Energy Balance of Earth — B.2 Greenhouse Effect — IB Physics Notes

2 Lesson 2 of 4 · B.2 Greenhouse Effect

Energy Balance of Earth

Earth stays at roughly the same temperature year after year because it radiates away exactly as much energy as it absorbs. This lesson builds that balance into a single equation — and uses it to explain why the Earth is warm enough to live on.

🌍 Hook – why is the Earth exactly the right temperature?

Mercury bakes at over 400°C by day; Mars freezes at −60°C on average. Earth sits at a gentle 15°C — comfortable for liquid water and life. That is not luck. It is the outcome of a balance: Earth absorbs solar radiation at exactly the rate it radiates thermal energy back into space. Get that balance's inputs even slightly wrong — through a change in albedo or emissivity — and the equilibrium temperature shifts.

Photovoltaic solar panels absorbing and reflecting incident thermal radiation

Fig 1 — 580 W of thermal radiation strikes a 1.86 m × 2.12 m solar panel perpendicularly. Some is reflected (the panel's albedo), the rest absorbed.

💡 Diagnostic question — if the Sun suddenly got hotter, would the Earth's temperature keep rising forever? No. A higher T means the Earth radiates away thermal energy faster (power ∝ T⁴), so it quickly reaches a new, higher equilibrium temperature where absorption = emission again.

📐 Emissivity, recapped

Before building the full balance, recall that a real surface radiates less efficiently than a perfect black body of the same temperature. Emissivity, ε, compares the two:

Emissivity (no units, 0 ≤ ε ≤ 1) $$\varepsilon = \frac{\text{power radiated per unit area}}{\sigma T^4}$$

Rearranged, this is how emissivity is measured in practice: record the actual power radiated per unit area of a surface at known temperature T, then divide by $\sigma T^4$ — the black-body value.

🔢 Sample calculation. A surface radiates $1.3\times10^{16}$ W over an area $1.4\times10^{14}\ \text{m}^2$, at $T = 210\ \text{K}$:
$\varepsilon = \dfrac{1.3\times10^{16}/1.4\times10^{14}}{5.67\times10^{-8}\times210^4}$
$= 0.84$

⚖️ Modelling a planet's energy balance

Over any extended time interval, a planet in thermal equilibrium satisfies:

$$\text{radiant energy received} = \text{radiant energy emitted}$$

The planet only intercepts sunlight over its disc-shaped cross-section, $\pi r^2$ — the "shadow" it casts. But it radiates from its entire curved surface, $4\pi r^2$. Putting intensity, albedo, and emissivity together:

Planetary energy balance $$S\pi r^2(1-\alpha) = \varepsilon\sigma(4\pi r^2)T^4$$

where S is the solar constant and α is the albedo. Note $\pi r^2$ cancels from both sides — the planet's radius does not actually affect its equilibrium temperature.

For Earth: $S = 1.36\times10^3\ \text{W m}^{-2}$, $r = 6.4\times10^6\ \text{m}$, average emissivity (Earth + atmosphere) $\varepsilon = 0.61$, average albedo $\alpha = 0.315$:

$$1.36\times10^3\times(1-0.315) = 0.61\times(5.67\times10^{-8})\times4T^4$$ $$T = 286\ \text{K}\ (13^{\circ}\text{C})$$

The Earth's actual mean surface temperature is 288 K (15°C) — so this simple model is reasonably accurate.

📖 Three things follow immediately: (1) ε and α fundamentally set a planet's temperature, and both take different values for a planet without an atmosphere. (2) human activity is measurably changing Earth's ε and α. (3) small changes in T can drive large — mostly harmful — changes in climate.

🧭 Nature of science – using computers to expand human knowledge

Predicting the future is usually far more likely to go wrong than right, because almost every real situation has too many variables and unknown factors — this is true even for events governed purely by physics, like next week's weather, let alone events shaped by human behaviour.

Mathematical modelling represents a situation with equations and numbers, but nearly always carries simplifications and assumptions that leave predictions uncertain. Modern computers can handle the enormous datasets and calculations that climate modelling requires — far beyond what any person could manage — but a computer's predictions are only as good as the input data and the specific tasks humans set it. One check on accuracy: run a model on a known past situation and see whether it correctly predicts what actually happened next. Predicting the past is always much easier than predicting the future.

🏡 The greenhouse effect – energy transfers within the system

So far we treated "Earth + atmosphere" as a single system. To understand the mechanism properly, we need the energy transfers between the surface and the atmosphere — the greenhouse effect.

An ornate glass greenhouse used to grow plants in a colder climate

Fig 2 — Glasshouses like this became popular over the last two centuries in colder countries, for growing plants native to hotter climates.

The Sun's surface radiates at 5780 K, mostly as visible light and short-wavelength infrared. This radiation is transmitted through the greenhouse glass and warms the interior; some longer wavelengths are reflected by the glass, and much of the visible light exits again after bouncing off surfaces inside.

The warmed ground, pots and plants absorb infrared and heat up, then transfer thermal energy to the air by conduction; convection carries the warm air upward, but it mostly stays inside the greenhouse. Crucially, the contents of the greenhouse are much cooler than the Sun (≈300 K), so they re-radiate at much longer infrared wavelengths — and these longer wavelengths cannot pass back out through the glass. Thermal energy is effectively trapped inside.

Diagram of visible and short-wave infrared entering a greenhouse through glass, while long-wave infrared radiated from the warm interior is trapped

Fig 3 — Visible and short-wave IR pass through the glass and warm the interior; the interior's own re-radiated long-wave IR cannot pass back out.

🤔 Discussion — how valid is comparing this to the Earth's atmosphere? Is "greenhouse effect" a useful term, or could it mislead? (There is no single correct answer — discuss after finishing this topic.)

📊 Sun vs Earth: two very different spectra

The Sun's black-body spectrum (5780 K) peaks at short, visible wavelengths. The Earth's much cooler surface (288 K) radiates a far lower-power spectrum, peaking at much longer infrared wavelengths. Compared to the Sun, Earth's radiation is at much lower power, with much longer wavelengths.

Most of the Sun's radiation passes straight through the atmosphere to reach the surface. But because Earth's outgoing radiation sits at different, longer wavelengths, a much smaller proportion of it is able to pass back out through that same atmosphere. Even so, over time the surface temperature adjusts until: total power received (Earth + atmosphere) = total power radiated (Earth + atmosphere).

🔎 Graph mastery

Graph type	Axes (X, Y)	Shape	What to read
Black-body spectra (Sun vs Earth)	Wavelength /10⁻⁷ m (log scale) · relative intensity	Tall, narrow peak at short wavelength for the Sun (5780 K); much smaller, broader peak at long wavelength for Earth (288 K)	The two peaks cannot be directly compared in height — Earth's is exaggerated for visibility. Read off which curve dominates at a given wavelength
Solar spectrum reaching sea level	Wavelength /10⁻⁶ m · relative intensity	Smooth black-body curve (top of atmosphere) with a jagged, dipped curve beneath it (sea level)	The dips are wavelengths absorbed by atmospheric gases before reaching the ground; UV, visible and IR regions are marked

Black-body spectra comparing the Sun at 5780 kelvin and the Earth at 288 kelvin on a logarithmic wavelength scale

Fig 4 — The Sun's peak sits at short (visible) wavelengths; Earth's sits far to the right, at long infrared wavelengths.

Graph comparing the Sun's spectrum outside the atmosphere with the spectrum reaching sea level, showing absorption dips

Fig 5 — Absorption bands in the atmosphere remove specific wavelengths before sunlight reaches sea level.

🔄 Energy flow through the Earth's atmosphere

Averaged over the whole planet and over an extended time, the mean intensity of solar radiation arriving at Earth is 340 W m⁻². With an average albedo of 0.315, only 68.5% of that — 233 W m⁻² — is actually absorbed, split between the atmosphere and the surface.

Fig 6 — The whole Earth, its surface, and its atmosphere each receive and emit thermal energy at the same overall rate. Energy moves between surface and atmosphere by radiation and convection.

🔬 ATL research skill — find published data for this energy flow diagram online. Do sources agree? Which look most reliable or up to date? Do different websites trace back to the same original source?

💨 Greenhouse gases

Greenhouse gases are gases in the Earth's atmosphere that absorb and re-emit infrared radiation, thereby affecting Earth's temperature. The principal ones are water vapour (H₂O), carbon dioxide (CO₂), methane (CH₄) and nitrous oxide (N₂O) — each with both natural and human-activity origins. Atmospheric concentrations of CO₂, CH₄ and N₂O have been increasing significantly in recent years.

Earth's atmosphere formed over millions of years through volcanic and biological processes, and through collisions with comets and asteroids. By volume, air is approximately 78% nitrogen, 21% oxygen and 0.9% argon, with naturally occurring traces of other gases including CO₂ and water vapour. The greenhouse gases matter disproportionately: they absorb and then re-emit infrared radiation in all directions, an effect explained at the molecular level either by a resonance model or in terms of molecular energy levels.

✍️ Worked example – the Moon's average emissivity

Calculate the average emissivity of the Moon. Assume an average albedo of 0.12 and an average surface temperature of 274 K (noting the Moon's actual surface temperature varies enormously, so "average" is not a well-defined quantity).

Identify: the Moon is airless, but the same energy-balance equation applies with the Moon's own albedo, temperature, and an unknown emissivity to find.

Data: $S = 1.36\times10^3\ \text{W m}^{-2}$, $\alpha = 0.12$, $T = 274\ \text{K}$.

Equation:
$1.36\times10^3\times(1-0.12)$
$= \varepsilon\times(5.67\times10^{-8})\times4\times274^4$

Answer: mean emissivity $\varepsilon = 0.94$.

The Moon shown as a simple flat illustration for the worked example on lunar emissivity

Fig 7 — The Moon has no atmosphere, so its albedo and emissivity are set by the lunar surface alone.

🌊 Real-world: the ice–albedo feedback loop

If the Earth's surface temperature rises, snow and ice melt faster — a reasonable expectation, since a warmer surface transfers thermal energy to ice more quickly. But it works both ways: as snow and ice melt, they expose darker ocean or land underneath, which has a lower albedo. That darker surface absorbs more solar radiation, which raises the surface temperature further — melting still more ice. Average ocean albedo is higher in winter and closer to the poles, both because more of the surface is ice-covered and because sunlight strikes at a shallower, more reflective angle.

✓ Examiner tip: name this explicitly as positive feedback — a change that reinforces itself, rather than opposing itself (negative feedback).

🚀 HL Extension – temperatures of planets around distant stars

The same energy-balance equation applies to any planet orbiting any star. The equilibrium surface temperature depends on: the star's luminosity/temperature, the planet's orbital distance (which sets its "solar constant" via the inverse-square law), the planet's albedo, its emissivity, whether it has an atmosphere (and the atmosphere's greenhouse gas content), and any internal heat sources.

Challenge: two planets orbit at the same distance from identical stars, but Planet X has a thick, cloudy atmosphere and Planet Y has none. Which is warmer? It depends on the balance of two competing effects: a thick cloudy atmosphere usually raises albedo (reflecting more sunlight, cooling the planet) but also raises the effective emissivity term's denominator effect by trapping outgoing IR (warming the planet) — Venus's extreme atmosphere shows the greenhouse effect winning decisively.

⚠️ Trap corner – common mistakes

Misconception	Correct view
A planet absorbs sunlight over its whole surface, $4\pi r^2$	It only intercepts sunlight over its cross-sectional "shadow" area, $\pi r^2$; it emits from the full surface, $4\pi r^2$
Albedo and emissivity are the same property	Albedo (α) describes reflection of incoming shortwave radiation; emissivity (ε) describes how efficiently a surface radiates thermal energy — they are independent
The model predicting T = 286 K is "wrong" because the real value is 288 K	A simple model using averaged ε and α is expected to be approximate; 286 K vs 288 K counts as a good agreement
The "greenhouse effect" traps energy the exact same physical way a garden greenhouse does	Real greenhouses mostly work by preventing convective heat loss (trapped warm air); the atmospheric greenhouse effect is dominated by selective infrared absorption/re-emission by gases, not blocked convection
Radius affects a planet's equilibrium temperature in this model	$\pi r^2$ cancels algebraically from both sides of the balance equation — temperature depends only on S, α and ε, not on r

⚠️ Most common slip: using the wrong area — $\pi r^2$ (cross-section, for absorption) instead of $4\pi r^2$ (full sphere, for emission), or vice versa.

ε = P/A ÷ σT⁴ Sπr²(1−α) = εσ(4πr²)T⁴ solar constant S = 1360 W m⁻² Earth: ε ≈ 0.61, α ≈ 0.315 T ≈ 286 K matches observed 288 K ice–albedo positive feedback greenhouse gases: H₂O, CO₂, CH₄, N₂O r cancels — radius doesn't set T

Practice check — multiple choice

1. In the Earth's energy balance equation, $\pi r^2$ is used for the absorbed power and $4\pi r^2$ for the emitted power because

A the Earth absorbs sunlight over its whole spherical surface

B only the sunlit disc faces the Sun, but the whole sphere radiates in all directions

C the Earth's surface area changes with the seasons

D albedo only applies to half the Earth

2. Earth's average albedo of 0.315 means that

A 31.5% of incoming solar radiation is absorbed

B 68.5% of incoming solar radiation is absorbed, and 31.5% is reflected

C 31.5% of outgoing infrared is trapped by the atmosphere

D the Earth's emissivity is 0.315

3. Why can't sunlight-warmed air inside a greenhouse simply radiate its heat straight back out through the glass?

A Glass is opaque to all radiation

B The warm interior re-radiates at much longer infrared wavelengths, which the glass does not transmit

C Convection removes all the heat before it can radiate

D All the Sun's radiation is absorbed by the glass itself

4. A rise in surface temperature melts ice, exposing darker land/ocean, which absorbs more sunlight and raises the temperature further. This is an example of

A negative feedback

B an isolated system

C positive feedback

D a violation of energy conservation

5. Which combination gives the HIGHEST equilibrium surface temperature in the energy balance model?

A High solar constant, high albedo, high emissivity

B High solar constant, low albedo, low emissivity

C Low solar constant, high albedo, high emissivity

D High solar constant, high albedo, low emissivity

Practice check — fill in the blanks

Complete the statements:

A planet absorbs sunlight over its area, $\pi r^2$, but radiates from its full area, $4\pi r^2$. The average intensity of sunlight arriving at Earth is called the . The fraction of incoming radiation reflected is the , and the efficiency with which a surface radiates compared to a black body is its . Gases such as CO₂ and water vapour that absorb and re-emit infrared radiation are called . When melting ice exposes darker surfaces that absorb more sunlight and warm the Earth further, this is called feedback. Within the atmosphere, energy also moves between surface and air by , not just radiation.

Exam‑style questions

Exam question · Calculate / State / Describe

Q1. 580 W of thermal radiation is incident perpendicularly on photovoltaic solar panels of dimensions 1.86 m × 2.12 m. (a) Calculate the intensity of the radiation. (b) If the panel's albedo is 0.21, determine the total rate at which the panel absorbs thermal radiation. (c) Describe how the albedo of the panels is minimised. (d) State two reasons why it is best if the panels are perpendicular to the incident radiation. [6 marks]

✓ (a) $I = 580/(1.86\times2.12) = 147\ \text{W m}^{-2}$.
✓ (b) absorbed rate $= 580\times(1-0.21) = 458\ \text{W}$.
✓ (c) an anti-reflective coating and/or a dark, textured surface reduces the fraction reflected.
✓ (d) perpendicular incidence maximises the effective area exposed to the beam (no cosθ reduction), and reflection (albedo) increases at large angles of incidence, so keeping the beam perpendicular minimises reflection losses.

Exam question · State / Explain

Q2. (a) State why it would be reasonable to expect that, if the temperature of the Earth's surface were to rise, snow and ice would melt quicker. (b) Conversely, explain why it would also be reasonable to expect that, if large amounts of snow and ice were to melt, the temperature of the Earth's surface would rise. [4 marks]

✓ (a) A warmer surface transfers thermal energy to the ice/snow at a greater rate, so it reaches its melting point and changes state faster.
✓ (b) Melting ice and snow exposes darker land or ocean underneath, which has a lower albedo.
✓ That darker surface absorbs a greater fraction of incoming solar radiation, raising the surface temperature further — a positive feedback loop.

Exam question · Explain

Q3. Explain why the average albedo of ocean water will tend to increase (a) in winter (b) closer to the poles. [3 marks]

✓ (a) More of the ocean surface is covered by ice/snow in winter, which has a much higher albedo than open water.
✓ (b) Near the poles the Sun's rays strike the water at a much shallower (more oblique) angle.
✓ Reflection off a water surface increases sharply at shallow angles of incidence, so more of the light is reflected rather than absorbed.

Exam question · Discuss

Q4. Discuss why the emissivity of clouds makes them an important factor affecting the average emissivity of Earth. [3 marks]

✓ Clouds cover a very large and constantly changing fraction of the Earth's surface area at any given time.
✓ Cloud emissivity varies considerably with cloud type, thickness and altitude, so it is not a fixed value.
✓ Because clouds cover so much area, even modest changes in average cloud cover or type shift the whole-Earth average emissivity noticeably, which is why cloud behaviour is one of the largest uncertainties in climate models.

Exam question · Calculate

Q5. A brick wall is 2.34 m high and 3.80 m long. The bricks have an emissivity of 0.72. At what rate is thermal energy radiated away from the wall if its temperature is 18°C? [3 marks]

✓ $A = 2.34\times3.80 = 8.89\ \text{m}^2$, $T = 291\ \text{K}$.
✓ $P = \varepsilon\sigma AT^4 = 0.72\times(5.67\times10^{-8})\times8.89\times291^4$.
✓ $P \approx 2.6\times10^3\ \text{W}$ (about 2.6 kW).

Exam question · Show that

Q6. Show that the total power radiated away from the Earth's surface is approximately $1\times10^{17}\ \text{W}$. Assume surface temperature 288 K and Earth radius $6.4\times10^6\ \text{m}$. [3 marks]

✓ $A = 4\pi r^2 = 4\pi(6.4\times10^6)^2 = 5.15\times10^{14}\ \text{m}^2$.
✓ Using the average Earth emissivity $\varepsilon = 0.61$: $P = \varepsilon\sigma AT^4 = 0.61\times(5.67\times10^{-8})\times5.15\times10^{14}\times288^4$.
✓ $P \approx 1.2\times10^{17}\ \text{W} \approx 1\times10^{17}\ \text{W}$, as required.

Exam question · Determine

Q7. To determine a value for the Earth's surface temperature if it never had an atmosphere, use estimated values of ε = 0.9 and albedo = 0.3. [3 marks]

✓ $1.36\times10^3\times(1-0.3) = 0.9\times(5.67\times10^{-8})\times4T^4$.
✓ $T^4 = 952/(2.04\times10^{-7}) = 4.66\times10^9$.
✓ $T \approx 261\ \text{K}$ (about −12°C).

Exam question · Determine

Q8. Determine a value for the surface temperature of the Earth if it never had an atmosphere and behaved as a perfect black body. [2 marks]

✓ A perfect black body has $\varepsilon = 1$ and $\alpha = 0$: $1.36\times10^3 = 1\times(5.67\times10^{-8})\times4T^4$.
✓ $T^4 = 6.00\times10^9$, so $T \approx 278\ \text{K}$ (about 5°C).

Exam question · Calculate

Q9. Calculate the average surface temperature of Mars, assuming an average emissivity of 0.95 and an average albedo of 0.21. Its solar constant is 590 W m⁻². [3 marks]

✓ $590\times(1-0.21) = 0.95\times(5.67\times10^{-8})\times4T^4$.
✓ $T^4 = 466/(2.15\times10^{-7}) = 2.16\times10^9$.
✓ $T \approx 216\ \text{K}$ (about −57°C).

Exam question · Predict

Q10. Suppose climate change resulted in a 5% change in both the emissivity and the albedo of the Earth. Predict a new value for the maximum average surface temperature. [4 marks]

✓ To get the maximum possible temperature, both changes must act to warm the planet: a 5% decrease in albedo ($0.315\to0.299$) and a 5% decrease in emissivity ($0.61\to0.580$).
✓ $1.36\times10^3\times(1-0.299) = 0.580\times(5.67\times10^{-8})\times4T^4$.
✓ $T^4 = 953/(1.31\times10^{-7}) = 7.25\times10^9$.
✓ $T \approx 292\ \text{K}$ (19°C) — about 6 K warmer than the original 286 K.

Exam question · State

Q11. State the factors which will affect the surface temperatures of planets orbiting a distant star. [4 marks]

✓ The star's luminosity/surface temperature, and the planet's orbital distance (which together set its "solar constant" via the inverse-square law).
✓ The planet's albedo.
✓ The planet's emissivity.
✓ Whether it has an atmosphere, and that atmosphere's greenhouse gas content; also any significant internal heat source.

Score: 0 pts

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