 Solution

Assume a circular orbit of radius $a$, a spherical planet, bolometric Bond albedo $A_B$, negligible internal heat, unit thermal emissivity and complete redistribution of absorbed heat over the sphere. The stellar luminosity is $L_s=4\pi R_s^2\sigma T_s^4$, and the planet absorbs the incident radiative flux through its projected cross-section:Thermal reradiation is $P_{\mathrm{emit}}=4\pi R_p^2\sigma T_{\mathrm{eq}}^4$. Equating these gives the planetary equilibrium temperatureThe planet's radius cancels. A non-black thermal emissivity $ϵ$ divides the absorbed flux by $ϵ$ in the fourth-power balance. For uniform dayside-only reradiation, the emitting area is $2\pi R_p^2$ and $T_{\mathrm{day}}=2^{1/4}{T}_{\mathrm{eq}}$. With no local redistribution, the substellar point has $T_{\mathrm{sub}}=\sqrt{2}{T}_{\mathrm{eq}}$, while other points depend on incidence angle. These are different temperature conventions, not contradictory formulas.

The planetary equilibrium temperature is not the surface greenhouse temperature or the internal effective temperature of a planet. If intrinsic cooling matters and both powers escape through the same emitting area, the total effective temperature obeys $T_{\mathrm{eff}}^4=T_{\mathrm{int}}^4+T_{\mathrm{eq}}^4$.