Astronomy Assignment #11
Text Problems:
Unit Number
57
58
Questions for Review
4, 5, 6, 7, 9
7, 8, 9
Problems
10
12, 13, 14, 17, 18
Problems in Unit 58
11. Alpha Centauri is actually two stars that orbit each other. Recent measurements show that Alpha
Centauri A has an angular diameter of 0.0085 arcsec, while Alpha Centauri B has an angular diameter of
0.0060 arcsec. The pairs are at a distance of 4.4 lyr. What are their diameters compared to the Sun.
This is a fairly messy problem. There is one short cut we can use. After we calculate the radius of Alpha
Centauri A, we can simply calculate the radius of Alpha Centauri B using the ratio of their angular sizes, Alpha
Centauri B will be a factor of (60/85) smaller than Alpha Centauri A.
A
L
To determine the angular size of Alpha Centauri A we need to use the angular size formula
,


2D
360
where A is the angular diameter of the star, L is its true diameter, and D is its distance. We need to solve for
L. We will have to change 4.4 light years into meters and 0.0085 arcsec into degrees to get a meaningful
answer.
A
L


2D
360

 1  
 .0085"

 3,600"  





A
m


 L  2D 
 2   4.4 ly   9.46  1015    


ly  
360
360



L  2  4.16  10
16
2.36  10 6
m
 1.71  10 9 m

360
The diameter of Alpha Centauri A is 1.71 x 109 meters. The Sun’s diameter is 1.39 x 109 meters as
determined from the table in the text’s appendix. Thus, Alpha Centauri A is slightly larger than the Sun
with a diameter of 1.23 solar diameters.
Alpha Centauri B is (60/85) = 0.706 times smaller than Alpha Centauri A. based on the ratio of their
angular sizes (and the fact that they are at the same distance). So Alpha Centauri B is slightly smaller
than the Sun with a diameter of 0.867 solar diameters.
12. A star with the same color as the Sun is found to produce a luminosity 81 times larger. What is its
radius, compared to the Sun’s?
The following three problems all use the Stefan-Boltzmann Law L  4R 2  T 4 . We can avoid painful
and error prone calculations, by using a ratio technique as illustrated below.
2
4
2
4
LStar  4RStar
 TStar
and LSun  4RSun
 TSun
Take the ratio of the two equations above…
2
4
 RStar
LStar 4RStar
 TStar



2
4
LSun 4RSun
 TSun
 RSun
L
  Star
 LSun
  RStar
  
  RSun



2
T
  Star
 TSun






2
T
  Star
 TSun



4
4
Now we have an expression where only ratios are used…much simpler that working with the full
equation. This problem gives us the ratio of temperatures (1, since the stars are the same color they are
also the same temperature) and the ratio of the luminosities (81). We need to solve for the ratio of radii.
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun



4
 LStar 


LSun  81
 RStar 


 
 4  81
4
R
1
 TStar 
 Sun 


 TSun 
The ratio of the radii squared is 81, so the ratio of the radii must be 81  9 . The radius of the star in
question is 9 times the radius of the Sun. A note is worthy here: We expected this star to be larger
because it was the same temperature as the Sun, by quite a bit more luminous. What this problem is
trying to illustrate, is the luminosity depends on the square of the stellar radius, not just on the radius
alone.
2
13. A star with the same radius as the Sun is found to produce a luminosity 81 times larger. What is its
surface temperature compared to the Sun?
As in the preceding question
L
  Star
 LSun
 TStar

 TSun
  RStar
  
  RSun



2
T
  Star
 TSun



4
 LStar 


LSun  81


 
 2  81
2
1
 RStar 



 RSun 
4
TStar 4
 81  3
TSun
This star is only three times hotter than the Sun. Notice how important temperature is to luminosity. A
factor of three increase in temperature increases the luminosity by a factor or 81!

14. The surface temperature of Arcturus is about half as hot as the Sun’s, but Arcturus is about 100
times more luminous that the Sun. What is its radius, compared to the Sun’s?
As before
2
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun



4

100  100  1,600
 
4
0.0625
1

 
2
The ratio of the radii squared is 1600, so the ratio of the radii must be 1,600  40 . The radius of
Arcturus is 40 times the radius of the Sun. A note is worthy here.
 RStar

 RSun
2
17. A star is five times as luminous as the Sun and has a surface temperature of 9,000 K. What is its
radius compared to that of the Sun?
This is a classic Stefan-Boltzmann Problem.
2
4
2
4
LStar  4RStar
 TStar
and LSun  4RSun
 TSun
Take the ratio of the two equations above…
2
4
R
LStar 4RStar
 TStar

  Star
2
4
LSun 4RSun  TSun  RSun
L
  Star
 LSun
  RStar
  
  RSun
2
  TStar
  
  TSun
2
R
5   Star
 RSun
  9,000 K 
  

  5,800 K 
R
5   Star
 RSun

  1.554

R
5   Star
 RSun

  5.80


2
  TStar
  
  TSun






4
4
4
2
2
RStar
5

 .928
RSun
5.80
The star has a radius just slightly smaller than the Sun at 0.928 solar radii.
18. Using the information that Betelgeuse has an angular size of 0.05 arcsec and is 130 pc from Earth,
show how you can calculate its diameter in km.
A
L
, where A = 0.05” = 1.39x10-5, and


2D
360
D = 130 pc = 130 x (3.08x1016m) = 4.00x1018m.
This is an angular size problem. So we use
3
A
L


2D
360



A
1.38  10 5
L

2

D

 2  (4.00  1018 m)  9.63  1011 m
360 
360 
The diameter of Betelgeuse is 9.63 x 1011 m or 692 solar diameters.
4