Atmospheric physics 15: Emissivity of the atmosphere

Classical Mechanics Level pending

The assumption, that the atmosphere completely absorbs the heat radiation of the earth, leads to a much stronger greenhouse effect, than we observe in reality. In fact, there are several windows in the absorption spectrum of the atmosphere, so that only a part of the thermal radiation is absorbed. For a more realistic model of the greenhouse effect, it is therefore assumed, that there is a gray atmosphere, that has an emissivity ε < 1 \varepsilon < 1 .

What is the value of the emissivity ε \varepsilon of earth's atmosphere, so that the surface temperature is equal to the actual mean temperature T s = 288 K T_\text{s} = 288 \,\text{K} ?

Details and Assumptions:

  • The emissivity ε \varepsilon indicates the proportion of the incident radiation, that is absorbed by the atmosphere. Thus, the earth's surface emits the intensity ( 1 − ε ) σ T s 4 (1 - \varepsilon) \sigma T_\text{s}^4 directly into space, while the part ε σ T s 4 \varepsilon \sigma T_\text{s}^4 is absorped.
  • The emissivity ε \varepsilon also describes the thermal radiation efficiency of the atmosphere. Thus, the atmosphere radiates the intensity ε σ T a 4 \varepsilon \sigma T_\text{a}^4 as heat radiation into space and onto earth's surface.
  • For the sunlight, the atmosphere is still transparent, while the earth's surface has an albedo α = 0.3 \alpha = 0.3 for visible light.
ε ≈ 0.53 \varepsilon \approx 0.53 ε ≈ 0.86 \varepsilon \approx 0.86 ε ≈ 0.14 \varepsilon \approx 0.14 ε ≈ 0.38 \varepsilon \approx 0.38 ε ≈ 0.77 \varepsilon \approx 0.77

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

In contrast to the last problem, the radiation of the atmosphere must be modified to I a = ε σ T a 4 I_\text{a} = \varepsilon \sigma T_\text{a}^4 . In addition, we get the transmission I t,s = ( 1 − ε ) σ T s 4 I_\text{t,s} = (1 -\varepsilon) \sigma T_\text{s}^4 through the atmosphere as another term. The energy balance is therefore I in = I r + I t,s + I a I in + I a = I r + I s \begin{aligned} I_\text{in} &= I_\text{r} + I_\text{t,s} + I_\text{a} \\ I_\text{in} + I_\text{a} &= I_\text{r} + I_\text{s} \end{aligned} This equations can be rewritten to 1 − α 4 S 0 = ( 1 − ε ) σ T s 4 + ε σ T a 4 1 − α 4 S 0 = σ T s 4 − ε σ T a 4 \begin{aligned} \frac{1-\alpha}{4} S_0 &= (1- \varepsilon) \sigma T_\text{s}^4 + \varepsilon \sigma T_\text{a}^4 \\ \frac{1-\alpha}{4} S_0 &= \sigma T_\text{s}^4 - \varepsilon \sigma T_\text{a}^4 \end{aligned} Thus we obtain a linear system of equations for the two unknowns T s 4 T_\text{s}^4 and T a 4 T_\text{a}^4 , that can be solved by matrix inversion: [ 1 − ε ε 1 − ε ] [ T s 4 T a 4 ] = ( 1 − α ) S 0 4 σ [ 1 1 ] ⇒ [ T s 4 T a 4 ] = ( 1 − α ) S 0 4 σ 1 ε ( 2 − ε ) [ ε ε 1 − ( 1 − ε ) ] [ 1 1 ] = ( 1 − α ) S 0 4 ( 2 − ε ) σ [ 2 1 ] \begin{aligned} \left[ \begin{array}{cc} 1 - \varepsilon & \varepsilon \\ 1 & - \varepsilon \end{array} \right] \left[ \begin{array}{c} T_\text{s}^4 \\ T_\text{a}^4 \end{array} \right] &= \frac{(1- \alpha) S_0}{4 \sigma} \left[ \begin{array}{c} 1 \\ 1 \end{array}\right] \\ \Rightarrow \qquad \left[ \begin{array}{c} T_\text{s}^4 \\ T_\text{a}^4 \end{array} \right] &= \frac{(1- \alpha) S_0}{4 \sigma} \frac{1}{\varepsilon (2 - \varepsilon)} \left[ \begin{array}{cc} \varepsilon & \varepsilon \\ 1 & -(1- \varepsilon) \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \end{array}\right] \\ &= \frac{(1- \alpha) S_0}{4 (2 - \varepsilon) \sigma} \left[ \begin{array}{c} 2 \\ 1 \end{array}\right] \end{aligned} For the surface temperature therefore applies T s = [ 1 2 − ε ( 1 − α ) S 0 2 σ ] 1 / 4 = { [ ( 1 − α ) S 0 4 σ ] 1 / 4 ≈ 255 K ε = 0 [ ( 1 − α ) S 0 2 σ ] 1 / 4 ≈ 303 K ε = 1 T_\text{s} = \left[ \frac{1}{2 - \varepsilon} \frac{(1 - \alpha) S_0}{2 \sigma} \right]^{1/4} = \begin{cases} \left[ \frac{(1 - \alpha) S_0}{4 \sigma} \right]^{1/4} \approx 255\,\text{K} & \varepsilon = 0 \\ \left[ \frac{(1 - \alpha) S_0}{2 \sigma} \right]^{1/4} \approx 303\,\text{K} & \varepsilon = 1 \end{cases} The solution contains the two preceding solutions as special cases for ε = 0 \varepsilon = 0 and ε = 1 \varepsilon = 1 . If the general equation is solved for the emissivity, the result is ε = 2 − ( 1 − α ) S 0 2 σ T s 4 = 2 − ( 1 − 0.3 ) ⋅ 1370 2 ⋅ 5.67 ⋅ 1 0 − 8 ⋅ 28 8 4 ≈ 0.77 \varepsilon = 2 - \frac{(1 - \alpha) S_0}{2 \sigma T_\text{s}^4} = 2 - \frac{(1 - 0.3) \cdot 1370 }{2 \cdot 5.67 \cdot 10^{-8} \cdot 288^4} \approx 0.77

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