A2 - radiation balance

for temperature stability, the total heat received from the sun must be balanced by the total heat radiated
the solar flux at earth's equator:

S = σ T_{⊙}^{4} {(\frac{R_{⊙}}{d_{E a r t h - S u n}})}^{2} = 1370 {Wm}^{- 2}

H_{i n} = S π R_{E}^{2} (1 - α)

where, $α$ is the bond albedo, which is the fraction of flux reflected, $\sim 30 %$ for earth

the heat leaving the earth: $H_{o u t} = 4 π σ R_{E}^{2} T_{E}^{4}$
assuming radiation balance, the average temperature of the earth can be calculated as:

T_{E} = {(\frac{S (1 - α)}{4 σ})}^{\frac{1}{4}} \approx 255 K = 18 ° C

the mean surface temperature is actually $288 K$ , which is due to the greenhouse effect
at any latitude $θ$ (angle from the centre between the equator and the point), the distance from the axis is $R_{E} \cos θ$ (side-on view)

with head-on view, it will be along a line of length $2 R_{E} \cos θ$ and width $w$
with top-down, it will be a circle with radius, $R_{E} \cos θ$
so the radiation in would be $S^{'} \cos θ 2 R_{E} \cos θ w$ ö
the radiation out would be $σ T_{θ}^{4} 2 π R_{E} \cos θ w$

T_{θ} = {(\frac{S^{'} \cos θ}{π σ})}^{1 / 4}

the slab model

a graphical method for analysing radiation balance that can be used to explain the GHG effect
solar flux at equator: $S 1370 W m^{- 2}$ (ignore albedo)
average solar flux: $\frac{S}{4} = 342 W m^{- 2} \equiv 100$ units (half due to day/night, and half from curved hemispherical surface of the earth incident surface of $π R_{E}^{2} \to 2 π R_{E}^{2}$ surface of illuminated hemisphere)
with $T_{s u r f a c e} = 288 K$ , which corresponds to $144$ units