The Science Behind the Moon Project

In the Moon Project, you will be collecting data on the position of the Moon.  You will transform this raw data into information you can use to develop a picture of the Sun-Earth-Moon System. This page will lay down the foundations of the Moon Project and define the terminology that will be used throughout the semester. Please keep in mind that you can find this same information in the AST 1001 lab manual as well.


Now visualize the sky in terms of the celestial sphere; remember, we're inside it. Imagine that you're facing south, looking up at the sky. Each day or night, celestial objects rise in the east (to your left), cross the sky along the curved paths shown in Figure 4, and set in the west (to your right). Of course this behavior is really caused by Earth's eastward rotation.

Figure.4: Panoramic view of the sky facing southward.

The sketch in Figure 4 represents a huge curved view. The top edge is almost at the zenith (straight overhead) and the sides are directly to your left and right. Stars and other celestial objects follow the curved paths parallel to the celestial equator as they move across the sky each night or day.

This is an invisible line extending North-South across the sky, from the south point on the horizon and passing straight overhead. Stars cross the meridian from left to right if you're facing south. A star is highest in the sky at the time when it is on the meridian.

Here is one example of finding the meridian:
Stand in Northrup Mall.
Look towards Coffman Memorial Union. At this point, you are facing South.
Point towards the sky straight above your head, then slowly bring your arm down and point to Coffman Memorial Union. This is your meridian.
On the East of the meridian is the Village Wok, Moos Tower, and Downtown St. Paul.
On the West of the meridian is the Metrodome, Willey Hall, and Downtown Minneapolis.

The concept of meridian is very important when you take your Moon observation.

Another way of thinking of meridian is this: It is the projection of a longitude line out into space.

The celestial equator is the projection of the Earth's equator out into space . This is similar to the concept of astronomical meridian (see above). Our view of the celestial equator's location in the sky is also shown in Figure 4. It extends from directly east on the horizon to directly west on the opposite horizon. As seen from the Twin Cities, the celestial equator is tilted upward at an angle of 45° from the southern horizon.

In this project we'll try to estimate the Moon's orbital rate more accurately, in more or less the same way that ancient astronomers did. We will need to relate the daily motion of the Moon across our sky to its elongation. To do this we will measure the angle on the sky between the Moon and due South (remember we are imaging the arc across the sky from North to South - or meridian). The Moon's rotation angle from the meridian is called Hour Angle or H.A., see Figure 7.

Figure 7: Hour Angle of a star as seen on the Southern sky.

In this case the hour angle appears in the map approximately as an east-west angular distance. You can estimate it by the methods described earlier. In Figure 7 the star's H.A. is close to + 40°, probably between two and three 'hand widths' across the sky from the meridian.

If you stand and wait a while, you will notice the star appears to move across the sky to the West. Thus, the H.A. of a star, planet, the Sun, or the Moon continuously increases by about 15 degrees per hour (360° in 24 hours) as it moves westward across the sky. We reckon hour angle to be negative on the left side of the meridian, i.e., in the eastern half of the sky and positive on the right side of the meridian, i.e., in the western half of the sky. It is this parameter, the H.A. of the Moon you will be measuring.

To figure out the Sun's Hour Angle, note the following:

A simple formula that agrees with both of these facts is:

Sun's H.A = ( CST in hours - 12 (noon) ) × 15°/hr

Note that the Sun's H.A. is negative (East) before noon and a positive value (West) after noon.

Example: What is the Sun's hour angle at 9:30 a.m. Daylight Saving time on July 4 ?

9:30 CDT is 8:30 CST or 8.5 hours. Therefore the Sun's H.A. at the specified time is

( 8.5 h - 12 h ) × 15° = - 3.5 × 15° = -53°

The negative sign indicates that the Sun is east of the meridian. In other words, if you were looking directly South at 9:30 AM on July 4, the Sun would be 53° East (to your left) of the meridian along the ecliptic.

Standing on the Earth, you could measure the H.A. of both the Moon and the Sun. The difference between these two angles must be the Elongation angle of the Moon. This can be seen more clearly in Figure 8.

Figure 8: The Hour Angle of the Sun and the Moon for one particular configuration.

In Figure 8 we have chosen the time of day to be just after Sunset so that the Sun's hour angle is about +95 degrees and we placed the Moon at +30 degrees. Note that the Moon's elongation, the angle between the Sun and the Moon, is the difference between these two hour angles. If you get a negative number, just add 360 degrees.

Moon's Elongation = H.A. of the Sun - H.A. of the Moon

This formula can be used to find the Moon's elongation (angular distance from the Sun), even at night when we cannot see the Sun. This is possible because we know where the Sun is if we know the time of day. At midnight, for example, the Sun is on the other side of the Earth (H.A is about 180°), out of view. At sunset the Sun is on the Western Horizon (H.A. is about 90°). We can easily calculate its hour angle at any given time!

Another way of interpreting elongation is this:  The Moon's elongation is the Moon's orbital position around the Earth from a bird's eye (top-down) view. Check out Figure 3 on page 48 of your AST 1001lab manual for a nice picture.


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