A Lunar Occultation

On the evening of Thursday, October 2nd, 2003, the Moon will pass in front of the star tau Sgr. This event can provide dramatic evidence concerning the angular sizes of stars, and yield information useful in estimating the Moon's distance.

An occultation occurs when a nearby celestial object passes in front of a more distant one and completely hides it from view. In a lunar occultation, the Moon passes in front of a star or planet. Lunar occultations of faint stars occur all the time, but they're hard to see because the Moon's light swamps dim objects. On the evening of October 2nd, 2003 at 18:57 (6:57 pm), observers in Hawaii will see the Moon occult tau Sgr, a 3rd magnitude star in the constellation of Sagittarius. This is the best and most easily observed occultation visible from Hawaii this semester.

Fig. 1 shows simulated images of the Moon and stars in Sagittarius on 10/02/03, 18:40. At that time, the Moon will be just past first quarter, but already dazzling enough to obscure all but the brightest stars. In these images, tau Sgr appears just to the left of the Moon; it may not be visible to the unaided eye, but you should have no trouble seeing it with binoculars.


Fig. 1. The Moon and stars in Sagittarius on 10/02/03, 18:40, simulated using Celestia. tau Sgr is the star just to the left of the Moon. Left: wide-field view, showing the `teapot'. Right: view through 10×50 binoculars.


To watch this occultation, you need an observing site with a good view toward the south. You will need binoculars to see tau Sgr with the Moon so close in the sky. Also bring a watch, set as accurately as possible, and the chart included with this handout.

IMPORTANT: If you are going to be on another island on Thursday night, please let me know. Predicted times differ by a few minutes on the neighbor islands.

If possible, begin looking at about 18:30. With binoculars, you may already be able to see tau Sgr to the west of the Moon; the angular separation between them will be about half of the Moon's angular diameter, which is 0.5°. (Note: the angular diameter of an object is just the angular separation between opposite sides of the object; thus the angle from your eye to the left and right sides of the Moon is 0.5°.) As the Moon moves eastward in its orbit, its dark side will advance toward tau Sgr and finally cover up the star at 18:57 (6:57 pm). When this happens, the star's light will be cut off in a fraction of a second. Your main goal is to be looking at the star at that exact moment!

There are two different measurements we want you to make:

  1. At about 18:50, draw the Moon on the chart included with this handout. Try to show the Moon's position with respect to tau Sgr and any other stars you can see as accurately as possible. The Moon's angular diameter of 0.5° corresponds to 1 cm on the scale of this chart, so draw the globe of the Moon as a circle 1 cm in diameter, and shade the dark side so that the direction of sunlight is clear.

  2. When you see tau Sgr disappear behind the Moon, start counting seconds, and look at your watch as soon as you are sure the star is really gone. Subtract the number of seconds you counted from the reading on your watch to get an accurate time for the star's disappearance.

Please be sure to record your observing location along with the times you measure. Observers in different places will see the star disappear at slightly different times as the Moon's shadow sweeps across the island. If enough people can make accurate timings, we can estimate the speed of the Moon's shadow. To set your watch accurately, call 983-3211.

The star will reappear from behind the Moon roughly one hour later, at 19:56 (7:56 pm). It will probably be harder to tell the exact instant of reappearance, since the star emerges on the bright side of the Moon. Watching the star reappear is optional.


Just how fast is tau Sgr'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 wink out extremely fast - you will definitely not notice it fading out gradually.

We can use this fact to make a very rough estimate of the distance to tau Sgr. 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 tau Sgr 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 tau Sgr 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:

Here we use our guess for tau Sgr'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 . 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 actual distance to tau Sgr is about 120 l.y., so our simple estimate yields a distance about 220 times too small. 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 pair of binoculars!

From the fact that our calculation drastically underestimated the distance by such a large factor, it follows that tau Sgr actually disappears in much less than 0.1 sec. We can calculate how long it really takes by putting tau Sgr's correct distance D = 1.1×1015 km and diameter b = 1.4×107 km into the above equation, and solving for tau Sgr's angular diameter, . (Notice that tau Sgr is 10 times larger than the Sun -- it's a giant star!) We get  = 0.0000007°, an angle which is 20 times smaller than our upper limit. Consequently, the star actually takes only about 0.005 sec to disappear behind the Moon, as you can check by dividing 0.0000007° by 0.00015°/sec. This is much too fast for human perception, but electronic detectors can measure such rapid disappearances, and astronomers have used occultations to measure the angular diameters of stars.




Make the observations described above, and write a report on your work. This report should include, in order,

  1. a general motivation for the observations,
  2. a description of the observing site and equipment you used,
  3. a summary of your observations, and
  4. the conclusions you have reached.

In more detail, here are several things you should be sure to do in your lab report:

This report is due in class on October 7.

Joshua E. Barnes (barnes@ifa.hawaii.edu)

Last modified: September 30, 2003