability distribution of $\theta$ over one orbit is
$$
\mathcal{P}(\theta) \, d \theta = \frac{d\theta}{2\pi}
$$
and so the probability distribution with respect to $\lambda$ is given by
$$
\mathcal{P}(\lambda) = 2 \mathcal{P}(\theta) \left| \frac{d\theta}{d\lambda} \right| = \frac{1}{\pi} \left| \frac{d\theta}{d\lambda} \right|
$$
where the factor of 2 comes from the fact that there are two possible values of $\theta$ for every value of $\lambda$. Taking the appropriate derivative, we find that
$$
\frac{d \theta}{d \lambda} = \frac{d}{d\lambda} \left[ \sin^{-1} \left( \frac{\sin \lambda}{\sin \alpha} \right) \right] = \frac{\cos \lambda}{\sqrt{ \sin^2 \alpha - \sin^2 \lambda}}
$$
and so
$$
\boxed{ \mathcal{P}(\lambda) = \frac{\cos \lambda}{\pi \sqrt{ \sin^2 \alpha - \sin^2 \lambda}}.}
$$

This distribution has the shape found in [novotny's answer](https://astronomy.stackexchange.com/a/53729/10004); it is relatively flat in the region near $\lambda = 0$ but diverges sharply in the region where $\lambda \to \pm \alpha$. Physically, this is because when the moon is near the extremes of its range of ecliptic latitude, its ecliptic latitude is not changing very fast, and therefore it spends quite a lot of time at such values of $\lambda$. Conversely, when it is near the region where $\lambda \approx 0$, its ecliptic latitude is changing relatively quickly, and so it will not spend a lot of time in such regions, leading to a lower probability of being observed at such latitudes.

> **Comments:**
> **username_3**: You should be computing $d\lambda/d\theta$ as $\theta$ is the independent variable here. The $\sqrt{\sin^2\alpha - \sin^2\lambda}$ factor in the denominator doesn't quite make sense and leads to an infinity at $\lambda=\alpha$.  
> **username_2**: @username_3:  To go from $\mathcal{P}(\theta)$ to $\mathcal{P}(\lambda)$ you need to take the derivative of the inverse function $\theta(\lambda)$, which is (in this case) $\theta(\lambda) = \sin^{-1}(\sin\lambda/\sin\alpha)$.  The mathematical details can be found [here](https://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables), though it's a bit jargony. ...  
> **username_2**: ... As to the divergence in $\mathcal{P}(\lambda)$, it's a probability density, and has to be integrated over a range of $\lambda$ to get a measurable value.  So for example, the probablity of finding $\lambda$ between $\lambda_0$ and $\alpha$ works out to be$$\int_{\lambda_0}^\alpha\mathcal{P}(\lambda)\, d\lambda = \left[\frac{1}{\pi}\sin^{-1} \left( \frac{\sin\lambda}{\sin \alpha} \right)\right]_{\lambda_0}^\alpha$$ which works out to be finite even though the integrand is divergent.  Effectively, the divergence is sufficiently "weak" that the distribution is still integrable.  

---

### Answer by username_4 on 2023-05-24. Score: 0

Short Answer: why don’t you ask a Muslim? The Islamic calendar is lunar-centric, and New-moon timing is a big deal. The record for “Newest-moon” Observations is typically re-set by some group of Muslims. And they (usually) aren’t using any super-equipment with mega-funding, either.

Long(ish) Answer: the lunar theory includes the misalignment between the orbit plane and the ecliptic, but… the pull of the Sun and other planets causes that orbit to precess. Lunar theory thus includes the “major” and “minor standstills,” the cycle in which the orbit plane bobs about the ecliptic. In other words, an epicycle or metacycle. Thus, the criterion of ‘closer than Venus’ or not doesn’t have a single answer, but a sliding scale.

If you take an average over ~months, then the precession is a bit low and irrelevant. Take the current nodal state, and average together “X” number of “months” to get the declination. If, instead, you want a long-term or theoretically rigorous answer, then you need to take this (16 year? can’t remember off the top of my head) additional cycle into account as a higher-order term. (Plus the eccentricity, of course.)


---

