PX282 - H2 - spectrum of a planet

image: D.C. Catling & J.F. Kasting, Exoplanets: Habitability and Characterization

there are two components in the spectrum

the reflected component

the luminosity:

L_{r e f l e c t} = \frac{L_{⊙}}{4 π a^{2}} A_{G} π R_{p}^{2}

where,
$a$ is the orbital separation of the planet from the sun, so the fraction on the RHS is the solar flux at the planet
$A_{G}$ is the geometric albedo, which depends on the viewing angle
$R_{p}$ is the radius of the planet, so $π R_{p}^{2}$ is its cross-sectional area

for the earth, $A_{G} ≃ 0.3$ , so $L_{r e f l e c t} ≃ 5 \times 10^{16}$ W

the thermal component

by equating heating and cooling, assuming no significant source of internal heat:

L_{i n} = \frac{L_{⊙}}{4 π a^{2}} π R_{p}^{2} (1 - A_{B}) = 4 π \frac{R_{⊙}^{2} σ T_{⊙}^{4}}{4 π a^{2}} π R_{p}^{2} (1 - A_{B})

where, $A_{B}$ is the bond albedo (angular average)

assuming efficient heat redistribution, eg. rapid rotation, or atmospheric lag
(this assumes that the planet is emitting from its entire surface area, whereas only receives sunlight on half of its surface at a time)

L_{o u t} = 4 π R_{p}^{2} σ T_{p}^{4}

equating the incoming and outgoing luminosities, the equilibrium temperature of a planet is:

T_{p} = (1 - A_{B})^{1 / 4} {(\frac{R_{⊙}}{2 a})}^{1 / 2} T_{⊙}

note: it does not depend on the size of the object
for earth: $T_{p} = 255$ K $= - 18 °$ C
surface temperature can be higher due to the greenhouse effect
$L_{t h e r m a l} ≃ 1.2 \times 10^{17}$ W
from wien's displacement law: $λ_{p e a k} = \frac{0.0029}{T_{p}} ≃ 11 μ$ m for earth