Atmospheric physics 13: Earth's energy budget

The earth has a cross-sectional area of $A = \pi R_\text{E}^2$ , so that all rays incident on this surface are absorbed or reflected. (This cross-sectional area also equal to the earth's shadow.) Thus, the incident sunlight corresponds to the total power $P_\text{in} = \pi R_\text{E}^2 S_0$ Reflected light and thermal radiation are emitted from the earth. The reflected light corresponds to an power $P_\text{r} = \alpha P_\text{in}$ , with the albedo $\alpha$ of the earth's surface. Thermal radiation is emitted from the whole surface $S = 4\pi R_\text{E}^2$ of the earth. Therefore, the outgoing energy flows results $\begin{aligned} P_\text{r} &= \alpha \pi R_\text{E}^2 S_0 \\ P_\text{th} &= 4 \pi R_\text{E}^2 \sigma T_\text{s}^4 \end{aligned}$ In equilibrium, incoming and outgoing flows must be the same $\begin{aligned} & & P_\text{in} &= P_\text{r} + P_\text{th} \\ \Rightarrow & & \frac{S_0}{4} &= \alpha \frac{S_0}{4} + \sigma T_\text{s}^4 \\ \Rightarrow & & T_\text{s} &= \left[\frac{(1 - \alpha) S_0}{4 \sigma} \right]^{1/4} \\ & & &= \left[\frac{(1 - 0.3) \cdot 1370}{4 \cdot 5.67 \cdot 10^{-8}} \right]^{1/4} \,\text{K} \approx 255 \,\text{K} \approx -18 ^\circ\,\text{C} \end{aligned}$ Note: Due to the spherical shape of the earth, each square meter of the surface receives on average $I_\text{in} = \frac{1}{4} S_0$ as irradiance or $I_\text{in} - I_\text{r} = \frac{1}{4} (1 - \alpha) S_0$ a net solar intensity. The factor $\frac{1}{4}$ reflects the ratio of cross section to the surface