Key Idea: This topic follows the Sun's energy from source to planet: how its intensity spreads and weakens with distance, the black-body laws that fix how much power a hot body radiates and at which colour, and how albedo and the greenhouse effect decide how warm a planet ends up. It is examined on both papers — quick Paper 1A multiple-choice (inverse-square scaling, how a black-body curve shifts, the absorbed fraction) and longer Paper 2 structured questions ('show that' the solar constant or ~240 W m⁻², a black-body radius comparison, the greenhouse mechanism in words).
📋 Key formulas
Four of these are given in the data booklet (look for the booklet badge). The inverse-square and S ÷ 4 forms are just the given equations applied to a sphere.
- intensity — radiation power per unit area (W m⁻²)
- radiation power passing through the area (W)
- area the power spreads over (m²)
- intensity at distance d (W m⁻²)
- total power radiated by the source (W)
- distance from the source (m)
- luminosity — total power radiated (W)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
- surface area of the body (m²; a sphere has A = 4πr²)
- absolute surface temperature (K — kelvin)
- wavelength of peak (brightest) emission (m)
- absolute surface temperature (K — kelvin)
- fraction of incident sunlight reflected (no unit, 0 to 1)
- power reflected/scattered back to space (W)
- power arriving from the Sun (W)
⚖️ The four ideas side by side
| Idea | Key relationship | What to remember |
|---|---|---|
| Intensity & solar constant | I = P/A, and I = P/4πd² over a sphere | Double the distance → one quarter the intensity (d is squared). Solar constant S = 1.36 × 10³ W m⁻² is an intensity at Earth's distance, not a power. |
| Stefan-Boltzmann | L = σAT⁴ | Total power ∝ T⁴ — double the kelvin temperature → power × 16. Compare two bodies by dividing one L by the other so σ cancels. |
| Wien's law | λₘₐₓ T = 2.9 × 10⁻³ m K | Peak wavelength and T are inversely related — hotter = shorter, bluer peak. Divide the constant by T to get λₘₐₓ. |
| Albedo & energy balance | absorbed = (1 − albedo) × S/4 | Albedo is the reflected fraction; absorbed = 1 − albedo. Whole-planet average is S ÷ 4 ≈ 240 W m⁻² absorbed. |
🌍 Where the Sun's average energy goes (Earth)
| Step | What happens | Intensity (W m⁻²) |
|---|---|---|
| Arriving (average) | Solar constant shared 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 − 0.30) = 0.70, warms the planet | 0.70 × 340 = 240 |
| Re-radiated | At a steady temperature, Earth radiates this same amount away as infrared | 240 |
Sunlight is captured on the disc Earth shows the Sun (area πr²) but shared over the whole sphere (4πr²): πr² ÷ 4πr² = 1/4. Earth's surface (~290 K) is far cooler than the Sun (~5800 K), so by Wien's law it radiates at a much longer (infrared) peak wavelength — and it is that outgoing infrared the greenhouse gases trap.
✏️ Worked exam-style questions
A star behaves as a black body. Its radius is 9.0 × 10⁸ m and its surface temperature is 4.5 × 10³ K. Taking σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and treating the star as a sphere (A = 4πr²), determine the total power it radiates.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A planet's star has a surface temperature of 7.25 × 10³ K. (a) Find the wavelength at which the star radiates most strongly. (b) At the planet the star's intensity is 2.0 × 10³ W m⁻². A moon orbits three times further from the star than the planet. Find the intensity at the moon.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A spacecraft near Earth's orbit, where the intensity is the solar constant S = 1.36 × 10³ W m⁻², carries a solar array of area 12 m² that is 25% efficient. Determine the useful electrical power the array produces.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
(a) A planet receives an average solar intensity of S ÷ 4 = 340 W m⁻² and has an albedo of 0.32. Show that it absorbs about 230 W m⁻². (b) Outline how greenhouse gases keep its surface warmer than this balance alone would suggest.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
🧠 Quick self-check
Tap each card to reveal the answer.
Double your distance from a source — what happens to the intensity? It drops to a quarter (× 1/4), because I ∝ 1/d² (the inverse-square law). Triple the distance → one ninth.
Double a black body's kelvin temperature — what happens to its radiated power? It rises by a factor of 16 (= 2⁴), because L = σAT⁴ depends on T to the fourth power.
A hotter black body peaks at a ... wavelength? A shorter (bluer) wavelength. Wien's law: λₘₐₓ and T multiply to a constant, so they move opposite ways.
If the albedo is 0.30, what fraction of sunlight is absorbed? 0.70 — the absorbed fraction is (1 − albedo). Albedo itself is the reflected fraction.
Why is the whole-planet average intensity taken as S ÷ 4? Sunlight lands on the disc Earth shows the Sun (πr²) but is shared over the whole sphere (4πr²): πr² ÷ 4πr² = 1/4.
Which radiation do greenhouse gases trap, and why those gases? The outgoing infrared the warm surface emits — not the incoming sunlight. CO₂/CH₄/H₂O/N₂O bonds resonate at infrared frequencies; N₂/O₂ bonds do not.
🎯 Exam tips
Exam Tips
- Intensity is W m⁻²: divide power by the AREA it spreads over. A source radiating in all directions spreads over a sphere, so A = 4πd² (the 4π is easy to drop).
- Inverse-square scaling beats finding the Sun's power: I₂ = I₁ (d₁/d₂)². Double d → quarter I; triple d → ninth I.
- Always put T in KELVIN (K = °C + 273) before Stefan-Boltzmann or Wien. Power goes as T⁴ (×16 for a doubled T), not T.
- Compare two black bodies by writing L = σAT⁴ for each and dividing — σ (and 4π for spheres) cancels, leaving a clean ratio of radii² and T⁴.
- Albedo is the REFLECTED fraction; the absorbed (or transmitted) fraction is 1 − albedo. The whole-planet average uses S ÷ 4, but the Sun directly overhead uses the full S.
- Greenhouse effect: sunlight passes IN; the gases trap the OUTGOING infrared. The two-mark mechanism is 'absorb the surface's infrared' + 're-emit some back down'.
- Solar-panel useful output = incident intensity × panel area × efficiency, with the efficiency written as a decimal.