HomeLearning HubIB DP PhysicsB.2 The greenhouse effect
B.2

The greenhouse effect

Theme B · The particulate nature of matter · SL and HL

In the old syllabus this sat inside Unit 8, Energy production. It now stands on its own inside the matter theme, and it carries more weight than it used to.

🎯What you need to be able to do

  • Use the solar constant and calculate the intensity received at a planet.
  • Apply albedo and emissivity, and distinguish a black body from a grey body.
  • Explain how greenhouse gases absorb infrared radiation, in terms of molecular resonance.
  • Construct a simple energy balance model of a planet.
  • Distinguish the natural greenhouse effect from the enhanced greenhouse effect.

📚The physics

Start with the incoming energy. The Sun radiates in all directions, so at a distance \(d\) the power is spread over a sphere of area \( 4\pi d^{2} \) and the intensity is

\[ I = \frac{L}{4\pi d^{2}} \]

where \(L\) is the Sun’s luminosity. At Earth’s orbit this gives the solar constant, about 1360 W m\(^{-2}\).

Then correct it twice, and both corrections are easy to forget. First, the Earth intercepts sunlight as a disc of area \( \pi r^{2} \) but it radiates from its whole sphere of area \( 4\pi r^{2} \). The ratio is \( 1/4 \), so the average incoming intensity over the whole surface is only about 340 W m\(^{-2}\). Second, some of that is reflected straight back.

Albedo is the fraction of incident radiation reflected: for Earth as a whole roughly 0.30. So the absorbed intensity is \( 340 \times (1 - 0.30) \approx 238 \) W m\(^{-2}\). Fresh snow has an albedo near 0.85 and open ocean near 0.06, which is why melting ice accelerates warming — a bright reflector is replaced by a dark absorber, so more energy is absorbed, so more ice melts. That feedback loop is worth being able to state in one sentence.

Emissivity \(e\) is the ratio of the power a body radiates to the power a perfect black body at the same temperature would radiate, so Stefan’s law becomes \( P = e\sigma A T^{4} \). A black body has \( e = 1 \); anything real is a grey body with \( e < 1 \). Earth’s effective emissivity is about 0.61.

Why greenhouse gases behave differently from the rest of the air. The atmosphere is mostly nitrogen and oxygen, which are diatomic molecules of two identical atoms with no dipole moment — they barely interact with infrared at all. Greenhouse gases such as CO\(_2\), H\(_2\)O, CH\(_4\) and N\(_2\)O have vibrational modes whose natural frequencies lie in the infrared. Incoming radiation at those frequencies drives the molecules at resonance, so it is strongly absorbed and then re-radiated in all directions, including back downwards.

This is why the wavelength matters. The Sun, at about 5800 K, peaks in the visible by Wien’s law, and the atmosphere is largely transparent there, so sunlight reaches the ground. The Earth, at about 288 K, re-radiates in the infrared — exactly where greenhouse gases absorb. Energy comes in through an open window and tries to leave through a closed one. That asymmetry is the greenhouse effect.

Natural versus enhanced. The natural greenhouse effect is not a problem; without it Earth’s mean surface temperature would be about −18 °C rather than about +15 °C, and the oceans would be ice. The enhanced greenhouse effect is the additional warming from human-added greenhouse gases. Getting this distinction right is worth marks, and getting it wrong suggests the mechanism has not been understood.

✏️Worked example

Estimate Earth’s mean surface temperature from an energy balance, taking the solar constant as 1360 W m\(^{-2}\), albedo 0.30, emissivity 1, and \( \sigma = 5.67 \times 10^{-8} \) W m\(^{-2}\) K\(^{-4}\).

Power absorbed = solar constant × disc area × (1 − albedo) \( = 1360 \times \pi r^{2} \times 0.70 \).

Power radiated = \( \sigma T^{4} \times \) sphere area \( = \sigma T^{4} \times 4\pi r^{2} \).

Set them equal. The \( \pi r^{2} \) cancels from both sides — the planet’s size is irrelevant, which is a satisfying result in itself. That leaves

\[ \sigma T^{4} = \frac{1360 \times 0.70}{4} = 238\ \text{W m}^{-2} \]

So \( T^{4} = 238/(5.67 \times 10^{-8}) = 4.20 \times 10^{10} \), giving \( T = 255 \) K, or about −18 °C.

The answer is wrong, and that is the point. The real mean surface temperature is about 288 K. The 33 K discrepancy is the natural greenhouse effect, which this model leaves out entirely. A model that fails by a known, explainable amount teaches more than one that quietly fits.
The trap. Dividing by 4 is the step most often missed. If you forget it you get 361 K, about 88 °C, and conclude the planet is boiling.

🔭See it happen

PhET — The Greenhouse Effect simulation. Add and remove greenhouse gases and watch the surface temperature follow.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. The solar constant at Earth is 1360 W m\(^{-2}\). Mars orbits at 1.52 AU. Find the intensity of sunlight at Mars.
Intensity falls as the inverse square of distance: \( I = 1360/1.52^{2} = 1360/2.31 = 5.9 \times 10^{2} \) W m\(^{-2}\).
2. A planet receives an average 340 W m\(^{-2}\) over its whole surface and has an albedo of 0.45. Find the intensity it absorbs.
\( 340 \times (1 - 0.45) = 340 \times 0.55 = 187 \) W m\(^{-2}\).
3. Earth absorbs about 238 W m\(^{-2}\) and has an effective emissivity of 0.61. Using \( P = e\sigma T^{4} \), find the predicted surface temperature.
\( T^{4} = \dfrac{238}{0.61 \times 5.67 \times 10^{-8}} = 6.88 \times 10^{9} \), so \( T = 288 \) K, about 15 °C. Including the emissivity is what closes the 33 K gap that the black-body model on this page leaves open.
4. Explain why nitrogen and oxygen, which make up most of the atmosphere, are not greenhouse gases.
They are diatomic molecules made of two identical atoms, so they have no dipole moment and their vibrational modes do not couple to infrared radiation. Greenhouse gases such as CO\(_2\), H\(_2\)O and CH\(_4\) have vibrational modes whose natural frequencies lie in the infrared, so they absorb strongly at exactly the wavelengths the Earth radiates.
5. State the ice–albedo feedback loop in one or two sentences.
Fresh snow and ice have a high albedo (about 0.85) while open ocean has a very low one (about 0.06). Warming melts ice, replacing a bright reflector with a dark absorber, so more energy is absorbed, so more warming occurs and more ice melts — a positive feedback that amplifies the original change.
6. Distinguish the natural greenhouse effect from the enhanced greenhouse effect.
The natural greenhouse effect is the warming produced by greenhouse gases that have always been present; without it Earth’s mean surface temperature would be about −18 °C rather than about +15 °C, and the oceans would be frozen. The enhanced greenhouse effect is the additional warming caused by greenhouse gases added by human activity. The natural effect is not a problem — it is what makes the planet habitable.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — The Greenhouse Effect simulation
  • NASA — Global Climate Change, vital signs and evidence