Math 5490
Homework Set 1 Solutions
Fall 2014
Set 1
1.
Venus and Mars
a. Compute the solar flux in Watts per square meter for Venus and Mars. You may assume that
the orbit of Venus is at a distance of 1.08  1011 meters from the Sun while Mars is at a distance
of 2.28  1011 meters.
b. Assume that Venus and Mars are replaced in their orbits by perfect black bodies. What would
their surface temperatures be?
c. Assume that Venus is replaced by an otherwise perfect black body, but with an albedo of 0.71.
What would the surface temperature be? The actual albedo of Venus is about 0.71, and the
surface temperature is approximately 737K. Discuss any discrepancy between this value and
your computed value.
d. Assume that Mars is replaced by an otherwise perfect black body, but with an albedo of 0.17.
What would the surface temperature be? The actual albedo of Mars is about 0.17, and the
surface temperature is approximately 213K. Discuss any discrepancy between this value and
your computed value.
Solution. Recall the formulas from class (9/8/2014).
The solar flux F at a distance r from the sun is given by
2
r 
F  6.33  10  S  W/m 2 ,
r
7
where 6.33  107 W/m 2 is the solar flux at the surface of the sun, and where
rS  6.96  108 m is the solar radius.
The surface temperature of a black body is given by
Q  T 4 ,
where Q  F / 4 is the average planetary surface flux,   5.67  108 W/m 2 K 4 is the
Stefan-Boltzmann constant, and T is the surface temperature in Kelvin. Therefore,
T  Q   .
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If the planet has an albedo of  , then the surface temperature becomes
T   (1   )Q   .
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a. For Venus, r  1.08  1011 m , so the flux is F  2629 W/m 2 .
For Mars, r  2.28  1011 m , so the flux is F  590 W/m 2 .
b. For Venus, Q  F / 4  657 , so the surface temperature would be
T  (Q  )1 4  328 K , ( 55 oC or 131 o F ).
Set 4
Page 1 of 3
Math 5490
Homework Set 1 Solutions
Fall 2014
For Mars, Q  F / 4  147 , so the surface temperature would be
T  (Q  )1 4  226 K , ( 47 oC or 53 o F ).
c. For Venus, T   (1   )Q     0.29  657 /    241 K , ( 32 oC or 26 o F ).
Venus has a very thick atmosphere, consisting mostly of carbon dioxide, giving a
“runaway greenhouse” effect and allowing the surface temperature to reach 737 K
( 464 oC or 867 o F ).
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d. For Mars, T   (1   )Q     0.83  147 /    216 K ( 58 oC or 72 o F ). Mars
has a very thin atmosphere with an insignificant greenhouse effect. The actual
surface temperature of 213 K is consistent with this model.
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2.
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Work Exercise 8, Section 2.9 from the textbook (repeated here for your convenience).
Solution. Recall the formulas from class (9/8/2014).
(i) We use equation (2.8) from the text:
(1   )Q   T 4
with   0.3 ,   0.6 , and Q  F 4 , where Fmin  1365.5 and Fmax  1367 . Solving for
 4  0.7  1365.5 
temperature, we have Tmin  
8 
 0.6  5.67  10 
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 289.50 K , while
 4  0.7  1367 
Tmax  
 289.58 K . Subtracting the two, we have a temperature
8 
 0.6  5.67  10 
variation of about 0.08 K .
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Set 4
Page 2 of 3
Math 5490
Homework Set 1 Solutions
Fall 2014
(ii) Rewriting equation (2.15), we have
1
Fmin  A  BTmin
4
Subtracting, we have
and
1
Fmax  A  BTmax .
4
1
F  BT , which yields
4
T 
1
0.7
F 
1.5  0.13 K
4B
4  2.09
(iii) It takes longer than eleven years for the surface temperature to respond to changes
in insolation.
Set 4
Page 3 of 3