Stefan-Boltzmann Law Problems
1. A star with the same color as the Sun is found to produces a luminosity 81 times larger. What is
its radius compared to the Sun?
Since we are asked to compare this star to the Sun we will use a ratio technique that saves a lot
of numerical work. Below I will decode the problem by writing out ht given information, state
the Stefan-Boltzmann Law and derive the ratio we need.
Given TStar = TSun (since they have the same color)
LStar = 81 LSun
Determine RStar compared to RSun
Stefan Boltzmann Law:
L  4R 2  T 4
2
2
4
2
4
2
4
 RStar
  TStar
  RStar   TStar 
LStar 4RStar
 TStar
RStar
 TStar




 





2
4
2
4
2
4
LSun
4RSun  TSun
RSun  TSun  RSun   TSun   RSun   TSun 
2
 R  T 
L
 Star   Star    Star 
LSun  RSun   TSun 
4
4
Now, that we have a simple expression with ratios, we’ll fill in the known ratios and find
the unknown ratio
2
R 
4
81   Star   1
 RSun 
2
R 
  Star   81
 RSun 
RStar
9
RSun
The star has a radius 9 times that of the Sun.
2. 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?
The same approach as in the previous problem is useful here. If you re-derived the equation of ratios
using the Stefan-Boltzmann Law for the star and the Sun you would get
2
LStar  RStar   TStar 
 


LSun  RSun   TSun 
4
Just as before.
4
T 
R
L
In this problem Star  81 and Star  1 , so  Star   81 . Therefore the temperature of the star
RSun
LSun
 TSun 
4
is 3 (3 =81) times the temperature of the Sun.
3. The surface temperature of Arcturus is about half as hot as the Sun’s, but Arcturus is about 100
times more luminous than the Sun. What is its radius, compared to the Sun’s?
Again, we use a ratio approach where you start with the Stefan-Boltzmann relation, derive the
ratio expression and then fill in the known ratios and solve for the unknown ratio. Kjjust as was
done in the first problem.
2
LStar  RStar   TStar 
 


LSun  RSun   TSun 
4
2
 R  1
T
L
1
In this problem Star  100 and, Star  so 100   Star     . Therefore
LSun
TSun 2
 RSun   2 
R
So, Star  40 and the radius of the star is 40 times the radius of the Sun.
RSun
4
2
 RStar 

  1600 .
 RSun 
4. If a star’s surface temperature is 30,000 K, how much power does a square meter of its surface
radiate?
This is a very simple question IF you understand the Stefan-Boltzmann Law, because the term
T4 is just the power radiated per square meter, clearly stated on page 467. So the answer to this
question is just…
This star radiates T4 = 5.67x10-8 W/m2K4 ·(30,000 K)4 = 4.6x1010 W/m2 or 46 billion watts per
square meter from its surface.
5. Betelgeuse is roughly the same temperature as Proxima Centauri, the closest star to the Earth
aside the Sun. However, it is 109 times brighter. How much larger in radius is Betelgeuse than
Proxima Centauri?
This question is the most involved of th3ese practice problems because you are not given enough
information in the question to find the answer directly. The Stefan-Boltzmann Law is a
relationship between luminosity, temperature and radius. In this problem we are asked to
determine the radii ratio of the stars given that the two stars have roughly the same temperature,
but we are not given luminosity ratios. We are, however, given their brightness ratio.
The challenge now becomes how we can use the ratio of brightness to determine the ratio of
luminosities. This can be done only if we know the distances to the two stars because brightness
L
is related to luminosity by distance as in B 
.
4D 2
So this problem has three parts: (1) Look up the distances to the stars, (2) calculate the ratio of
luminosities, and finally calculate the ratio of radii.
Part 1: From Appendix Table 10 Proxima Centauri is 4.23 ly away and from Appendix Table 9
Betelgeuse is 500 ly away.
Part 2: Using the definition of brightness derive an expression for the ratio of the two stars
luminosities.
B
L
4D 2
BBetel.
BPr ox.
LBetel
2
4DBetel
LBetel 4DPr2 ox LBetel




2
LPr ox
LPr ox
LPr ox
4DBetel
2
4DPr ox
L
 4.23 
10  Betel  

LPr ox  500 
LBetel
 1.40  1013
LPr ox
D
  Pr ox
 DBetel




2
2
9
Part 3: Use the ratio of luminosities with the temperature ratio to determine the ratio of radii
using the Stefan-Boltzmann Law L  4R 2  T 4 . See the first problem to review the derivation
of the following ratio formula. Fill in the particular values for this problem and solve for the
ratio of the radii
LStar  RStar 


LSun  RSun 
2
T 
  Star 
 TSun 
2
4
R 
4
1.40  10   Star   1
 RSun 
R
 Star  1.40  1013  3.74  106
RSun
13
Finally, Betelgeuse has a radius that is about 3.7 million times larger than the radius of Proxima
Centauri.
6. A star is five times as luminous as the Sun and has a surface temperature of 98,000 K. What is
its radius, compared to that of the Sun?
Again, this is a Stefan-Boltzmann problem (luminosity, radius and temperature) that will be most
easily solved using a ratio approach. See the first problem to review how to arrive that this
equation
2
LStar  RStar   TStar 
 


LSun  RSun   TSun 
4
This equation should not be memorized. Only the Stefan-Boltzmann Law needs memorizing.
Use your algebra skills to derive the equation each time you need it. Filling in the known ratios
yields
2
 R   98,000 K 
5   Star   

 RSun   5,800 K 
4
We just have to solve for the ratio of radii.
2
 R   98,000 K 
5   Star   

 RSun   5,800 K 
4
2
R 
4
5   Star   16.9
R
 Sun 
2
R 
5   Star   81,510
 RSun 
2
 RStar 
5
5


 R   81,510  6.13  10
 Sun 
RStar
 6.13  10 5  0.0078
RSun
So the radius of the star is only 0.0078 the radius of the Sun. This is just slightly smaller than the Earth
that is 1/109 = 0.0092 the radius of the Sun. This “star” that is slightly smaller than the Earth is not
really a star, but a white dwarf stellar remnant; the dead exposed core of a blown up star, not producing
anymore energy, but only cooling off.