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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. The whole topic is one equation — energy in equals energy out — applied to a planet. What makes it worth a page is that the simple version of that equation gives an answer that is wrong by 33 K, and the reason it is wrong is the greenhouse effect itself.

🎯What you need to be able to do

  • Use the solar constant, and explain why the average intensity over a planet’s surface is a quarter of it.
  • Define and use albedo, and describe what makes a surface’s albedo high or low.
  • Define and use emissivity, and distinguish a black body from a grey body.
  • Construct an energy balance model of a planet and solve it for the equilibrium temperature.
  • Explain, in terms of molecular resonance, why greenhouse gases absorb infrared and nitrogen and oxygen do not.
  • Explain why the wavelength of the incoming radiation differs from that of the outgoing radiation, and why that asymmetry matters.
  • Name the main greenhouse gases and their sources.
  • Distinguish the natural greenhouse effect from the enhanced greenhouse effect.

☀️The incoming energy, and the factor of four

The Sun radiates in all directions, so at a distance \(d\) its power is spread over a sphere of area \( 4\pi d^{2} \) and the intensity is the apparent brightness from B.1:

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

At the Earth’s orbital distance this evaluates to about 1360 W m\(^{-2}\), a number called the solar constant, \(S\). It is the power falling on one square metre held square-on to the Sun, just outside the atmosphere.

Almost nowhere on Earth is that the actual intensity, for two reasons that are both geometry. The planet presents a disc of area \( \pi r^{2} \) to the incoming beam, so that is what it intercepts. But it radiates from its whole sphere, area \( 4\pi r^{2} \), and it spins, so the incoming energy ends up spread over four times the area that caught it.

Parallel rays of sunlight arriving at a planet. The shadow the planet casts shows that it intercepts the beam over a flat disc of area pi r squared, while the sphere it radiates from has area four pi r squared. The ratio of the two areas is one to four, so an incoming intensity of 1360 watts per square metre becomes an average of 340 watts per square metre spread over the whole surface.
The planet catches sunlight with a disc and re-radiates from a sphere. Since \( 4\pi r^{2} / \pi r^{2} = 4 \), the average incoming intensity is \( S/4 \), whatever the planet’s size.
\[ \text{average incoming intensity} = \frac{S}{4} = \frac{1360}{4} = 340\ \text{W m}^{-2} \]
The single most-missed step in this topic. Forget the factor of four and the energy balance below gives 360 K, about 87 °C, and you will conclude that the Earth is boiling. The four is not a fudge factor: it is the ratio of a sphere’s area to that of its own shadow. If you can draw the disc and the sphere, you will never lose it.

🪝Albedo: how much is thrown straight back

Not all of that 340 W m\(^{-2}\) is absorbed. The albedo of a surface is the fraction of the incident power that is scattered back:

\[ \text{albedo} = \frac{\text{total scattered power}}{\text{total incident power}} \]

It is a ratio, so it has no units, and it runs from 0 (a perfect absorber) to 1 (a perfect reflector).

An incoming beam of radiation striking a surface and splitting into a reflected part and an absorbed part, with albedo defined as the reflected fraction. Beside it, a bar chart of typical albedo values: fresh snow about 0.85, thick cloud about 0.7, sea ice about 0.5, desert sand about 0.4, grassland about 0.25, forest about 0.15 and open ocean about 0.06, with the Earth's global average of about 0.30 marked across them.
Bright, white and rough scatters; dark and smooth absorbs. The Earth’s average of about 0.30 is a weighted mixture of all of these, and it is not a constant of nature.

A planet’s overall albedo therefore changes with the seasons, the latitude, the time of day and above all the cloud cover. Roughly 0.30 of the sunlight arriving at the Earth is scattered back to space, so the absorbed intensity is

\[ (1 - \alpha)\frac{S}{4} = 0.70 \times 340 = 238\ \text{W m}^{-2} \]

Because ice is bright and water is dark, albedo is also the mechanism behind one of the clearest positive feedback loops in the climate system.

A closed loop of four boxes with arrows running clockwise: temperature rises, so ice melts, so the albedo falls because dark ocean has replaced bright ice, so more of the incoming radiation is absorbed, so the temperature rises further. A label identifies this as positive feedback because the loop reinforces the change that started it.
Nothing in the loop is controversial physics — each arrow is one line of reasoning — and that is exactly why it is worth being able to state in one sentence.

⬜Emissivity: real bodies are not black bodies

B.1 treated everything as a perfect black body. Real surfaces radiate less than that. The emissivity \(e\) of a body is the ratio of the power it radiates to the power a black body at the same temperature would radiate:

\( e = \dfrac{\text{power radiated by the body}}{\text{power radiated by a black body at the same }T} \)
Two spectral curves plotted against wavelength at the same temperature. The upper curve is a black body with emissivity one; the lower curve, of the same shape but scaled down, is a grey body with emissivity about 0.6. The area under each curve is the total power radiated, and the ratio of the two areas is the emissivity. Beside them, Stefan's law is written twice: P equals sigma A T to the fourth for a black body, and P equals e sigma A T to the fourth for a real one.
A grey body radiates the same shape of spectrum as a black body at that temperature, scaled down by \(e\). So Wien’s law still gives the peak; only the total power changes.

Stefan’s law then picks up one extra factor:

\[ P = e\sigma A T^{4} \]

A black body has \( e = 1 \) by definition; every real surface is a grey body with \( e < 1 \). The Earth’s effective emissivity, treating the whole planet as one radiating surface at its mean surface temperature of 288 K, works out at about 0.61 — and that number is not an input to the model below, it is a measure of how much the atmosphere holds back.

⚖️The energy balance model of a planet

A planet at a steady temperature must be radiating exactly as fast as it absorbs. That single sentence is the whole model:

A planet drawn as a circle with two arrows. An incoming arrow labelled one minus albedo times S over four, equal to 238 watts per square metre, represents absorbed solar radiation; an outgoing arrow labelled e sigma T to the fourth represents the radiation the planet emits. The two are set equal to give the equilibrium condition, and solving it with emissivity one gives 255 kelvin, which is 33 kelvin below the observed mean surface temperature of 288 kelvin.
Equilibrium is not “nothing is happening”: it is two large flows that happen to be equal. Change either one and the temperature moves until they balance again.
\[ (1 - \alpha)\frac{S}{4} = e\sigma T^{4} \]

✏️Worked example 1 — the temperature of a bare planet

Estimate the 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\(^{-4}\).

Write both sides for the whole planet before cancelling anything.

\[ \text{absorbed} = S\,\pi r^{2}(1-\alpha) \qquad \text{radiated} = \sigma T^{4}\,4\pi r^{2} \]

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

\[ \sigma T^{4} = \frac{S(1-\alpha)}{4} = \frac{1360 \times 0.70}{4} = 238\ \text{W m}^{-2} \]
\[ T^{4} = \frac{238}{5.67\times10^{-8}} = 4.20\times10^{9} \quad\Rightarrow\quad T = 255\ \text{K} = -18\ ^\circ\text{C} \]
The answer is wrong, and that is the point. The real mean surface temperature is about 288 K, or +15 °C. 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 — and if you instead put the measured 288 K into \( 238 = e\sigma T^{4} \) and solve for \(e\), you get \( e = 238/390 = 0.61 \), the effective emissivity quoted above. The two statements are the same fact seen twice.

✏️Worked example 2 — how much does albedo matter?

Ice loss lowers the Earth’s albedo from 0.30 to 0.25. With emissivity still 1, find the new equilibrium temperature.
\[ \sigma T^{4} = (1-0.25)\times 340 = 255\ \text{W m}^{-2} \quad\Rightarrow\quad T^{4} = 4.50\times10^{9} \]
\[ T = 259\ \text{K} \]
Four kelvin from a five-hundredths change in a dimensionless ratio. That is the feedback loop above with a number attached, and it is why albedo is not a detail. Note also that the fourth-power law works against runaway: a hotter planet radiates very much faster, so each extra watt absorbed buys a smaller and smaller temperature rise.

🌐Where the energy actually goes

The two-arrow model is enough to pass an exam question, but it hides the fact that the surface and the atmosphere are two separate systems, each of which must balance on its own. Following the flows makes the greenhouse mechanism visible before any molecules are mentioned.

An energy budget diagram of the Earth in watts per square metre, rounded so that every part balances. Of 340 arriving, 100 is reflected by clouds and the surface, 80 is absorbed by the atmosphere and 160 by the surface. The surface emits 400 of infrared, of which 390 is absorbed by the atmosphere and 10 escapes straight to space through the atmospheric window, and loses a further 100 by evaporation and convection. The atmosphere radiates 230 out to space and 340 back down to the surface. The surface therefore receives 160 plus 340 and loses 400 plus 100, and the top of the atmosphere receives 340 and loses 100 plus 230 plus 10.
Rounded numbers, chosen so that all three balances close exactly. The one to notice is the 340 W m\(^{-2}\) of back radiation: the surface receives more energy from the atmosphere than it does from the Sun.

Three checks, and they all have to work:

top of atmosphere\( 340 = 100 + 230 + 10 \)
surface\( 160 + 340 = 400 + 100 \)
atmosphere\( 80 + 390 + 100 = 230 + 340 \)

The surface is warm not because sunlight is intense — only 160 W m\(^{-2}\) of it gets down there — but because it sits under an atmosphere that returns 340 W m\(^{-2}\) to it. Remove the greenhouse gases and that return flow largely disappears.

🔬Why greenhouse gases, and not the rest of the air

The atmosphere is 78% nitrogen and 21% oxygen, and neither absorbs infrared to any useful degree. Both are diatomic molecules made of two identical atoms: the charge is distributed symmetrically, the molecule has no dipole moment, and a vibration cannot change one. An electromagnetic wave has nothing to push on.

The greenhouse gases — water vapour, carbon dioxide, methane and nitrous oxide — are different. Their molecules can vibrate in ways that change the distribution of charge, so the oscillating electric field of a passing wave can drive them. And crucially, the natural frequencies of those vibrations lie in the infrared.

Left: a nitrogen molecule of two identical atoms, symmetric with no dipole moment, with an infrared wave passing straight through it unabsorbed. Right: a carbon dioxide molecule shown in its bending and asymmetric stretching modes, each of which separates the centres of positive and negative charge and so creates an oscillating dipole that an infrared wave can drive. Below, a resonance curve of absorbed energy against driving frequency shows a sharp peak where the driving frequency matches the molecule's natural frequency of vibration.
This is the forced-oscillation picture from C.4, applied to a molecule: drive an oscillator at its natural frequency and the energy transfer is enormous.

So an infrared photon of the right frequency drives a greenhouse-gas molecule at resonance, is strongly absorbed, and the energy is re-radiated a moment later — but in a random direction. Roughly half of it heads back down. That is the whole mechanism.

🔭The open window and the closed one

The mechanism only matters because the radiation going out is at completely different wavelengths from the radiation coming in, and Wien’s law says why.

Sun, 5800 K\( \lambda_{\max} = \dfrac{2.9\times10^{-3}}{5800} = 0.50\ \mu\text{m} \)
Earth, 288 K\( \lambda_{\max} = \dfrac{2.9\times10^{-3}}{288} = 10\ \mu\text{m} \)
Two black-body curves on a logarithmic wavelength axis from 0.1 to 100 micrometres, each scaled to its own peak so both are visible. The Sun's curve at 5800 kelvin peaks at 0.5 micrometres in the visible; the Earth's curve at 288 kelvin peaks near 10 micrometres in the infrared, and the two barely overlap. Below the curves, bars mark where the atmosphere absorbs: it is largely transparent across the visible, strongly absorbing across most of the infrared through water vapour and carbon dioxide bands, with a gap between about 8 and 13 micrometres known as the atmospheric window.
Each curve is scaled to its own peak, so this shows where the two spectra sit, not their relative size. Sunlight arrives where the atmosphere is transparent and leaves where it is not.

Sunlight arrives mostly as visible light, and the atmosphere is largely transparent there, so it reaches the ground and warms it. The warmed ground re-radiates at around 10 µm — right where carbon dioxide and water vapour absorb. Energy comes in through an open window and has to leave through a closed one. The narrow gap between about 8 and 13 µm where the atmosphere is reasonably transparent is called the atmospheric window, and it is the 10 W m\(^{-2}\) that escapes directly in the budget above.

🏭The gases, and natural versus enhanced

The four main greenhouse gases set out with their natural and human sources. Water vapour comes from evaporation from the oceans and is the largest natural contributor, but its concentration is controlled by temperature rather than directly by emissions. Carbon dioxide comes from respiration, decay, volcanoes and above all the burning of fossil fuels and deforestation. Methane comes from wetlands, ruminant livestock, rice paddies, landfill and fossil fuel extraction, and is a far stronger absorber per molecule than carbon dioxide. Nitrous oxide comes from soil bacteria and from nitrogen fertilisers.
Concentration and strength are different things: methane is present in tiny amounts compared with carbon dioxide but absorbs far more strongly per molecule.

Keep the two effects distinct, because examiners test the distinction directly:

  • The natural greenhouse effect is not a problem — it is the reason the planet is habitable. Without it the mean surface temperature would be the 255 K calculated above rather than 288 K, and the oceans would be ice.
  • The enhanced greenhouse effect is the additional warming caused by the greenhouse gases human activity has added, principally carbon dioxide from burning fossil fuels. It shifts the balance point, so the planet settles at a higher temperature.
Do not write that the greenhouse effect is bad. A script that says “greenhouse gases trap heat and this causes global warming” earns very little: it neither names the mechanism nor separates the natural effect from the enhanced one. The answer that scores is the chain — short-wavelength radiation in through a transparent atmosphere, absorption at the surface, re-radiation at long wavelength, resonant absorption by greenhouse-gas molecules, re-radiation in all directions including downwards, so the surface sits at a higher temperature than the bare balance predicts.

🔭See it happen

PhET — The Greenhouse Effect simulation. Add and remove greenhouse gases and watch the surface temperature follow. The version with the photon view is the useful one: you can watch infrared photons leave the ground, get absorbed, and come back down, which is the step most written answers skip.

📝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\(^{-2}\). Note that the Sun’s luminosity never had to be looked up — only the ratio of the distances matters.
2. Explain why the average intensity of solar radiation over a planet’s whole surface is a quarter of the solar constant.
The planet intercepts the beam over its cross-sectional area, a disc of area \( \pi r^{2} \), but the energy is distributed over the whole sphere of surface area \( 4\pi r^{2} \) as the planet rotates. The ratio is \( \pi r^{2} : 4\pi r^{2} = 1 : 4 \), so the average intensity is \(S/4\). It does not depend on the radius.
3. 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, and its equilibrium temperature if it behaves as a black body.
Absorbed: \( 340 \times (1 - 0.45) = 187 \) W m\(^{-2}\). Then \( T^{4} = 187/(5.67\times10^{-8}) = 3.30\times10^{9} \), so \( T = 240 \) K. Higher albedo, colder planet — as expected, since more of the arriving energy is thrown straight back.
4. A body of surface area 2.0 m\(^{2}\) at 350 K has an emissivity of 0.40. Find the power it radiates.
\( P = e\sigma AT^{4} = 0.40 \times 5.67\times10^{-8} \times 2.0 \times 350^{4} \). With \( 350^{4} = 1.50\times10^{10} \), \( P = 6.8 \times 10^{2} \) W. A black body of the same size and temperature would radiate \( 1.7\times10^{3} \) W — the emissivity is exactly the ratio of the two.
5. Explain why nitrogen and oxygen, which make up 99% of the atmosphere, are not greenhouse gases.
Both are diatomic molecules made of two identical atoms, so the charge distribution is symmetric and the molecule has no dipole moment; vibrating does not create one. An infrared wave’s oscillating electric field therefore has nothing to couple to, and the radiation passes through. Greenhouse gas molecules can vibrate in modes that separate the centres of positive and negative charge, and the natural frequencies of those vibrations lie in the infrared, so they absorb strongly by resonance.
6. Explain why the atmosphere is transparent to incoming solar radiation but not to outgoing radiation from the surface.
The two are at very different wavelengths, by Wien’s law. The Sun’s surface is at about 5800 K, so its radiation peaks at about 0.5 µm, in the visible, where the atmosphere absorbs very little. The Earth’s surface is at about 288 K, so it re-radiates with a peak near 10 µm, in the infrared, which is precisely where the vibrational resonances of carbon dioxide and water vapour lie. Energy enters through a transparent window and tries to leave through an opaque one.
7. State the difference between the natural and the enhanced greenhouse effect.
The natural greenhouse effect is the warming produced by the greenhouse gases that have always been present; it raises the mean surface temperature from about 255 K to about 288 K and makes the planet habitable. The enhanced greenhouse effect is the additional warming caused by the extra greenhouse gases released by human activity, principally carbon dioxide from burning fossil fuels. The mechanism is identical; only the quantity of absorbing gas differs.
8. Describe the ice–albedo feedback loop and say whether it is positive or negative feedback.
A rise in temperature melts ice; ice has a high albedo (about 0.5–0.85) and the ocean or ground beneath it has a low one (about 0.06–0.25); so the average albedo falls; so a larger fraction of the incident radiation is absorbed rather than reflected; so the temperature rises further. The loop reinforces the change that started it, so it is positive feedback.
9. In the energy budget, the surface emits about 400 W m\(^{-2}\) but only receives about 160 W m\(^{-2}\) directly from the Sun. Explain how this is possible without violating conservation of energy.
The surface also receives back radiation from the atmosphere, about 340 W m\(^{-2}\), which is larger than the direct solar input. Its total input is therefore \( 160 + 340 = 500 \) W m\(^{-2}\), and its total output is \( 400 \) W m\(^{-2}\) of radiation plus about \( 100 \) W m\(^{-2}\) carried away by evaporation and convection — also 500. Nothing is created: the atmosphere is returning energy that originally came from the surface, having absorbed it and re-radiated it in all directions.
10. A planet with no atmosphere has an albedo of 0.12 and receives a solar constant of 2600 W m\(^{-2}\). Estimate its mean surface temperature.
\( \sigma T^{4} = (1-0.12)\times\dfrac{2600}{4} = 0.88 \times 650 = 572 \) W m\(^{-2}\). So \( T^{4} = 572/(5.67\times10^{-8}) = 1.01\times10^{10} \) and \( T = 317 \) K, about 44 °C. With no atmosphere there is no greenhouse correction to add, so unlike the Earth calculation this estimate should be roughly right — though a slow-rotating planet would be far hotter on the day side and colder on the night side than any single average suggests.

🔗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.

  • NASA Earth Observatory — the Earth’s energy budget
  • HyperPhysics — greenhouse effect and atmospheric absorption
  • PhET — The Greenhouse Effect