<p>One should consider the problem as relative position between EARTH and MOON only, irrespective of EARTH's tilt w.r.t. ecliptic. I am not able to put up a sketch here but would explain as--
Thus if we consider the geometry between EARTH and MOON, then as per my calculation as :
The C to C distance  = 383,000 km @ 5.145 deg inclined, Horizontal distance will be = 383000 x COS( 5.145) = 381,452 km. AND the vertical distance above north pole = 383,000 x SIN ( 5.145)- 6317 (radius of EARTH)- 1737 (radius of MOON) = 26,338 km.
Thus highest inclination above NORTH POLE tangent plane  is
Arc TAN ( 26338 / 381452 ) = 3.944 deg.</p>

CC BY-SA 4.0
2020-01-18T23:05:17.257Z
3.944 is the wrong answer. The calculation by Mike G is correct. The problem with your calculation is that 5.145 is with respect to the ECLIPTIC. The angle between the Moon and EQUATOR (which equals the HORIZON at north pole) is 23.44+5.145=28.6 to 23.44-5.145=18.3 degrees depending on where the ascending node of the Moon is located on the ecliptic.
JohnHoltz
CC BY-SA 4.0
2020-01-19T00:08:13.557Z
Will it mean we can see almost 28.6 deg beyond the south pole of Moon from Equator of Earth when Moon is at highest elevation above North Pole?
Mohan Mone
CC BY-SA 4.0
2020-01-19T15:49:16.263Z
The question is how high will the Moon appear above the horizon if you are viewing from the North Pole (or South Pole) of the Earth. The question is not about how much of the Moon can be seen by viewing from the Earth's pole versus viewing it from the equator.
JohnHoltz
CC BY-SA 4.0
2020-01-18T22:53:55.090Z
2020-01-18T22:53:55.090Z
Mohan Mone
<p><span class="math-container">${R_e}/{R}$</span> is about 0.017 radian or 1&deg;.
Using your trigonometry and <a href="https://en.wikipedia.org/wiki/Orbit_of_the_Moon" rel="noreferrer">these values</a>:</p>

<ul>
<li>Lunar orbit inclination = 5.15&deg; (varies between 5.0&deg; and 5.3&deg;)</li>
<li>Lunar perigee = 362600 km, apogee = 405400 km</li>
<li>Earth polar radius = 6357 km</li>
<li>Ecliptic obliquity = 23.44&deg;</li>
</ul>

<p>I get 27.7&deg; at perigee and 27.8&deg; at apogee.
At that altitude, <a href="https://en.wikipedia.org/wiki/Atmospheric_refraction" rel="noreferrer">atmospheric refraction</a> is only about 0.03&deg;.
The highest apparent altitude I found with <a href="https://ssd.jpl.nasa.gov/horizons.cgi" rel="noreferrer">JPL HORIZONS</a> was 27.91&deg; on 1987-09-15 around 17:15 UT.</p>

<p>Due to <a href="https://en.wikipedia.org/wiki/Lunar_precession" rel="noreferrer">nodal precession</a>,
the Moon doesn't reach &plusmn;28.6&deg; geocentric declination every year, only near a major <a href="https://en.wikipedia.org/wiki/Lunar_standstill" rel="noreferrer">lunar standstill</a> every 18 or 19 years, e.g. in 2006 or 2025.
Near a minor lunar standstill, e.g. in 2015, the Moon's declination is limited to &plusmn;18.3&deg;.</p>

CC BY-SA 4.0
2019-03-19T13:59:50.260Z
For other extreme declination dates, see [Können and Meeus 1972](http://adsabs.harvard.edu/abs/1972JBAA...82..192K).
Mike G
CC BY-SA 4.0
2019-03-19T06:04:47.250Z
2019-03-19T06:04:47.250Z
Mike G
<p>The question is in the title. I'd like to find out the maximum angle the moon ever makes above the horizon at the North Pole. By "the horizon at the North Pole," I mean the tangent plane to the earth at that point.</p>

<p>Note that I'm not asking for the highest angle that the moon ever <strong>appears</strong> to make above the horizon. Appearances are distorted by refraction, etc., although I'm not sure by how many degrees. Though I'd be interested in the exact distortion: is there a data repository where I can look up the highest observed angle? </p>

<p>I would make the following prediction. The moon's orbital plane is inclined to the earth's equatorial plane at about <span class="math-container">$23.5+5\approx 28.5$</span> degrees. Let <span class="math-container">$R_e$</span> denote the radius of the earth and <span class="math-container">$R$</span> denote the radius of the moon's orbit around the moon. A little geometry for the highest point suggests</p>

<p><span class="math-container">$$\theta_{\text{max}}=\tan^{-1}\left(\frac{R\sin(28.5^{\circ})-R_e}{R\cos(28.5^{\circ})}\right)$$</span></p>

<p>But because <span class="math-container">$R&gt;&gt;R_e$</span> (about 385,000 compared to 6300 km), <span class="math-container">$\theta_{\text{max}}$</span> should be just a little less than <span class="math-container">$28.5^{\circ}$</span>.</p>

<p>Is this right?</p>

CC BY-SA 4.0
2019-03-19T14:54:57.507Z
Well, distortion depends on the atmosphere .  Consider the phenomenon of mirages - the moon could conceivably appear to be anywhere.
Carl Witthoft
<the-moon><observational-astronomy>
What is the highest angle the moon ever makes above the horizon at the North Pole?
curious21
# What is the highest angle the moon ever makes above the horizon at the North Pole?

(Asked by: username_0 on 2019-03-18)

The question is in the title. I'd like to find out the maximum angle the moon ever makes above the horizon at the North Pole. By "the horizon at the North Pole," I mean the tangent plane to the earth at that point.

Note that I'm not asking for the highest angle that the moon ever **appears** to make above the horizon. Appearances are distorted by refraction, etc., although I'm not sure by how many degrees. Though I'd be interested in the exact distortion: is there a data repository where I can look up the highest observed angle?

I would make the following prediction. The moon's orbital plane is inclined to the earth's equatorial plane at about $23.5+5\approx 28.5$ degrees. Let $R\_e$ denote the radius of the earth and $R$ denote the radius of the moon's orbit around the moon. A little geometry for the highest point suggests

$$\theta\_{\text{max}}=\tan^{-1}\left(\frac{R\sin(28.5^{\circ})-R\_e}{R\cos(28.5^{\circ})}\right)$$

But because $R>>R\_e$ (about 385,000 compared to 6300 km), $\theta\_{\text{max}}$ should be just a little less than $28.5^{\circ}$.

Is this right?

**Tags:** the-moon, observational-astronomy
> **Comments:**
> **username_1**: Well, distortion depends on the atmosphere .  Consider the phenomenon of mirages - the moon could conceivably appear to be anywhere.  

---

## 2 answers

### Answer by username_4 on 2019-03-19. Score: 11 | Accepted answer

${R\_e}/{R}$ is about 0.017 radian or 1°.
Using your trigonometry and [these values](https://en.wikipedia.org/wiki/Orbit_of_the_Moon):

* Lunar orbit inclination = 5.15° (varies between 5.0° and 5.3°)
* Lunar perigee = 362600 km, apogee = 405400 km
* Earth polar radius = 6357 km
* Ecliptic obliquity = 23.44°

I get 27.7° at perigee and 27.8° at apogee.
At that altitude, [atmospheric refraction](https://en.wikipedia.org/wiki/Atmospheric_refraction) is only about 0.03°.
The highest apparent altitude I found with [JPL HORIZONS](https://ssd.jpl.nasa.gov/horizons.cgi) was 27.91° on 1987-09-15 around 17:15 UT.

Due to [nodal precession](https://en.wikipedia.org/wiki/Lunar_precession),
the Moon doesn't reach ±28.6° geocentric declination every year, only near a major [lunar standstill](https://en.wikipedia.org/wiki/Lunar_standstill) every 18 or 19 years, e.g. in 2006 or 2025.
Near a minor lunar standstill, e.g. in 2015, the Moon's declination is limited to ±18.3°.

> **Comments:**
> **username_4**: For other extreme declination dates, see [Können and Meeus 1972](http://adsabs.harvard.edu/abs/1972JBAA...82..192K).  

---

### Answer by username_2 on 2020-01-18. Score: -1

One should consider the problem as relative position between EARTH and MOON only, irrespective of EARTH's tilt w.r.t. ecliptic. I am not able to put up a sketch here but would explain as--
Thus if we consider the geometry between EARTH and MOON, then as per my calculation as :
The C to C distance = 383,000 km @ 5.145 deg inclined, Horizontal distance will be = 383000 x COS( 5.145) = 381,452 km. AND the vertical distance above north pole = 383,000 x SIN ( 5.145)- 6317 (radius of EARTH)- 1737 (radius of MOON) = 26,338 km.
Thus highest inclination above NORTH POLE tangent plane is
Arc TAN ( 26338 / 381452 ) = 3.944 deg.

> **Comments:**
> **username_3**: 3.944 is the wrong answer. The calculation by Mike G is correct. The problem with your calculation is that 5.145 is with respect to the ECLIPTIC. The angle between the Moon and EQUATOR (which equals the HORIZON at north pole) is 23.44+5.145=28.6 to 23.44-5.145=18.3 degrees depending on where the ascending node of the Moon is located on the ecliptic.  
> **username_2**: Will it mean we can see almost 28.6 deg beyond the south pole of Moon from Equator of Earth when Moon is at highest elevation above North Pole?  
> **username_3**: The question is how high will the Moon appear above the horizon if you are viewing from the North Pole (or South Pole) of the Earth. The question is not about how much of the Moon can be seen by viewing from the Earth's pole versus viewing it from the equator.  

---

