## Friday, July 15, 2011

### Astronomy Solutions

We now look at the solutions to the two astronomy problems given at the end of the blog, 'Tackling Simple Astronomy Problems(2)':

http://brane-space.blogspot.com/2011/07/tackling-simple-astronomy-problems-2.html

Those problems were:

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?

Solutions:

(1) The declination diagram constructed for London, latitude 51.5 N, is shown in Fig. 1 with all the relevant angles shown - derived from the basic principles of plane geometry.

The most southerly star by declination is also marked on the southern horizon. Since CE (the celestial equator) and NCP (the north celestial pole) must be 90 degrees apart, and CE defines the circle for 0 degrees declination, it is evident that the most southerly declination is: - 38.5 degrees, or 38.5 degrees south of CE.

The circumpolar stars would be all those observed from London which over time decribe a circle around the NCP (also verified by a time exposure photo- see e.g. that appended)

Thus the condition for circumpolarity would be:

51.5 =< d =< 0

where d denotes declination, and the sign =< denotes "less than or equal to".

The declination of all stars that would pass through the Londoner's zenith is gien by the angle delta extending from the CE (at 0 degrees) to z.

This is:

(90 - 38. 5) deg = + 51. 5

In other words, all object with a declination of +51.5

(2) This solution requires a declination diagram for Barbados (lat. 13 deg N) which is shown in Fig. 2

The star of interest is Alphecca with declination +26 deg 50’.

The maximum altitude a, attained by Alphecca, is easily computed from:

a = (90 - z)

where z denotes the zenith distance (or distance from the observer's zenith).

We know that all objects with declination d = 13 deg pass through the zenith, therefore the zenith distance z for an obect of declination +26 deg 50’ must be:

z = (26 deg 50’ - 13 deg) = 13 deg 50’.

Note this is also the 'meridian zenith distance'.

Therefore, the maximum altitude, a:

a = (90 deg - 13 deg 50’) = 76 deg 10'