The movement of the moon

Question

I was in Cuba from January 15th to the 22nd this year and there was a quarter moon. If I use a compass to explain the orientation of the quarter moon, it started in the north position around 4 pm. By 7 pm, it was in the east position, and by 11 pm, it was in the south position. Normally the quarter moon would start in the east and end in the east as it travels across the sky.

Could someone please explain this phenomenon?

Thanks!

Answer 1

I think there might have been a problem with your compass readings, or "in the north" really means just a little north of east.

Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see this answer for a bit about the Moon's motion).

• it started in the north position around 4 pm.
• By 7pm, it was in the east position,
• and by 11 pm, it was in the south position.

I think there might have been something affecting your compass. The Magnetic declination there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like

1. iron or other ferromagnetic material,
2. mechanical compass from the wrong hemisphere sticking due to dip angle
3. or if it was an electronic compass, incorrect degaussing/calibration.

The other possibility is that by "in the north" you really mean only "slightly north of east" or "east-north-east".

---

I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.

The Large black dots represent 7PM local time, and each dot to the west is one more hour.

Some Python:

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files
ts    = load.timescale()
de421 = load('de421.bsp')
earth = de421['earth']
moon  = de421['moon']

Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,
                                 longitude_degrees = -75.813917,
                                 elevation_m       = 10.)

hours = np.arange(16, 23.1, 0.5) + 5

days   = (15, 17, 19, 21)

altazs, lines, linez= [], [], []
for day in days:
    times = ts.utc(2019, 1, day, hours)
    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()
    alt, az    = [thing.degrees for thing in (alt, az)]
    alt[alt<0] = np.nan
    altazs.append((alt, az))
    r = (1 - alt/90.)
    theta = rads * az
    lines.append((r, theta))
    x, y = [r*f(theta) for f in (np.sin, np.cos)]
    linez.append((x, y))

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    for alt, az in altazs:
        plt.plot(hours, alt)
        plt.plot(hours[::2], alt[::2], 'ok')
    plt.subplot(2, 1, 2)
    for alt, az in altazs:
        plt.plot(hours, az)
        plt.plot(hours[::2], az[::2], 'ok')
    plt.show()

th = np.linspace(0, twopi, 201)
cth, sth = [f(th) for f in (np.cos, np.sin)]
if True:
    plt.figure()
    plt.plot(cth, sth, '-k', linewidth=1.5)
    plt.plot([0], [0], 'or', markersize=8)
    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')
    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')

    plt.text(-0.05,  0.85, 'N', fontsize=16)
    plt.text( 0.9,  -0.05, 'E', fontsize=16)
    plt.text(-0.95, -0.05, 'W', fontsize=16)
    plt.text(-0.05, -0.90, 'S', fontsize=16)

    for x, y in linez:
        linewidth= 1
        plt.plot(x, y)
        plt.plot(x[:-1:2], y[:-1:2], '.k')
        plt.plot(x[:1], y[:1], 'ok')
    plt.show()

Answer 2

One picture tells more than thousands of words.

Check this out:

• Moon rising (near East):

• Moon setting (near West):

That picture can you show, that Moon rises in the East and sets in the West (in fact just little more to the North - 25°).
It's probably just a technical problem with a compass because Moon cannot rise in the North. (The earth always revolves around its axis, which runs along the north-south direction.)

(This picture is from the site "theskylive.org" for location Cuba and date 16/17th January 2019.)

Answer 3

To avoid confusion with compass directions on the ground, let's express the orientation of the Moon by analogy to an imaginary clock dial.

Relative to the celestial equator and pole, the bright limb of a first quarter moon is always oriented between 2 and 4 o'clock depending on the time of year. In astronomical terms, its position angle is between 240° and 300° counterclockwise from celestial north. On the evening of 2019-01-15, JPL HORIZONS says the Moon's sub-Sun position angle was about 252°, or 3:36 on an equatorially mounted clock dial.

Relative to the horizon and zenith, an observer in the Arctic would see the Moon in a similar orientation, near 3 o'clock all evening. In general, however, the celestial equator is inclined to the horizon at an angle of 90° minus the observer's geographic latitude. As celestial objects cross the sky parallel to the equator, they roll through a range of parallactic angles. This answer illustrates it with the Sun, and this answer gives a standard formula.

Moon and equatorial grid rendered by Stellarium

Observing from Havana (23°N) on January 15-16, the Moon's parallactic angle went from -67° at 1:40pm, to 0° while crossing the meridian at 7:50pm, to +67° at 2:00am. Subtracting this from the bright limb's position angle, we get angles of 319°, 252°, and 185° counterclockwise from the zenith, or 1:22, 3:36, and 5:50 on a vertical clock dial. From a more northerly latitude, the rolling effect would be smaller; from the southern hemisphere, it would be reversed.