Question

How does a planet's distance from the sun relate to its average surface temperature?

Answer 1

The sunlight reaching a planet falls off with distance as 1/r^2 (r = distance from the Sun). Because a planet's radiative equilibrium temperature depends on the fourth root of the absorbed sunlight, the equilibrium temperature scales as r^(-1/2). In other words, if you double the distance, the equilibrium temperature drops by a factor 1/√2 ≈ 0.71.

Quantitatively, a simple zero-atmosphere equilibrium temperature is
Teq = [ (S0 (1 − A)) / (4 σ r^2) ]^(1/4)
where
- S0 ≈ 1361 W/m^2 is the solar constant at 1 AU,
- A is the planetary Bond albedo (fraction of sunlight reflected),
- σ is the Stefan–Boltzmann constant,
- r is distance in AU if S0 is the 1 AU constant (or r in meters with the appropriate S at that distance).

Example (assuming Earth’s albedo and no greenhouse):
- Earth (1 AU): Teq ≈ 255 K (−18 °C). The actual global mean surface temperature is ≈288 K because of the greenhouse effect.
- Mars (1.52 AU): Teq ≈ 207 K (actual ≈210 K; thin atmosphere so little greenhouse warming).
- Venus (0.72 AU): Teq ≈ 300 K, but actual surface ≈737 K because of a tremendous greenhouse effect.

So: distance is the primary driver (temperature falls roughly as r^−1/2), but albedo and especially atmospheric greenhouse effects, plus rotation, heat transport, and internal heat sources, can strongly change a planet’s actual surface temperature.