Spherical Trigonometry from Vector Dot Products Applied To The Celestial Sphere

 The approach to the celestial sphere in many standard college courses (for Astronomy majors), often begins with the analysis of spherical triangles such as depicted below:

                        

Fig. 1: Spherical triangle with vector directions to reference circle

The fundamental approach also requires examining the relationship between angles derived as functions of vector dot products and cross products - which we will examine in this post.

Examples:

(i^ x j^)  =  sin c m =   sin c (⊥ i   +  j)

(i^ x k^)  =  sin b u =  sin b (⊥ i   +  k)

Applying the vectors shown:

(i^ x j^) · (i^ x  k^)  = sin c  sin b cos A

= i^· [ j^ x  (i^ x  k^)]  = i^· [( j^· k^)i^   - ( i^· j^)k^]  =

cos a - cos b cos c

Such that the fundamental formula of spherical trigonometry is arrived at:

cos a =  cos b cos c +  sin b sin c cos A

Which can be applied to any spherical system, whether for Earth or the celestial sphere.

Further:  sin A  = sin a sin B/ sin b

Two companion formulas are also easy to retrieve:

i) cos b = cos a cos c + sin c sin a cos B

ii)  cos c = cos a cos b + sin a sin b cos C

These in tandem set the stage for application to a spherical reference frame;

Fig. 2: Celestial sphere showing basic planes, orientations.

The line EC follows the z-axis from the Earth's center directed outwards as shown.  The plane DBA is called the "fundamental plane" to which EC is normal (i.e. at 90^o ). Similarly, EA and EB mark the x and y axes, respectively.    We now apply this to the celestial sphere:

This will allow use of our fundamental formula which we call the "law of cosines" for spherical triangles.   Now, we use Fig. 3, for a celestial sphere application, in which we use the spherical trig relations to obtain an astronomical measurement.

Using the angles shown in Fig. 3 each of the angles for the law of cosines (given above) can be found. They are as follows:

cos a = cos (90^o - d)

where d = declination

cos b = cos (90 ^o - Lat)

where 'Lat' denotes the latitude. (Recall from Fig. 1 if φ is polar distance (which can also be zenith distance) then φ = (90 - Lat))

cos c = cos z

where z here is the zenith distance.

sin b = sin (90 deg - Lat)

sin c = sin z

and finally,

cos A = cos A

Where A is the azimuth.

Example Problem:

Let's say we want to find the declination of the star if the observer's latitude is 45 ^o N, the azimuth of the star is measured to be 60 ^o, and its zenith distance z = 30 ^o. Then one would solve for cos a:

cos a = cos (90 ^o - d)=

cos (90 ^o - Lat) cos z + sin (90 ^o - Lat) sin z cos (A)

cos (90 ^o - d) =  cos (90 ^o - 45 ^o) cos 30 ^o

+ sin (90 ^o - 45 ^o) sin 30 ^o cos 60 ^o

And:

cos (90 ^o - d) = cos (45 ^o) cos 30 ^o

+ sin (45 ^o) sin 30 ^o cos 60 ^o

We know, or can use tables or calculator to find:

cos 45 ^o = Ö2 / 2

cos 30 ^o = Ö3/ 2

sin 45 ^o = Ö2/ 2

sin 30 ^o = ½

cos 60 ^o = ½

Then: 

cos (90 - d)= {(Ö2/ 2 )( Ö3/ 2)} + {Ö2/ 2} Ö (½) }

cos (90 - d)= Ö6/ 4 + Ö2/ 8 = {2Ö6 + Ö2}/ 8

cos (90 - d) = 0.789

Suggested Problems:

1) Show that: 

a) (i^ x j^)  =  sin c (⊥ i   +  j)

b) (i^ x k^)  =  sin b u =  sin b (⊥ i   +  k)

2) In a spherical triangle ABC, C  =  90 ^o  , a = 119 ^o  30'  and B =  52 ^o.5.  Calculate the values of b, c and A.

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