Atmospheric physics 14: Greenhouse effect

For the solar radiation, we must take into account the factor $\frac{1}{4}$ in the intensity, since only a part of the earth is illuminated at a time, and we consider the average solar irradiance over the entire earth's surface. (For details, see the previous problem.) The corresponding energy flows thus result $I_\text{in} = \frac{1}{4} S_0$ and $I_\text{r} = \frac{1}{4} (1 - \alpha) S_0$ .

For the energy balance of earth and atmosphere, we must set up two equations, each corresponding to the transport space-atmosphere and atmosphere-surface. Downward (positive) flows are on the left side and upwards (negative) flows are the right side of the equation: $\begin{aligned} I_\text{in} &= I_\text{r} + I_\text{a} & & \text{(space-atmosphere)} \\ I_\text{in} + I_\text{a} &= I_\text{r} + I_\text{s} & & \text{(atmosphere-surface)} \end{aligned}$ Here, $I_\text{a} = \sigma T_\text{a}^4$ and $I_\text{s} = \sigma T_\text{s}^4$ respectively denote the thermal radiation of atmosphere and surface. In the second equation, $I_\text{a}$ appears as a source term, because the thermal radiation of the atmosphere causes additional heating of the earth's surface. We can rewrite the equations as follows $\begin{aligned} \frac{1- \alpha}{4} S_0 &= \sigma T_\text{a}^4 &\Rightarrow \quad T_\text{a} &= \left[ \frac{(1 - \alpha) S_0}{4 \sigma} \right]^{1/4} \\ \frac{1- \alpha}{4} S_0 + \sigma T_\text{a}^4 &= \sigma T_\text{s}^4 & \Rightarrow \quad T_\text{s} &= \left[ \frac{(1 - \alpha) S_0}{2 \sigma} \right]^{1/4} = 2^{1/4} T_\text{a} \end{aligned}$ While the temperature of the atmosphere corresponds to the value $T_\text{a} = 255\,\text{K}$ , we determined in the last problem, the temperature of the surface is larger by a factor $2^{1/4}$ , so that $T_\text{s} = \left[ \frac{(1 - 0.3) 1370}{2 \cdot 5.67\cdot 10^{-8}} \right]^{1/4} \,\text{K} \approx 1.189 \cdot 255\,\text{K} \approx 303\,\text{K} \approx + 30^\circ\,\text{C}$