ECM3730: Mathematics of Climate Change
ASSESSED PROBLEM SHEET 1
1
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
2
3
4
5
6
Effective
Measured
Distance
Solar
Radiative
Surface
from Sun Constant Planetary Temperature Temperature
(m)
(W m-2)
Albedo
(K)
(K)
5.79E+10
9191
0.12
435
452
1.08E+11
2641
0.77
228
760
1.50E+11
1370
0.30
255
288
2.28E+11
593
0.24
211
230
7.80E+11
51
0.51
102
134
1.43E+12
15
0.46
77
88
2.87E+12
4
0.56
52
59
4.50E+12
2
0.51
43
48
5.91E+12
1
0.50
37
37
7
Effective
Emissivity
(leaky
greenhouse
model)
0,29
1,98
0,77
0,58
1,32
0,80
0,80
0,76
-0,07
ECM3730 Mathematics of Climate Change
ASSIGNMENT 1
17/02/2010
1a)
�
�4
Fsun = σ · Tsun = 6.37 × 107 Wm−2
� r �2
� r �2
Sc = Fsun ·
= σ · (Tsun )4 ·
b
b
1b) Tsun = 5790 K; r = 6.955 × 108 m
• Mercury:
Sc = 6.37 × 107 Wm−2 · 1.44 × 10−4 = 9191.29 Wm−2
• Venus:
Sc = 6.37 × 107 Wm−2 · 4.15 × 10−5 = 2641.72 Wm−2
• Mars:
Sc = 6.37 × 107 Wm−2 · 9.31 × 10−6 = 592.74 Wm−2
• Jupiter:
Sc = 6.37 × 107 Wm−2 · 7.95 × 10−7 = 50.65 Wm−2
• Saturn:
Sc = 6.37 × 107 Wm−2 · 2.37 × 10−7 = 15.07 Wm−2
• Uranus:
Sc = 6.37 × 107 Wm−2 · 5.87 × 10−8 = 3.74 Wm−2
• Neptune:
Sc = 6.37 × 107 Wm−2 · 2.39 × 10−8 = 1.52 Wm−2
• Pluto:
Sc = 6.37 × 107 Wm−2 · 1.38 × 10−8 = 0.88 Wm−2
1
2a) Planetary Energy Balance
Sc
(1 − αi ) · Si
= σTe4 ⇒ Te4 =
4
4σ
�
�0.25
(1 − αi ) · Si
⇒ Te =
4σ
(1 − αi )
2b)
• Mercury:
�
(1−0.12)·9191.29
4σ
Wm−2
�0.25
= 434.56K
• Venus:
�
(1−0.77)·2641.72
4σ
Wm−2
�0.25
= 227.51K
Te =
Te =
• Mars:�
Te =
(1−0.24)·592.74
4σ
Wm−2
�0.25
= 211.11K
• Jupiter
�:
Wm−2
�0.25
= 102.28K
• Saturn:
�
Wm−2
�0.25
= 77.40K
Te =
Te =
(1−0.51)·50.65
4σ
(1−0.46)·15.07
4σ
• Uranus:
�
Wm−2
�0.25
= 51.90K
• Neptune:
�
Wm−2
�0.25
= 42.57K
(1−0.50)·0.88 Wm−2
4σ
�0.25
= 37.32K
Te =
Te =
• Pluto:�
Te =
(1−0.56)·3.74
4σ
(1−0.51)·1.52
4σ
2
3a) Effective Emissivity ε
• Top of atmosphere:
• Surface:
•
σTe4 = (1 − ε)σTs4 + εσTa4
⇒
εσTa4 = σTe4 − (1 − ε)σTs4
σTe4 + εσTa4 = σTs4
⇒
εσTa4 = σTs4 − σTe4
⇒ σTe4 − (1 − ε)σTs4 = σTs4 − σTe4
• Hence
2σTe4 = σTs4 + (1 − ε)σTs4
2σTe4 = σTs4 + σTs4 − εσTs4
εσTs4 = 2σTs4 − 2σTe4
� �4
Te
ε=2−2
Ts
3b)
• Mercury: �
�4
434.56K
ε = 2 − 2 452K
= 0.29
• Venus:
�
�4
227.51K
ε = 2 − 2 760K
= 1.98
• Earth : �
�4
255K
ε = 2 − 2 288K
= 0.77
• Mars:
�
�4
211.11K
ε = 2 − 2 230K
= 0.58
• Jupiter: �
�4
102.28K
ε = 2 − 2 134K
= 1.32
• Saturn: �
�4
77.40K
ε = 2 − 2 88K
= 0.80
• Uranus: �
�4
51.90K
ε = 2 − 2 59K
= 0.80
3
• Neptune: �
�4
42.57K
ε = 2 − 2 48K
= 0.76
• Pluto:
�
�4
37.32K
= −0.07
ε=2−2
37
3c) Ts = 273 − 373K
• Earth: Te = 255K
� �4
�
�4
Te
255K
– Ts = 273K ⇒ ε = 2 − 2 Ts
= 2 − 2 273K
= 0.48
� �4
�
�4
– Ts = 373K ⇒ ε = 2 − 2 TTes
= 2 − 2 255K
= 1.56
373K
⇒ Earth would be habitable over an effective atmospheric emissivity ε = 0.48−1.56
• Mars: Te = 211K
� �4
�
�4
Te
211K
– Ts = 273K ⇒ ε = 2 − 2 Ts
= 2 − 2 273K
= 1.29
� �4
�
�4
Te
211K
– Ts = 373K ⇒ ε = 2 − 2 Ts
= 2 − 2 373K
= 1.80
⇒ Mars would be habitable over an effective atmospheric emissivity ε = 1.29−1.80
3d) Unrealistic calculated effective emissivities:
• Venus: I assume that the unrealistic value is connected to the planet’s extremely
dense atmosphere, which consists mainly of carbon dioxide together with thick
clouds of sulfur dioxide; it has the strongest greenhouse effect in the solar system,
and therefore the leaky greenhouse model is probably not applicable.
• Jupiter: I would say that the leaky greenhouse model does not work here as Jupiter
is a gas giant, and its atmosphere does not contain greenhouse gases.
• Pluto: ”As Pluto moves away from the Sun, its atmosphere gradually freezes and
falls to the ground. As it edges closer to the Sun, the temperature of Pluto’s solid
surface increases, causing the ices to sublimate into gas. This creates an antigreenhouse effect; much like sweat cools the body as it evaporates from the surface
of the skin, this sublimation cools the surface of Pluto.”
(http://en.wikipedia.org/wiki/Pluto)
4
4a)
• Top of the atmosphere:
σTe4 = (1 − ε)σTs4 + εσTa4
• Surface:
σTe4 + σTa4 = σTs4
⇒ε=1
2σTe4 = σTs4 ;
⇒
σTe4 = σTa4
2σTa4 = σTs4
4b)
• Kirchoff’s law: the emissivity of an object has to equal its absorptivity.
4
• the (i − 1)th-layer emits a temperature Ti−1
, in equal parts down to the earth
�
�4
�
�4
1
1
and up into space; hence 2 Ti−1 is emitted upwards, and 2 Ti−1 is emitted
downwards which then encounters the i-th layer.
4
• Similarly the (i + 1)th-layer emits a temperature Ti+1
, in equal parts down to the
�
�4
�
�4
earth and up into space; hence 12 Ti+1 is emitted downwards, and 12 Ti−1 is
emitted upwards which then encounters the i-th layer.
�
�4 �
�4
1
1
4
• Therefore, Ti = 2 Ti−1 + 2 Ti+1
�
�0.25
4
4
⇒ Ti = 0.5(Ti+1
+ Ti−1
)
4c)
• The temperature of the first (=highest) layer is equal to the effective temperature,
as shown in 4a):
T14 = Te4
• Second layer:
• Third layer:
T24 = Te + T1 = 2Te4
T34 = Te + T24 = 3Te4
• N-th layer:
TN = N · Te
• Hence, the temperature for the surface which can be seen as the ”(N + 1)-layer”:
Ts4 = (N + 1)Te4
⇒
Ts = (N + 1)0.25 · Te
4d) Venus: Te = 228K, Ts = 760K
Ts4 = (N + 1)Te4 ⇒ (N + 1) =
⇒N =
Ts4
Ts4
⇒
N
=
−1
Te4
Te4
7604
− 1 = 122
2284
5