## Thursday, August 15, 2019

### Spherical Astronomy Problem Solution

The problem again:

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

Solution:

From the celestial pole geometry (e.g. in Fig. 3 of post from Aug. 12):

90o = z +  max

Where  max    is the altitude at meridian transit (hence a maximum)

But if:   z =   max   = 45o

Then:   90o -   z    =     90o -     max    =  f (latitude)

For which your latitude is inserted.

But    d (decl.) =    z + φ

For example, take Barbados' latitude of 13 deg north.

Then:

d (decl.) = z +   f = (- 45o)   + 13o    = - 32o

Since the zenith distance z  plus altitude (max) must equal 90 degrees and we know CE (celestial equator) defines 0 degrees declination, then in this case (Barbados' location) the  star’s altitude of  max = 45 deg shows it to be SOUTH of CE. How much? 90 deg - 45 deg = 45 deg.   (The (-ve) sign implies direction below (south) of CE.)

The diagram below shows the geometry of the situation: At the zenith the declination is + 13o     on the basis of projection.  Also, the north celestial pole (NCP)  must be 13 degrees  above the northern horizon. The star S is   45o    above the observer's southern horizon. We know the celestial equator (CE) is 0 o  declination.  Then the position of the star S must be 32 degrees south of it, or as shown in the computation,  d  =  - 32o, which would be its declination.

As we can infer:  Z  CE =  13o

CE   S  =   32 o

The altitude max    =   45o     as shown

The zenith distance z =  Z CE +  CE S =   13   +  32 o  =   45 o

Thus,  90o =   z +  max