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 |

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.

There are two different measurements we want you to make:

- 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.
- 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 1.4×10^{6} 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

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:

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* = 1.1×10^{15} km and diameter
*b* = 1.4×10^{7} 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.

- International
Occultation Timing Association
Provides information on upcoming occultations by the Moon, planets, and asteroids.

- Chart for Moon: GIF
file or Postscript.
The GIF file should be printed at 100 dpi to get a scale of 2 cm per degree.

- Which side of Oahu will see the shadow first, the east side or the west side?
- When the Moon is waxing (before it's full), the side of the
Moon which moves forward across the stars is
**dark**. What about when the Moon is waning (after it's full); which side leads, the dark side or the bright side? Would it be as easy to see an occultation if the bright side came first? - Why does our estimate yield the
**smallest**distance tau Sgr could possibly have? (Hint: we've underestimated the actual diameter of this star, and overestimated the time it takes to disappear behind the Moon.) - Why can we use the parallax formula for our estimate of the distance to tau Sgr? (Hint: parallax involves constructing a very skinny triangle. Which way does the triangle point in this estimate?).

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

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

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

- Explain what an occultation is, using
**your own words**. - State the location of your observing site; you don't have to give an actual street address, but provide enough information to pin down your position to the nearest half-mile or so.
- Report the time when tau Sgr disappeared, and estimate the accuracy of your measurement.
- Include the attached chart with your measurement of the Moon's position.

**This report is due in class on October 7.**

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

Last modified: September 30, 2003

`http://www.ifa.hawaii.edu/~barnes/ASTR110L_F03/lunaroccultation.html`