Lunar Occultations

A lunar occultation occurs when the Moon passes in front of a star (or other object). Such events are interesting for several reasons; they provide a way to measure the speed of the Moon in its orbit, and also to get some information on the distances to stars.

This table provides data for lunar occultations visible from Honolulu during Astr110L class meetings in Fall 2005. Predicted times are rounded to the nearest minute. The altitude and azimuth columns give the Moon position at that time. The illumination column gives the percentage of the Moon's disk which is illuminated; the trailing `+' sign indicates the moon is waxing. The magnitude column gives the apparent magnitude of the star; events involving stars of magnitude 8.0 or fainter are difficult observe.

Day Date Time
Mag. Comments
Tues. 09/13/05 19:41 38 156 80+ 8.0
Tues. 09/13/05 21:05 43 179 80+ 7.2
Thur. 09/15/05 19:40 28 124 95+ 7.3
Thur. 10/06/05 19:18 14 237 14+ 8.1
Thur. 10/06/05 19:43 9 240 14+ 5.0 bright star, low altitude
Tues. 10/11/05 19:23 45 169 66+ 7.8
Weds. 10/12/05 20:24 50 170 77+ 7.4
Thur. 10/13/05 19:25 43 132 86+ 6.8
Thur. 10/13/05 20:02 48 142 86+ 7.4
Thur. 10/13/05 20:05 49 142 86+ 8.4
Tues. 11/08/05 20:54 39 219 52+ 7.2
Weds. 11/09/05 19:12 54 169 63+ 8.6
Thur. 11/10/05 19:27 58 151 74+ 8.2
Thur. 11/10/05 19:54 60 163 74+ 7.6
Thur. 11/10/05 20:00 61 166 74+ 8.4
Thur. 11/10/05 20:04 61 168 74+ 4.9 begins at 18:42
Thur. 12/08/05 19:54 65 204 59+ 7.6


If we observe an occultation from two or more locations separated by at least a few km along an East-West direction, we can estimate the speed of the Moon's shadow as it sweeps across the Earth. This works best if the Moon is fairly high to the South when it occults the star; in other words, the Moon's azimuth should be close to 180°. In the table above, azimuths near this value are shown in boldface; they are best for this purpose, but any observation will be better than none!

For example, consider the occultation on 10/12/05 at 20:24 HT. The occultation will occur about 20 sec earlier at Kapiolani Park than at Sandy Beach. The East-West separation of these two sites is about 15 km, implying the Moon's shadow moves at West to East at 0.75 km/sec relative to the Earth's surface. To this we should add the Earth's rotational velocity, which is 0.44 km/sec at Oahu's latitude. The result is 1.19 km/sec, as compared to the Moon's mean orbital speed of 1.02 km/sec. Much of the ~16% discrepancy in these speeds can be resolved by noting that the Moon is close to perigee and therefore, by Kepler's second law, moving faster than average in its orbit. A better calculation might also take account of the North-South separation of Kapiolani Park and Sandy Beach and the North-South component of the Moon's motion, but such refinements don't add much to the basic concept and would be hard to implement in this class.


Just how fast is a star's light cut off by the edge of the Moon? If the star was large enough to appear as a disk, and not just a point of light, you'd see it fade out gradually as the Moon covered it up. In fact, the star will vanish in a split second - you will definitely not notice it fading out gradually.

We can use this fact to make a very rough estimate of the distances to stars. Let's say the star takes less than 0.1 sec to vanish (this is about the shortest time we can easily perceive). Let's also say that the star has the same diameter as our Sun, which is 1.4×106 km. These are the only assumptions used in this estimate; neither is very accurate, but they are OK for a very rough answer. In particular, they will serve to find the smallest distance the star could possibly have.

As seen from Earth, the Moon moves with respect to the stars at an average rate of 0.00015°/sec (360° in 27.3 days); in other words, each second it's position changes by 0.00015°. So, if the star takes less than 0.1 sec to fade out, it must have an angular diameter which is less than one-tenth of this angle, or <0.000015°. In other words, 0.000015° is an upper limit for the star's angular diameter - we don't know the true value, but we do know that it is less than 0.000015°. (This is about 20 times smaller than anything we can see with our telescopes; in fact, even the most powerful telescopes have trouble seeing detail this small!)

Now if we know how big the star really is, and we know how big it appears to be, we should be able to work out its distance. The equation required is the same one used for parallax distances:

Parallax equation.

Here we use our guess for the star's actual diameter (1.4×106 km) for the baseline b, and our upper limit for the star's angular diameter (0.000015°) as the angle theta. The result for the star's distance D is about 5.3×1012 km, or 0.55 l.y. (light-years); remember that this is the smallest possible distance, and the actual distance can be much greater. In fact, the nearest stars are nearly 10 times further away. Still, this is at least a rough figure for the distance to a star; it's pretty good for an estimate made using just a small telescope!

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Joshua E. Barnes (
Last modified: October 3, 2005
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