 Solution

The stellar luminosity is $L_{\star }=4\pi R_{\star }^2{\sigma }_{\mathrm{SB}}{T}_{\star }^4$. At separation $a$, the planet intercepts a disk of area $\pi R_p^2$. For Bond albedo $A_B$ and uniform emission over its full surface, energy balance givesThis planetary equilibrium temperature neglects intrinsic heat. Dayside-only or nonuniform emission changes the redistribution factor.

For the single-layer greenhouse model, assume a black infrared surface, an atmosphere transparent to incoming stellar radiation, and infrared absorptivity/emissivity $\alpha$. Let $T_s$ be surface temperature and $T_a$ layer temperature. Kirchhoff's law makes the layer's emission in each direction $\alpha {\sigma }_{\mathrm{SB}}{T}_a^4$. Its balance isso $T_a^4=T_s^4/2$ for $\alpha >0$. Surface balance includes the downward layer emission:ThereforeThe no-atmosphere limit is $T_s=T_e$, and a completely IR-absorbing single layer gives $T_s=2^{1/4}{T}_e$. The outgoing flux $(1-\alpha )\sigma T_s^4+\alpha \sigma T_a^4$ remains $\sigma T_e^4$. The layer delays escape by absorption and re-emission; it does not permanently retain a fraction $\alpha$ of every emitted photon, which would incorrectly predict a divergent temperature at $\alpha =1$.