The big idea: When sunlight hits a surface, some is reflected (bounced back) and the rest is absorbed (taken in and turned into heat).
Albedo is just the fraction reflected — a number between 0 and 1.
A bright, shiny surface has a high albedo (reflects a lot); a dark surface has a low albedo (absorbs a lot).
High albedo — reflects a lot
- fresh snow / ice ≈ 0.8
- thick cloud ≈ 0.7
- stays cooler — less absorbed
Low albedo — absorbs a lot
- dark ocean ≈ 0.06
- forest / asphalt ≈ 0.1–0.2
- warms more — more absorbed
Spot it: Albedo = the reflected fraction. So the absorbed fraction is just (1 − albedo).
Earth's average (whole-planet) albedo is about 0.30 — roughly 30% of sunlight is reflected straight back to space.
Albedo is a simple ratio — the power reflected divided by the power arriving:
- fraction reflected (no unit, between 0 and 1)
- power reflected/scattered back (W)
- power arriving from the Sun (W)
The two fractions: Reflected fraction = albedo.
Absorbed fraction = 1 − albedo.
So if 30% is reflected (albedo 0.30), then 70% is absorbed.
Why we use S ÷ 4 for the average: The solar constant S ≈ 1360 W m⁻² is the intensity on a surface facing the Sun. But sunlight only lands on the disc Earth shows the Sun (area πr²), while the whole sphere (area 4πr²) shares it out over a day.
So the average intensity over the whole planet is S ÷ 4 ≈ 340 W m⁻².
Worked example — absorbed intensity
Earth's average albedo is 0.30 and the average incoming solar intensity is S ÷ 4 = 340 W m⁻². Find the average intensity that Earth absorbs.
Solution
- Absorbed fraction = 1 − albedo:
- Absorbed intensity = absorbed fraction × incoming intensity:
- Work it out — keep the unit:
Final answer
Iabs ≈ 240 W m⁻². (The other 102 W m⁻² is reflected straight back to space.)
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How this is tested: Albedo turns up two ways.
- Paper 1A: quick MCQs — find the intensity that reaches the surface from the solar constant and an albedo, or judge what albedo depends on (it changes daily, with latitude, and with cloud cover). - Paper 2: a 'show that' on the energy balance — e.g. show the absorbed intensity is about 240 W m⁻², then use emissivity to find what the atmosphere re-radiates back to the surface.
Classic trap: using albedo as the absorbed fraction. Albedo is the reflected fraction — absorbed is (1 − albedo).
The energy-balance idea: At a steady temperature, Earth is in balance: the power it absorbs from the Sun equals the power it radiates away as infrared.
energy in = energy out → that is what keeps the temperature constant.
| Step | What happens | Intensity (W m⁻²) |
|---|---|---|
| Arriving (average) | The solar constant S spread over the whole globe: S ÷ 4 | 340 |
| Reflected | Albedo 0.30 → 30% bounced straight back to space | 0.30 × 340 = 102 |
| Absorbed | The rest, (1 − albedo) = 0.70, is taken in and warms Earth | 0.70 × 340 = 240 |
| Re-radiated | At a steady temperature, Earth radiates this same amount away | 240 |
IB-style question — intensity reaching the surface
With the Sun directly overhead, the incoming intensity at the top of the atmosphere equals the solar constant, 1360 W m⁻². The atmosphere has an albedo of 0.25 for this beam. Find the intensity of sunlight reaching the ground.
Solution
- Albedo = the fraction reflected, so the fraction that gets through = 1 − albedo:
- Intensity reaching the ground = transmitted fraction × incoming intensity:
- Work it out — keep the unit:
Final answer
I ≈ 1.0 × 10³ W m⁻² (1020 W m⁻²) reaches the ground; the other 340 W m⁻² is reflected.