2. A star of declination Dec 27 is currently culminating on the meridian at your

Question

2. A star of declination Dec 27 is currently culminating on the meridian at your observatory, located at a latitude of 20 N. a ao pts) Calculate the airmass of the star. b) 10 pts) If the magnitude of the star at zenith is estimated to be m 600mm) 12.5 and that the atmospheric extinction curve is given by figure 10.13 (page 341) of the reference book, calculate the magnitude of the star when it culminates on your meridian. c) 10 pts) Compare this last value to the magnitude you would measure if you were instead observing the star culminating from Mauna Kea.

Explanation / Answer

1. zenith distance=1/(sin)lat*sin dec+coslat*cosdec*cos h)

h is the Hour Angle of the object observed. h=S-alpha where S is local sideral time, alpha right ascertain

where z is the zenith distance

For small zenith angles, air mass X=secz

or X=Secz*(1-0.0012*(sec2z-1))

let h=0

So secz=1/(sin20*sin -27+cos20*cos-27*cos 0)= 1.4663

Now Air mass of the star

X=Secz*(1-0.0012*(sec2z-1))=1.4663*(1-0.0012*(1.4663^2-1))=1.4643

b) At any particular wavelength, the magnitude of the observed object at the surface of the earth,

m( )=m0( )+k( )X(z)

m(600nm)=12.5+0.25*1.4643=12.8661 here k( )=0.25 taken from figure

*** remember The path length through the atmosphere is known as the air mass. It is not the actual mass. relative parameter only

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