Greenhouse effect

The greenhouse effect is really a story about black-body radiation (from B.1), so we begin there.

A black body is an idealised object that absorbs all the radiation falling on it and, at a given temperature, is the best possible emitter. Any warm object glows with a continuous spread of wavelengths — a black-body spectrum — and two rules govern that spectrum:

• Hotter bodies radiate more power. The intensity (power per unit area) follows the Stefan–Boltzmann law, $I = \sigma T^4$, where $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$. Because of the fourth power, a small rise in temperature is a large rise in power.
• Hotter bodies radiate at shorter wavelengths. The peak of the spectrum shifts to shorter wavelength as temperature rises (Wien's displacement law, B.1).

Now apply this to two very different objects:

Body	Surface temperature	Where its radiation peaks	So it emits mainly…
The Sun	$\approx 5800\,\text{K}$	$\approx 500\,\text{nm}$	visible light
The Earth	$\approx 255\,\text{K}$	$\approx 10\,\mu\text{m}$	infrared radiation

This single contrast is the key to the whole topic. Sunlight arrives as visible light, most of which passes straight through the atmosphere and warms the ground. The warm ground then radiates energy back as infrared. Greenhouse gases are almost transparent to incoming visible light but strongly absorb outgoing infrared — so the two motions of energy behave completely differently, and that asymmetry is what keeps the surface warm.

The solar constant $S$ is the mean intensity (power per unit area) of solar radiation arriving at the top of Earth's atmosphere, measured on a surface perpendicular to the Sun's rays at Earth's average distance: $S \approx 1361\,\text{W m}^{-2}$

How much power does the whole Earth intercept? Not the intensity times the surface area — the Sun's rays are (very nearly) parallel, so the Earth intercepts sunlight as though it were a flat disc facing the Sun. The area of that disc (the cross-section) is $A_{\text{intercept}} = \pi R^2$ where $R$ is Earth's radius. So the total incoming solar power is $P_{\text{in}} = S \times \pi R^2.$

How much power does the Earth radiate? The Earth radiates from its entire spherical surface (day and night, all directions), whose area is $A_{\text{emit}} = 4\pi R^2.$

Here is the crucial point that examiners test relentlessly: the planet absorbs over $\pi R^2$ but emits over $4\pi R^2$. The ratio $\dfrac{4\pi R^2}{\pi R^2} = 4$ is where the famous factor of 4 comes from. If you spread the intercepted sunlight over the whole surface, the average incoming intensity is only $\dfrac{S}{4} \approx 340\,\text{W m}^{-2}$.

Two dimensionless numbers (each between 0 and 1) turn the black-body idea into real planets.

Albedo $\alpha$ — the fraction of incident radiation a surface reflects: $\alpha = \frac{\text{total power reflected}}{\text{total power incident}}.$ It has no units because it is a ratio of powers. A perfect mirror has $\alpha = 1$; a perfect black body has $\alpha = 0$. Earth's average albedo is about $0.30$, meaning roughly 30% of incoming sunlight is reflected (by clouds, ice, deserts and the atmosphere) and the remaining 70% is absorbed. That is why the absorbed power carries the factor $(1-\alpha)$: $P_{\text{absorbed}} = S(1-\alpha)\,\pi R^2.$ Bright surfaces raise the albedo: fresh snow reflects most of the light that hits it ($\alpha \approx 0.8$–$0.9$), whereas the open ocean is dark ($\alpha \approx 0.06$). This is why melting ice, by lowering the albedo, is a warming feedback.

Emissivity $\varepsilon$ — how good a surface is at radiating, compared with a perfect black body at the same temperature: $\varepsilon = \frac{\text{power radiated by the surface}}{\text{power radiated by a black body at the same temperature}}.$ A perfect black body has $\varepsilon = 1$; a shiny, poor emitter has a small $\varepsilon$. Emissivity multiplies the Stefan–Boltzmann law: $P_{\text{radiated}} = \varepsilon\,\sigma A T^4.$ For many planetary calculations the Earth is treated as a near-perfect emitter in the infrared, so $\varepsilon \approx 1$ unless a question tells you otherwise.

Keeping them straight: albedo is about incoming, reflected radiation; emissivity is about outgoing, radiated radiation. Both are just fractions — no units, never greater than 1.

This is the central derivation of B.2, and it is built from first principles you already have. The idea is simple: if a planet's temperature is steady, the power it absorbs from the Sun must exactly equal the power it radiates back to space. This is the energy balance.

Step 1 — power absorbed. The planet intercepts sunlight over its disc $\pi R^2$, but only the fraction $(1-\alpha)$ is absorbed: $P_{\text{in}} = S(1-\alpha)\,\pi R^2.$

Step 2 — power radiated. The planet radiates as a (near) black body from its whole surface $4\pi R^2$, with emissivity $\varepsilon$ and the Stefan–Boltzmann law: $P_{\text{out}} = \varepsilon\,\sigma\,(4\pi R^2)\,T^4.$

Step 3 — balance them. At the equilibrium temperature, $P_{\text{in}} = P_{\text{out}}$: $S(1-\alpha)\,\pi R^2 = \varepsilon\,\sigma\,4\pi R^2\,T^4.$

Step 4 — the $\pi R^2$ cancels (this is why the planet's actual size does not matter). Dividing both sides by $\pi R^2$: $S(1-\alpha) = 4\,\varepsilon\,\sigma\,T^4.$

Step 5 — solve for $T$: $\boxed{\,T = \left[\dfrac{S(1-\alpha)}{4\,\varepsilon\,\sigma}\right]^{1/4}\,}$

Put the numbers in ($S = 1361\,\text{W m}^{-2}$, $\alpha = 0.30$, $\varepsilon = 1$, $\sigma = 5.67\times10^{-8}$): $T = \left[\frac{1361(1-0.30)}{4(1)(5.67\times10^{-8})}\right]^{1/4} = \left[\frac{952.7}{2.268\times10^{-7}}\right]^{1/4} = \left(4.20\times10^{9}\right)^{1/4} \approx 255\,\text{K}.$

What the answer tells you. $255\,\text{K} = -18\,^{\circ}\text{C}$ — far below freezing. Yet Earth's true average surface temperature is about $288\,\text{K}$ ($+15\,^{\circ}\text{C}$). The $33\,^{\circ}\text{C}$ difference is exactly what the greenhouse effect provides: without it, the planet would be a frozen ball. The model also shows the levers clearly — a higher albedo lowers $T$ (less energy absorbed), while a lower effective emissivity raises $T$ (energy escapes to space less easily), which is precisely what adding greenhouse gases does.

Now the physics that gives the topic its name. The main greenhouse gases in Earth's atmosphere are:

• Water vapour ($\text{H}_2\text{O}$) — the largest natural contributor,
• Carbon dioxide ($\text{CO}_2$),
• Methane ($\text{CH}_4$),
• Nitrous oxide ($\text{N}_2\text{O}$).

(Note that $\text{CO}_2$ is not the only greenhouse gas — a very common misconception.)

Why these gases and not nitrogen or oxygen? A molecule can absorb infrared only if the vibration of its bonds changes its distribution of charge (its dipole). The simple diatomic molecules $\text{N}_2$ and $\text{O}_2$ — which make up 99% of the air — cannot do this, so they are transparent to infrared. Greenhouse-gas molecules have bonds that do interact with infrared.

The molecular mechanism, step by step:

1. Sunlight in. Incoming solar radiation is mostly visible; greenhouse gases barely absorb it, so it passes through the atmosphere and warms the surface.
2. Infrared out. The warm surface radiates energy back upward as infrared.
3. Resonant absorption. Each greenhouse-gas molecule has bonds that vibrate at particular natural (resonant) frequencies. When outgoing infrared radiation has a frequency matching one of these natural bond-vibration frequencies, the molecule absorbs that photon and vibrates more energetically — this is resonance.
4. Re-emission in all directions. The excited molecule quickly re-emits infrared radiation, but in all directions — including back down towards the surface.
5. Net warming. Because some of the outgoing infrared is returned to the surface instead of escaping to space, the surface receives extra energy and settles at a higher temperature than the bare energy balance predicts.

Say it precisely in the exam. Greenhouse gases do not simply "trap heat like a blanket". The correct physics is: they absorb infrared radiation at the natural (resonant) frequencies of their molecular bonds and re-emit it in all directions, so some returns to the surface. Naming that mechanism is what earns the marks.

It is essential to keep two ideas apart.

The natural greenhouse effect is the warming that has always existed because Earth's atmosphere naturally contains water vapour, carbon dioxide and other greenhouse gases. It raises the surface from the bleak $255\,\text{K}$ predicted by the bare energy balance to a life-friendly $\approx 288\,\text{K}$ ($+15\,^{\circ}\text{C}$). Without it, Earth would be frozen and uninhabitable — so the natural greenhouse effect is a good and necessary thing.

The enhanced greenhouse effect is the extra warming caused by rising concentrations of greenhouse gases, largely from human activity — burning fossil fuels ($\text{CO}_2$), agriculture and livestock ($\text{CH}_4$, $\text{N}_2\text{O}$) and deforestation. More greenhouse-gas molecules absorb a greater fraction of the outgoing infrared and return more of it to the surface, so:

$\text{more greenhouse gas} \;\Rightarrow\; \text{more outgoing infrared absorbed} \;\Rightarrow\; \text{lower effective emissivity to space} \;\Rightarrow\; \text{higher surface temperature.}$

This shifts the energy balance: to radiate the same total power out to space at a higher opacity, the surface must sit at a higher temperature. The result is global warming and climate change.

Feedbacks amplify it. A warmer surface melts ice, lowering the albedo (dark water absorbs more than bright ice), which warms things further. Warming also increases evaporation, adding more water vapour — itself a greenhouse gas. These positive feedbacks are why a modest forcing can produce a larger response.

International-mindedness. Because the atmosphere is shared, greenhouse-gas emissions in one country affect the whole planet, which is why the physics feeds directly into global agreements on emissions. Understanding the underlying energy balance is what lets scientists quantify how much warming a given rise in $\text{CO}_2$ will cause.