The moon rises at a different time each day, but that difference changes. Why?

Question

Let's say the moon rose at 5pm yesterday and 5:30pm today. The difference is 30 minutes – but that difference changes from day to day and can be anywhere between twenty-some to seventy-some minutes. It is also usually a different value than the difference in moonset time.

I plotted the progressions of the change in both over the course of 2022 (according to rise and set times given by timeanddate.com), and a pattern emerged:

(To be clear, this is a graph of the changes in rise and set time, not the actual rises and set times themselves.)

I'm curious as to what causes this pattern, why they mirror each other, and what causes the lull in the slope of each line. I thought it could be related to the moon's phase, but after adding the data for illumination, it seemed that there was no correlation.

Answer 1

Lunar theory is complicated. ;)

There are two main causes for the variation shown in your graph.

Firstly, the Moon's orbital plane is not aligned with the Earth's equator. It's tilted by ~5° to the ecliptic (the Earth's orbital plane), which is itself tilted by ~23° to the equatorial plane.

Secondly, the Moon's orbit is rather elliptical. It has an eccentricity of around 0.055, which is more than 3 times the eccentricity of Earth's orbit. The Moon's orbital speed is fastest when it's closest to Earth, and slowest when it's furthest from Earth.

These two effects combine to vary the Moon's angular speed across the sky relative to the background of stars. This is similar to what happens with the motion of the Sun, but more complicated. The variation in the Sun's speed is known as the Equation of Time; I have some explanation about that in this answer and graphs showing the combined effects of the eccentricity and orbital tilt.

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As I said above, with moonrise & moonset, things are more complicated than with sunrise & sunset and the relatively simple Equation of Time. As well as being more eccentric, the Moon's orbit has a couple of short precession cycles.

From Wikipedia

the orbit of the Moon undergoes two important types of precessional motion: apsidal and nodal.

Apsidal precession means that

the major axis of the Moon's elliptic orbit (the line of the apsides from perigee to apogee), precesses eastward by 360° in approximately 8.85 years. 

So the point in its orbit where the Moon is closest to the Earth changes over a 8.85 year cycle.

Another type of lunar orbit precession is that of the plane of the Moon's orbit. The period of the lunar nodal precession is defined as the time it takes the ascending node to move through 360° relative to the vernal equinox (autumnal equinox in Southern Hemisphere). It is about 18.6 years and the direction of motion is westward, i.e. in the direction opposite to the Earth's orbit around the Sun if seen from the celestial north. 

In other words, the plane of the Moon's orbit "wobbles" relative to the ecliptic plane. The wobble is fairly regular, but it's affected by the distance to the Sun, and also by the other planets. I have a 3D diagram of this motion in this answer for a very simplified system which just focuses on the nodal precession so it pretends the orbit's perfectly circular and the motion is uniform. There's a related diagram here.

Other related Wikipedia articles that you may find useful include Orbit of the Moon and Saros (astronomy)

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Here are some plots produced using data from JPL Horizons.

This plot shows the speed of the right ascension of the Sun for 2022, for each day at 00:00 UTC, relative to the centre of the Earth. The speed is in arc-seconds per hour. This graph is correlated with the variation in the length of the solar day. To compute rising and setting times we also need the declination, and the observer's location.

The speed is fastest near the solstices, and slowest near the equinoxes, due to the obliquity of the ecliptic. It's also fastest near perihelion and slowest near aphelion.

Here's the equivalent plot for the Moon. It combines the effects due to the Moon & Earth's orbit around their barycentre with the effects due to the Earth-Moon system's orbit around the Sun.

Here's the plotting script, running on the SageMathCell server, so you can make your own plots.

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Here's a similar plot to the one in the question, produced using Horizons. It shows the rising and setting time differences for 2022, from Sydney Observatory.

According to the Horizons manual, the rising & setting times should be accurate to within two minutes. The calculations include the effects of atmospheric refraction.

Each plotted point has a horizontal coordinate corresponding to the time of the event, with the vertical coordinate giving the time difference from the corresponding event on the previous day. As in the OP plot, the rising time curve is red, the setting time curve is blue.

Here's its plotting script. This script takes a while to run because it computes a year's worth of data with a one minute time step.

Answer 2

tl;dr The change in time between moonrise and moonset is different because of the changing declination of the Moon. When the declination is increasing, the time between moonrise is shorter than normal, but the time between moonset is longer than normal. As shown in Figure 1, the line connecting the Moon's position when rising is shallow to the horizon, so the change between rising is smaller. The same line when setting is steep to the horizon, so the change between setting is larger. Figure 1: Moon was at position 1 when rising on the first day (although shown some time after moonrise) and at position 3 when rising the next day. The Moon is at position 2 when setting on the first day and position 4 when setting on the next day (although shown some time before moonset).

Long story: The mathematical demonstration that the rise and set times change by different values is based on the following figure and definitions. Note that my numbers are based on a point source, no atmosphere, and the northern hemisphere. Those assumptions would change the actual numbers for how the rate changes, but it does not change the conclusion. Figure 2: Nomenclature used in the following equations.

• decl. is the declination of the object; the angle of the object above or below the celestial equation.
• RA is the Right Ascension; the angle of the object along the equator measured from the March equinox. 1h RA = 15 degrees.
• Meridian is the line from due south to the zenith (the point overhead) to due north.
• LST is the Local Sidereal Time; the LST equals the right ascension (RA) of an object on the meridian. LST increases at a uniform rate of approximately 1.0027 hours of RA per hour of time. (24 hours of RA in 23 h 56 m 4.091 s of time; the sidereal day)
• HA is the hour angle; the angle (in degrees or hours of time) between the meridian and the object, measured along the equator. That is, $HA = RA-LST$ when the object is east of the meridian, and $HA = LST-RA$ when the object is west of the meridian.
• For an object on the equator (decl.=0), the HA is 6 hours when it rises or sets. For an object with a positive declination, the HA is longer than 6 hours. For an object with a negative declination, the HA is less than 6 hours.

When rising lets begin with the Moon on the celestial equation so that $$HA = 6h = RA_1 - LST_1 $$ when ignoring the diameter of the Moon and refraction. Subscript "1" indicates the moonrise on the first day. On the next day, the Moon has moved eastward to position 3, so its RA has increased. Just to assign an approximate number, assume the Moon's RA increases at 13 degrees per day = $\frac{13}{360}t$ hours of RA per hour of time t, and the new HA on the next day has increased by $h(decl._3)$ hours. (h(decl.) is based on some function of the declination. Both the Moon's changing declination and h(decl.) are beyond the details here!) On the next day we have $$\begin{align} 6h+h(decl._3) & = RA_3-LST_3 \\ & = (RA_1+\frac{13}{360}t)-(LST_1+1.0027t-24d) \\ & = (RA_1-LST_1) +\frac{13}{360}t-1.0027t+24d \\ & = (6h) +\frac{13}{360}t-1.0027t+24d \end{align} $$ where d=1 for the next day, d=2 for the second day after, and so on, 24d puts the LST back in the range of 0 to 24 hours RA, and t is the number of hours that have elapsed since moonrise on the first day. Rearranging the terms enables the elapsed time t to be calculated from $$ 1.0027t-\frac{13}{360}t = 24d-h(decl._3) $$ When the declination increases, h(decl.) is a positive number. Therefore, the time between moonrises is smaller than if h(decl.) were zero.

When setting on the first day $$HA = 6h+h(decl._2) = LST_2-RA_2 $$ On the next day, the Moon has moved eastward to position 4 $$\begin{align} 6h+h(decl._4) & = LST_4-RA_4 \\ & = (LST_2+1.0027t-24d) - (RA_2+\frac{13}{360}t) \\ & = (LST_2-RA_2) +1.0027t-24d-\frac{13}{360}t \\ & = (6h+h(decl._2)) +1.0027t -24d -\frac{13}{360}t \end{align} $$ Rearranging the terms enables the elapsed time to be calculated by solving for t from $$ 1.0027t-\frac{13}{360}t = 24d+h(decl._4)-h(decl._2) $$ When the declination increases, $h(decl._4)-h(decl._2)$ is a positive number. Therefore, the time between moonsets is larger than if the declination change were zero.

The difference in the two equations is whether h(decl.) is added or subtracted. The changing slopes and "standstills" occur because the Moon's RA and decl. change at nonuniform rates.