Exploring Our Universe:
From the Classroom to Outer Space
I. Spectroscopy

Activity #8 Solutions

1. Look at the small right triangle at the bottom right part of Figure 4.  The length of the hypotenuse is v. The side adjacent to a is the magnitude of the component we need. Call this component x.

cos a = x/v.
Therefore x = v cos a.

sin (90o + a) = sin 90o cos a  + cos 90o sin a = cos a.
 
 

2. Look at the small right triangle in the upper left part of Figure 4. The length of the hypotenuse is v. The side opposite q is the magnitude of the component we need here. Call this component x.

sin  q = x/v.
Therefore x = v sin  q.
 
 

3.  The observed speed = the component of the cloud velocity along the line of sight - the component of the Sun velocity along the line of sight.
 
 

4.  Applying the Law of Sines to the large triangle:

sin (90o + a) / r-sun  =  sin  q / r-cloud
Solving for sin (90o + a) and substituting into the equation of exercise 3 gives the required expression.
 
 

5.  Applying the Law of Cosines to the large triangle:

(r-cloud)2 = d2 + (r-sun)2 - 2 d (r-sun) cos q

 d2 + (r-sun)2 - 2 d (r-sun) cos q - (r-cloud)2 = 0

d2 + b d + c = 0

where b =  - 2 (r-sun) cos q   and c =  (r-sun)2  - (r-cloud)2
Applying the quadratic formula and the identity cos2 q - 1 = - sin2 q :
d = (r-sun) cos q  ± [ -(r-sun)2 sin2 q + (r-cloud)2]1/2


6.  To find r-cloud substitute into:

observed speed = v sin q [(r-sun/r-cloud) - 1)]

r-cloud  = 1.5 x 1020 m

To find d, substitute into the equation found in exercise 5.
d = 1.2 x 1020 m or 3.6 x 1020 m.
Note that there are two possible solutions for this problem corresponding to two intersections between the circle of radius r-cloud and the line of sight from the Sun to the cloud.