B.2 · IB Physics
Greenhouse effect
Sunlight arrives, the surface warms and radiates infrared, and an absorbing atmosphere sends part of it back down. Change the greenhouse strength, albedo and solar constant and watch the equilibrium temperature settle.
Predict first
Start from the default settings. If you raise the albedo α from 0.30 to 0.60 and change nothing else, what happens to the equilibrium surface temperature?
Absorbed flux A = S(1 − α) ÷ 4 drops by a factor 0.40 ÷ 0.70, and T ∝ A^¼, so T falls from about 288 K to about 250 K. The fourth root makes the change modest.
Check it in the model: Move the Albedo, α slider from 0.30 to 0.60 and compare the Surface temperature readout (about 288 K before, 250 K after).
sunlight infrared · arrow widths show flux in W m⁻²
σT⁴ = A + ε σT⁴ ÷ 2
T = [A ÷ (σ(1 − ε/2))]^¼
back-radiation = ε σT⁴ ÷ 2
σ = 5.67×10⁻⁸ W m⁻² K⁻⁴
Without greenhouse (ε = 0):
Greenhouse warming:
Absorbed solar flux:
Surface emission σT⁴:
Back-radiation:
Infrared out to space:
Try this
- Pick "No greenhouse" and note the surface temperature, then pick "Earth today". By how many kelvin does the atmosphere warm the surface?
- Keep ε = 0.78 and raise the albedo α from 0.30 to 0.60. Compare the fall in temperature with what happens if you lower the solar constant S by the same fraction.
- Set ε = 0.95 and compare the surface emission σT⁴ with the absorbed solar flux. Where does the extra energy at the surface come from, and why does the flux out to space still equal the absorbed flux?
Common error: saying greenhouse gases "trap" heat or reflect the infrared. They absorb outgoing infrared and re-emit it in all directions, so about half returns to the surface; at equilibrium the power leaving the top of the atmosphere still equals the solar power absorbed.