On 06-Oct-2008 at 19:05 HST we saw the Moon occult, or come in front of, the bright star σ Sgr.
We observed the occultation of σ Sgr (mag. 2.1) by the Moon from the PSB parking lot. Conditions were nearly perfect, with minimal cloud cover and good seeing.
Visual observations were made with binoculars and an 8" telescope with a 32 mm eyepiece. In addition, we set up another 8" telescope with an 18 mm eyepiece coupled to a digital camera. The real-time video output of the camera was captured using a laptop and an video-digital conversion device. A short-wave broadcast of WWVH was simultaneously recorded to provide accurate time.
Fig. 1 shows σ Sgr disappearing behind the Moon's dark side at 19:05:25 HST. The digital camera outputs 15 frames per second; thus the time between frames is (1/15) sec. In the video, the star appears at its normal brightness in one frame, is much fainter in the next frame, and is invisible in the frame after that. From this we cannot determine precisely how long it took σ Sgr to be covered up by the Moon. However, if it took longer than (1/15) sec, we would expect the star to appear fainter than normal on two or more frames. That's not what the video shows, so we can say that the Moon covered the star in less than (1/15) sec. This is an upper limit; the actual time could be much less.
|Fig. 1. Three frames from the video of the occultation of σ Sgr. The time between frames is (1/15) sec.|
If we assume that σ Sgr is a star like our Sun, we can estimate how far away it is. The basic idea is that an occultation is a lot like a total solar eclipse; in both cases the Moon covers up a more distant star. In a solar eclipse, it takes the Moon about one hour to cover up the Sun. If we moved the Sun from its actual distance of 1 AU to 2 AU, it would appear half as big and the Moon would take half an hour to cover it up; if we moved it to 10 AU, it would appear one-tenth as big and the Moon would cover it in one-tenth of an hour.
So if σ Sgr is just a more distant version of the Sun, its distance D is related to the time t required for the Moon to cover it up:
|× 1 AU|
(If you're not sure about this equation, plug in t = 0.5 hour and notice that you get D = 2 AU. What if you plug in t = 0.1 hour?) Using t = (1/15) sec, we get
|× 1 AU||=||
|× 1 AU||= 54,000 AU|
Now remember that (1/15) sec is an upper limit for t, and the actual time may be much less. But D is inversely proportional to t, so the actual distance D may be much more than 54,000 AU. In other words, this value is a lower limit.
Now, how far is 54,000 AU? Recall that 1 AU, or one astronomical unit, is the average distance between the Earth and the Sun. On p. 304 of Stars & Planets you'll find that light takes 499 sec to travel 1 AU. So to travel 54,000 AU at the speed of light takes
|54,000 AU × 499 sec = 26,946,000 sec = 0.85 year|
It takes light 0.85 year to travel a distance of 54,000 AU, so this distance is 0.85 LY (light-years).
In fact, σ Sgr is much further away than this — about 225 LY, according to recent measurements. As noted above, the distance of 54,000 AU = 0.85 LY deduced using our occultation observation is only a lower limit. We would need a video camera recording 1000 or more frames per second to actually measure how long it took the Moon to cover σ Sgr, and the wave-like nature of light creates additional effects which would come into play if we tried to make such a measurement. Moreover, σ Sgr is not that similar to the Sun; it's probably about 5 times larger and 8 times more massive.
But even with nothing more fancy than your unaided eyes, you can observe that bright stars are covered by the Moon in times less than about 0.1 sec, and use the equation above to infer that these stars must be at least many tens of thousands of times further away than the Sun. As we've seen, this is far less than their true distances, but it does represent a first step in estimating the size of the Universe.
Provides information on upcoming occultations by the Moon, planets, and asteroids.
Light has wave-like properties which affect just how long it takes a star to vanish when the Moon covers it. This is a very readable discussion.
Joshua E. Barnes
(barnes at ifa.hawaii.edu)
16 October 2008