## Tuesday, July 12, 2011

### Tackling Simple Astronomy Problems (2)

Problems to do with the celestial sphere (Fig. 1) and locating object positions in the sky also formed a major part of the work in the CXC astronomy syllabus. The diagram of Fig. 1 is taken from an article I wrote in the 2nd Volume (issue No.4) of The Journal of the B.A.S. to do with celestial coordinate systems, and transforming from one to the other. The key aspect is the generation via projection of the larger coordinate system (from the Earth's latitude and longitude system), from which advanced students could make all needed conversions, say from the celestial system (R.A., Declination) to the horizontal (altitude and azimuth).

For the more pedestrian students it was enough to teach them the use of declination diagrams (Fig. 2) to work out simple problems. The basic elements are listed in the diagram. In effect, rather than having to work in the realm of spherical geometry, the student was asked merely to get comfortable with working using plane geometry!
See also my earlier blogs on spherical astronomy and making coordinate conversions, e.g.

http://brane-space.blogspot.com/2011/03/introducing-practical-astronomy.html

http://brane-space.blogspot.com/2011/03/more-spherical-astronomy.html

http://brane-space.blogspot.com/2011/03/spherical-astronomy-matrix-methods.html

Even so, it was a formidable challenge for most students, and teachers! What was most often found is that while the general principles, plane geometry are fairly clear…the applications seldom were. For example, take this example which seeks to find the star’s declination.

The altitude of a star as it transits your meridian is found to be 45 deg along a vertical circle at azimuth 180 deg, the south point. Find the declination of the star.

Since this was designed for students at latitude 13 degrees north, the key to the solution rests on recognizing that z, the zenith distance is negative. From the geometry of Fig. 2 one sees that:

90 deg = z + a or z = 90 deg - a = 90 deg - 45 deg = 45 deg

But since we require: z = φ (latitude) = 13 deg

Then z must have a negative value, or: (-45 edg), since:

d (decl.) = z + φ = (-45 deg) + 13 deg = -32 degrees

This makes sense by examining the right side of Fig. 2. Since the zenith distance z, plus altitude (a) must equal 90 degrees and we know CE (celestial equator) defines 0 degrees declination, then a star’s altitude of a = 45 deg shows it to be SOUTH of CE. How much? 90 deg - 45 deg = 45 deg.

But.... this is still 32 deg south of CE, hence must be negative in value. (Remember CE is only 13 degrees from the zenith point). Of course, most zenith diagrams in tests were deliberately drawn not to scale, in order to make sure students grasp the principles and really attend to the geometry.

As the activity progressed, most CXC teachers found they had to go slow, keep the examples very basic and avoid synthesis with other activity objectives.

Some of the more advanced students (e.g. S.T.A.R. members) needed to see what more could be done with azimuth (trickier than altitude- measured along the observer's horizon) and they were provided extra worksheets, e.g. in seminars. One of the favorite sub-activities was to find the azimuth of the Sun for non-Caribbean locations, such as London, say for the rising or setting times at winter and summer solstice.

In general:

cos (A) = sin(d)/ cos (φ)

where A is the azimuth of the Sun, d is its declination, and φ the observer’s latitude. Students are made aware that d may be obtainable from a table – though they ought to know it right off from the equinox and solstice positions (+/- 23.5 deg) at solstices, 0 degrees at equinoxes.

As an example, let φ = 51.5 degrees N, for London. Now, for the December (winter) Solstice the Sun is over the Tropic of Capricorn (lat. 23.5 S) so its declination is just (- 23.5) degrees. We have for the Sun's azimuth at sunrise on Dec. 21:

cos (A) = sin (-23.5)/ cos (51.5)

which gives approximately, 130 degrees.

Where is this on our directional reference circle for azimuth? We know that 180 degrees is due South so that this must be:

40 degrees SOUTH of due East. (90 + 40 = 130)

Now, on the longest day of the year (say June 21), the Sun is over the Tropic of Cancer at 23.5 N latitude, so the Sun's declination is + 23.5 degrees. Then the azimuth for that date is:

cos (A) = sin (23.5)/ cos (51.5)

And A = 50 degrees,

This would put the Sun's rising position North of due East, specifically 40 degrees North of due East. Based on the preceding results, between the shortest and longest day of the year one would be seeing the Sun move from south of due east to north of due east, by the amount of degrees difference indicated.

This also discloses that - on the longest day - one cannot be observing the Sun at true Due East! But rather forty degrees North of that position at rise time. On the shortest day, one will be seeing the Sun 40 degrees south of East, or approximately south to south-east.

The Caribbean student thus becomes aware that while for him the cardinal directions of the compass essentially comport with the azimuth directions, starting from 0o and separated by 90 degrees, this doesn’t apply to other much northerly locations.

Observational limits - determining these - are also easy to do with the use of declination diagrams such as in Fig. 2. Here are two easy problems for readers to try their hands at:

1) Draw a declination diagram for London (lat. 51.5 N). What would be the most southerly star visible by declination? Which stars would be circumpolar? What declination parallel would pass through your zenith?

2) What is the maximum altitude which would be attained by Alphecca (declination +26 deg 50’) at Barbados (latitude 13 deg N).

What would the meridian zenith distance of Alphecca be?

We'll look at these solutions in the next instalment.