December 2012
Constellation of the Month: Triangulum
To the west of the great square of Pegasus,
which we described last month, near the wishbone-shaped Perseus
(January, 2012) is Triangulum. It is a
compact triangle with somewhat faint stars, so it will be a little challenging
to find. Some consider it an especially
attractive constellation.
The big item of interest in
Triangulum is the Triangulum Galaxy, M33.
It is the 3rd largest galaxy in our little Local Group of
galaxies – after the Andromeda Galaxy (see Nov. 2012), and our own Milky
Way. When seen through binoculars or a
small telescope, like the Discovery Center's, it is about the size of the full
moon.
Which brings us to
the question we posed last month: How
many full moons, side-by-side, would it take to stretch across the sky in a
line (arc) from the eastern horizon to the western horizon? The answer is about 360 – which is a lot more
than most people guess. That,
coincidentally, is the number of degrees in a circle. Since the arc we're talking about is a
half-circle, the Moon is about ½ degrees in diameter, as it appears to us in
the sky. This, in an extremely unlikely
coincidence, is the same size as the disk of the Sun. That is why, in a full eclipse of the sun,
the Moon will almost exactly cover the Sun.
You
can prove this half-degree figure for yourself – and maybe win a bet or two, at
sunset (or moonset). When the bottom of
the circle of the sun touches the horizon in the West, how long will it take
for the Sun to completely set? The
answer is 2 minutes, which is a lot quicker than most people would guess.
Here's
how you can use that information to calculate with width of the Sun or Moon in
degrees of a circle. The Sun or Moon traverse
across the half-circle (180 degrees) of the sky in 12 hours, which is 720 minutes. 2 minutes is 1/360th of 720
minutes. 1/360th of 180 degrees
(the arc of the sky) is ½ degree.
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