Planetary Effective Temperature

Balance what a planet absorbs against what it radiates and the whole of its climate reduces to two numbers. It intercepts sunlight on a disc of area $\pi R^2$ and radiates from a sphere of area $4\pi R^2$, so the radius cancels entirely and only the ratio 4 survives in the denominator — a rock the size of Ceres and a world the size of Earth at the same distance have the same effective temperature. Feed it Earth's 1361 W/m² and a Bond albedo of 0.30 and it returns 254.6 K, the famous 255 K, which is −18.6 °C.

That number is not Earth's surface temperature, and the gap is the entire point: the surface averages about 288 K, so the greenhouse effect is worth some 33 K — roughly the difference between the planet we have and a frozen one. Venus makes the case unforgettably. Its clouds reflect about 77 % of the sunlight, so its effective temperature is near 227 K, colder than Earth's despite being far closer to the Sun, while its surface sits at about 737 K. Three cautions come with the model. Use the Bond albedo — the fraction of all incident energy reflected in every direction and at every wavelength — because a geometric albedo or a visible-band reflectance will give a confidently wrong answer; the usual figures are 0.30 for Earth, 0.25 for Mars, 0.77 for Venus, 0.12 for the Moon. The model assumes emissivity one in the infrared, close enough for rock and water. And it assumes the absorbed heat is spread over the whole sphere, which needs a reasonably fast rotation or a mobile atmosphere: a tidally locked world has a dayside far hotter than this and a night far colder, and its average is not this number. Run the arithmetic backwards on a measured surface temperature and it demands a negative albedo — impossible, and the cleanest demonstration there is that the shortfall has to be paid by something other than sunlight.