Thermal eclipse depth

For blackbody star and planet, define the planet-star contrastA complete exoplanet secondary eclipse removes the planet's light from the combined out-of-eclipse light. Its exact normalized depth is $d_{\lambda }=r_{\lambda }/(1+r_{\lambda })$. The common approximation $d_{\lambda }\simeq r_{\lambda }$ requires $r_{\lambda }\ll 1$. This thermal model omits reflected light and spatial variations.

For blackbodies with $T_p<T_s$, the contrast $r_{\lambda }=(R_p/R_s{)}^2(e^{x_s}-1)/(e^{x_p}-1)$ increases with wavelength, where $x_i=hc/(\lambda k_B{T}_i)$. Indeed $d\log r/d\log \lambda =G(x_p)-G(x_s)>0$, with $G(x)=x/(1-e^{-x})$ increasing for $x>0$ because $1-(1+x)e^{-x}>0$. The contrast approaches $(R_p/R_s{)}^2{T}_p/T_s$ in the Rayleigh-Jeans law limit. This behavior differs from the peaked absolute blackbody flux.

For blackbody star and planet observed at one common wavelength sufficiently long for the Rayleigh-Jeans law to hold for both,This follows by substituting $B_{\lambda }=2c{k}_{B}T/{\lambda }^4$ in the thermal eclipse depth. A radiative flux ratio alone does not determine the radius without a planetary brightness temperature and a stellar temperature. Spectral peaks interpreted with Wien's displacement law must refer to $F_{\lambda }$, not $F_{\nu }$ or $\lambda F_{\lambda }$.