<p>Imagine the crescent moon as a bow about to draw an arrow. The arrow will always point towards the sun.</p>
<p>The Moon’s orbit is tilted by about 5° relative to an imaginary line in the sky called the <em>ecliptic</em>; it’s the apparent path of the Sun in the sky over a full year. The ecliptic itself is tilted by about 23.5° with respect to Earth’s equator. While (on human timescales) the direction of this ecliptic/equator tilt is fixed, the lunar orbit/ecliptic tilt direction slowly rotates with respect to a fixed position (by convention, the <em>vernal equinox</em>, which is where the Sun seems to be located at the March equinox). The period of this movement is about 18.6 years, which means it takes about 18.6 years to do a full circle.</p>
<p>Let’s now imagine the tilt of the lunar orbit be so that the Moon is <em>below</em> the ecliptic in late March. During that time, the arrow shot by the bow will point “up,” because the Sun is on the ecliptic and the ecliptic is to the north of the Moon.</p>
<p>But 9.3 years later (18.6 ÷ 2), the Moon will be <em>above</em> the ecliptic in late March, so its bowl will point down and the arrow will go down towards the sun-on-the-ecliptic.</p>
<p>Finally, the <em>apparent</em> tilt of the ecliptic in the sky changes during the year. Earth’s north pole is tilted towards the constellation (group of stars) we call Cancer. In the summer, at noon, Cancer is very high in the sky—but the Sun is there at that time, so we don’t see the stars of Cancer. But in the winter, the sun is in Capricorn, which is below the celestial equator (the imaginary projection of the Earth’s own equator on the apparent celestial sphere), so it looks low in the sky, because the ecliptic is then low in the sky.</p>
<p>We are now at a time of the year and of the 18.6-year cycle where the waning crescent Moon is high in late March, so the arrow will shoot down; the Moon’s orbit brings it above the ecliptic, so it shoot even <em>lower</em>; and the ecliptic is highly tilted with respect to the horizon at sunset, so the arrow shoots even lower still.</p>
<p>All combined, we see the Moon as you have it in your photograph.</p>
<p>(Side detail: The bright “star” in the photograph is planet Venus.)</p>

CC BY-SA 4.0
2023-03-27T02:40:03.287Z
2023-03-27T02:40:03.287Z
Pierre Paquette
<p>I am in Maryland, and the crescent moon has been horizontal (cereal bowl bottom pointed toward the earth) for at least the past month or two. Now it’s kind of diagonal. I’ve read that this is only seen close to the equator during certain times of the year. Why would I be able to see this far north of the equator? <a href="https://i.sstatic.net/AAjRX.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AAjRX.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-03-27T18:58:13.397Z
Beautiful photo! Being closer to the equator, even more sideways in [this answer to *The moon in the first phase visible in the sky is tilted*](https://astronomy.stackexchange.com/a/49005/7982) from one year ago
uhoh
<the-moon>
Sideways crescent moon in Maryland
Meghan Thomas
# Sideways crescent moon in Maryland

(Asked by: username_0 on 2023-03-26)

I am in Maryland, and the crescent moon has been horizontal (cereal bowl bottom pointed toward the earth) for at least the past month or two. Now it’s kind of diagonal. I’ve read that this is only seen close to the equator during certain times of the year. Why would I be able to see this far north of the equator? [![enter image description here](https://i.sstatic.net/AAjRX.jpg)](https://i.sstatic.net/AAjRX.jpg)

**Tags:** the-moon
> **Comments:**
> **username_1**: Beautiful photo! Being closer to the equator, even more sideways in [this answer to *The moon in the first phase visible in the sky is tilted*](https://astronomy.stackexchange.com/a/49005/7982) from one year ago  

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## 1 answers

### Answer by username_2 on 2023-03-27. Score: 3

Imagine the crescent moon as a bow about to draw an arrow. The arrow will always point towards the sun.

The Moon’s orbit is tilted by about 5° relative to an imaginary line in the sky called the *ecliptic*; it’s the apparent path of the Sun in the sky over a full year. The ecliptic itself is tilted by about 23.5° with respect to Earth’s equator. While (on human timescales) the direction of this ecliptic/equator tilt is fixed, the lunar orbit/ecliptic tilt direction slowly rotates with respect to a fixed position (by convention, the *vernal equinox*, which is where the Sun seems to be located at the March equinox). The period of this movement is about 18.6 years, which means it takes about 18.6 years to do a full circle.

Let’s now imagine the tilt of the lunar orbit be so that the Moon is *below* the ecliptic in late March. During that time, the arrow shot by the bow will point “up,” because the Sun is on the ecliptic and the ecliptic is to the north of the Moon.

But 9.3 years later (18.6 ÷ 2), the Moon will be *above* the ecliptic in late March, so its bowl will point down and the arrow will go down towards the sun-on-the-ecliptic.

Finally, the *apparent* tilt of the ecliptic in the sky changes during the year. Earth’s north pole is tilted towards the constellation (group of stars) we call Cancer. In the summer, at noon, Cancer is very high in the sky—but the Sun is there at that time, so we don’t see the stars of Cancer. But in the winter, the sun is in Capricorn, which is below the celestial equator (the imaginary projection of the Earth’s own equator on the apparent celestial sphere), so it looks low in the sky, because the ecliptic is then low in the sky.

We are now at a time of the year and of the 18.6-year cycle where the waning crescent Moon is high in late March, so the arrow will shoot down; the Moon’s orbit brings it above the ecliptic, so it shoot even *lower*; and the ecliptic is highly tilted with respect to the horizon at sunset, so the arrow shoots even lower still.

All combined, we see the Moon as you have it in your photograph.

(Side detail: The bright “star” in the photograph is planet Venus.)


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