A planet at orbital radius 1.50 × 10¹¹ m receives luminosity 3.83 × 10²⁶ W from its star. Its Bond albedo is 0.30 and it radiates uniformly from its entire spherical surface as a blackbody. Derive and calculate equilibrium temperature. Find the new temperature if albedo falls to 0.25, and explain why the radius of the planet cancels. Evaluate two limitations when comparing the result with measured surface temperature. Use σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

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T = [L(1−A)/(16πσr²)]^(1/4) ≈ 255 K. With albedo 0.25, T ≈ 259 K. Planet radius cancels because absorbed power scales with disk area while emission scales with surface area. Stellar flux at the orbit is S = L/(4πr²). The planet intercepts disk area πR_p² and absorbs fraction 1−A, so P_abs = (1−A)L R_p²/(4r²). Uniform blackbody emission from area 4πR_p² is P_emit = 4πR_p²σT⁴. Equating powers cancels R_p². Thus T = [L(1−A)/(16πσr²)]^(1/4). Substitution with A = 0.30 gives approximately 255 K. With A = 0.25, temperature multiplies by [(0.75)/(0.70)]^(1/4), giving about 259 K. The fourth-root response makes the temperature change modest. A real atmosphere absorbs and re-emits infrared radiation, producing a greenhouse effect, and temperature is not spatially uniform because of day–night, latitude and transport differences. Non-black emissivity, spectral albedo, clouds and internal heating also limit the model.

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