Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: greenhouse effect, albedo and radiative forcing (203-SN3-RE)

Q: Why do we divide the solar constant by four?

The Earth intercepts sunlight as a disc, whose area is pi times the radius squared, but it radiates from its whole spherical surface, which is four times larger. Averaged over the globe and over a day, each square metre therefore receives a quarter of the solar constant, about 340 watts per square metre.

Q: What would the temperature of the Earth be without a greenhouse effect?

Setting the absorbed solar energy, with an albedo of 0.30, equal to the energy radiated by a blackbody gives about 255 kelvins, or minus 18 degrees Celsius. The real mean surface temperature is 15 degrees: the natural greenhouse effect therefore contributes about 33 degrees.

Q: What is radiative forcing?

It is the change in the energy balance at the top of the atmosphere, in watts per square metre, caused by a disturbance such as the addition of carbon dioxide or aerosols, before the temperature has had time to respond. A positive forcing means more energy comes in than goes out, and the planet warms until balance is restored.

Q: Why does the forcing of CO2 follow a logarithm?

Because the centre of the carbon dioxide absorption band, near 15 micrometres, is already saturated: the atmosphere absorbs almost everything there. The added gas only acts on the edges of the band, more and more weakly. That is why each doubling of the concentration adds the same forcing, about 3.7 watts per square metre.

The radiative budget of a planet fits in one equality, what comes in equals what goes out, but every term hides a factor that is easy to forget: the quarter from the geometry, the complement of the albedo, the logarithm of the forcing, the division by one minus the feedback factor.

This sheet gathers the pitfalls of 203-SN3-RE tests on the greenhouse effect and radiative forcing, the tree that chooses the relation according to the question, a model write-up and a complete calculation, from the Sun to the warming caused by a doubling of CO₂.

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The thread of the chapter

A planet is in equilibrium when it sends back to space exactly what it absorbs: any disturbance of the balance, a forcing, is paid for in degrees until the outgoing radiation catches up with the incoming radiation. Every pitfall forgets one term of the budget or confuses the forcing with its response.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

The budget of a planet

• Solar constant $S=\dfrac{L}{4\pi d^{2}}$ , that is $1361$ W/m² for the Earth. Mean flux per square metre of SURFACE: $\dfrac{S}{4}=340$ W/m².
• Absorbed flux: $(1-\alpha)\dfrac{S}{4}$ , with albedo $\alpha=0.30$ : $238$ W/m².
• Radiative equilibrium: $\sigma T_{e}^{4}=(1-\alpha)\dfrac{S}{4}$ , hence $T_{e}\approx 255$ K. The real surface is at $288$ K: the $33$ K gap is the greenhouse effect.
• Single-layer model with absorptivity $f$ : $T_{s}=T_{e}\left(\dfrac{2}{2-f}\right)^{1/4}$ ; opaque layer, $T_{s}=2^{1/4}T_{e}$ .

Two budgets not to be confused: at the top, 238 come in and 240 go out; at the surface, 390 are emitted because the atmosphere sends part of the infrared back.

Seen from space, the Earth emits what it absorbs, 240 W/m², from the altitude where its radiation escapes. The surface itself emits 390 W/m², because it also receives the radiation sent back by the atmosphere.

Two radiations that do not overlap

• The Sun radiates like a blackbody at $5772$ K: $\lambda_{max}=\dfrac{2.898\times 10^{-3}}{5772}=502$ nm, in the visible.
• The Earth radiates near $255$ K: $\lambda_{max}=11.4$ µm, in the infrared, a wavelength twenty-three times longer.
• The atmosphere lets most of the visible through and absorbs a large part of the infrared (water vapour, CO₂ near $15$ µm): this shift is what makes the greenhouse effect possible.

The two spectra, scaled to the same height, barely overlap: the atmosphere sorts energy by wavelength, letting the solar in and holding the terrestrial back.

Forcing, response and feedbacks

• Radiative forcing of CO₂: $\Delta F=5.35\ln\dfrac{C}{C_{0}}$ W/m², that is $3.71$ W/m² per doubling, whatever the starting point.
• Planck response: $\lambda_{P}=\dfrac{d(\sigma T^{4})}{dT}=4\sigma T_{e}^{3}=3.74$ W/(m²·K), hence $\Delta T_{0}=\dfrac{\Delta F}{\lambda_{P}}$ .
• Feedbacks of factor $g<1$ : $\Delta T=\dfrac{\Delta T_{0}}{1-g}$ .
• Response in time: $T(t)=\Delta T_{eq}\left(1-e^{-t/\tau}\right)$ , with $\tau$ of the order of a decade for the surface ocean.

The forcing is imposed all at once and stays constant, in W/m²; the warming follows with a lag and tends toward

\frac{\Delta T_{0}}{1-g}

, in kelvins: the cause and the effect.

The forcing is an imbalance in W/m², the warming a response in kelvins: confusing them is confusing cause and effect.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Forgetting the factor 4 between the solar constant and the mean flux

the whole question, and an Earth at 87 °C

What not to write

“ $\sigma T_{e}^{4}=(1-\alpha)S$ , so $T_{e}=\left(\dfrac{0.70\times 1361}{\sigma}\right)^{1/4}=360$ K.”

What to write

“The disc $\pi R^{2}$ intercepts, the sphere $4\pi R^{2}$ re-emits: $\sigma T_{e}^{4}=(1-\alpha)\dfrac{S}{4}$ , $T_{e}=255$ K.”

The beam that hits the Earth has the cross-section of the disc

\pi R^{2}

; the sphere that re-emits has

4\pi R^{2}

: every square metre receives

S/4

on average, and

T_{e}=255

K, not

360

Why: The solar constant is a flux on a surface perpendicular to the rays. Spread over the whole surface of the globe, day and night, poles and equator, it is divided by four.

2. Using albedo as the absorbed fraction

the whole question

What not to write

“ $\sigma T_{e}^{4}=\alpha\dfrac{S}{4}=0.30\times 340$ , so $T_{e}=206$ K.”

What to write

“Albedo is the REFLECTED fraction: the planet absorbs $(1-\alpha)\dfrac{S}{4}=0.70\times 340=238$ W/m², and $T_{e}=255$ K.”

Of the

340

W/m², the share

\alpha

leaves without heating anything:

238

remain absorbed (green). Taking

\alpha

as the absorbed share gives

102

and a planet at

206

K (red).

Why: A high albedo cools: snow, ice and clouds send light back. A formula in which the temperature rises with the albedo is upside down.

3. Treating the forcing of CO₂ as proportional to the concentration

2 marks, and a 40 % overestimate

What not to write

“From $280$ to $420$ ppm: $2.17$ W/m². So from $420$ to $560$ ppm, another $140$ ppm, another $2.17$ W/m².”

What to write

“ $\Delta F=5.35\ln\dfrac{560}{420}=1.54$ W/m²: the forcing grows like the logarithm, each added ppm counts a little less.”

The logarithmic curve flattens as the concentration increases; the dotted line, which extends the slope between 280 and 420 ppm, overestimates the forcing more and more.

Why: The centre of the CO₂ absorption band is already saturated: the extra gas only acts in the wings. It is the RATIO of concentrations that counts, and a doubling always gives 3.7 W/m².

4. Multiplying instead of dividing by the feedback factor

2 marks

What not to write

“ $\Delta T_{0}=0.99$ K and $g=0.60$ , so $\Delta T=0.99\times(1+0.60)=1.6$ K.”

What to write

“ $\Delta T=\dfrac{\Delta T_{0}}{1-g}=\dfrac{0.99}{0.40}=2.5$ K.”

Each round of feedback adds

0.6

times the previous one: the stack climbs to

2.48

K (green). Multiplying by

1+g

counts only the first round,

1.59

K (red).

Why: A feedback applies to its own response: the warming due to water vapour itself produces more water vapour. The sum $1+g+g^{2}+\ldots$ equals $\dfrac{1}{1-g}$ , not $1+g$ .

5. Confusing the radiation emitted by the surface with what leaves for space

3 marks on a reasoning question

What not to write

“The surface is at $288$ K, so the Earth loses $\sigma\times 288^{4}=390$ W/m² to space, far more than the $238$ absorbed: it is cooling.”

What to write

“The surface emits $390$ W/m², but the atmosphere absorbs most of it and sends some back to the ground: what escapes to space is $240$ W/m², the equivalent of a blackbody at $255$ K.”

The surface at

288

K emits

390

W/m², but what escapes leaves from higher up, near

5

km and

255

240

W/m², which balance the

238

absorbed.

Why: The budget at the top of the atmosphere and the budget of the surface are two different budgets. Only the first has to balance the absorbed sunlight; the second contains the radiation sent back by the atmosphere.

6. Assuming the equilibrium warming is reached immediately

2 marks, and an error by a factor of 8

What not to write

“Pinatubo imposes $-3.0$ W/m² for a year, with $\lambda=1.2$ W/(m²·K): the climate cools by $2.5$ K.”

What to write

“With $\tau=7.6$ years, in one year: $\Delta T=-2.5\left(1-e^{-1/7.6}\right)=-0.31$ K, the observed order of magnitude.”

The forcing is negative: the anomaly DROPS. If the forcing lasted, it would tend toward

-2.5

K over a few decades; after one year, the green point is only at

-0.31

K, an eighth of it.

Why: The ocean has to warm or cool along with the air, and its heat capacity is enormous: a brief forcing produces only a fraction of its equilibrium response. This is also why present-day warming lags behind the forcing.

Which method to choose

Which relation, depending on what the problem asks

The quantity sought

With no atmosphere, the ground itself sends back the

238

W/m². With an opaque layer, it also receives the

238

that the layer radiates downward: it must emit

476

, hence

303

If an equilibrium temperature with no atmosphere → $\sigma T_{e}^{4}=(1-\alpha)\dfrac{S}{4}$ , with $S$ corrected by $\dfrac{1}{d^{2}}$

Example: Mars: $S=\dfrac{1361}{1.524^{2}}=586$ W/m², $T_{e}=210$ K
If a surface temperature with a greenhouse effect → single-layer model: $T_{s}=T_{e}\left(\dfrac{2}{2-f}\right)^{1/4}$

Example: $f=1$ : $2^{1/4}\times 254.6=303$ K
If a forcing due to a gas → $\Delta F=5.35\ln\dfrac{C}{C_{0}}$ for CO₂

Example: doubling: $3.71$ W/m²
If a forcing due to a change in albedo → $\Delta F=-\dfrac{S}{4}\Delta\alpha$

Example: $\Delta\alpha=+0.01$ : $-3.4$ W/m²
If an equilibrium warming → $\Delta T=\dfrac{\Delta F}{\lambda_{P}(1-g)}$ or $\Delta T=s\,\Delta F$ if the sensitivity $s$ is given

Example: $0.80\times 3.71=3.0$ K
If a warming after a given time → $\Delta T_{eq}\left(1-e^{-t/\tau}\right)$

Example: $3.08\times(1-e^{-10/7.56})=2.26$ K

Read the sign of a forcing before any calculation: whatever holds back infrared or darkens the planet warms it; whatever reflects light cools it.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating an equilibrium temperature, written up for the marker

When to use it: The problem gives a luminosity or a solar constant, a distance and an albedo.

1 Bring the flux to the distance of the planet: $S=S_{Earth}\times\dfrac{L/L_{\odot}}{d^{2}}$ , with $d$ in astronomical units.
2 Write the budget, naming every factor: “absorbed flux $(1-\alpha)\dfrac{S}{4}$ equals emitted flux $\sigma T_{e}^{4}$ ”.
3 Isolate $T_{e}$ and calculate, in kelvins.
4 Compare with the observed temperature and interpret the gap as a greenhouse effect.

Concluding sentence

“The planet absorbs $(1-\alpha)\dfrac{S}{4}=238$ W/m² and must emit as much; $\sigma T_{e}^{4}=238$ gives $T_{e}=255$ K, which is $33$ K below the observed surface: this is the greenhouse effect.”

The trap: Writing the temperature formula straight away without the budget: a marker cannot award the reasoning marks to an isolated number, and the forgotten factor of 4 becomes invisible.

Marking: Typically 1 mark for the flux at the right distance, 2 for the budget written down, 1 for the calculation, 1 for the interpretation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The order of magnitude of the temperature

The equilibrium temperature of a terrestrial planet of the solar system lies between 200 and 400 K. Above that, look for the forgotten factor of 4; below it, for an albedo taken backwards.

$360$ K for the Earth: forgotten factor of 4.
The direction of the changes

More albedo cools, more greenhouse gas warms, farther from the star cools. Check every result against these three arrows.

A positive $\Delta T_{e}$ for $\Delta\alpha=+0.01$ betrays a lost sign.
The doubling as a landmark

A CO₂ forcing must stay below 3.7 W/m² as long as the concentration has not doubled, and the equilibrium warming of a doubling lies between 2 and 5 K.

$5$ W/m² for $420$ ppm: linear formula wrongly used.
Kelvins under the fourth power

Any temperature raised to the fourth power is in kelvins, and any equilibrium temperature is converted to °C only at the end.

$T_{e}=254.6$ K, or $-18.6$ °C.

The typical problem, taken apart

From the Sun to the warming of a doubling of CO₂

Data: $S=1361$ W/m², $\alpha=0.30$ , $\sigma=5.67\times 10^{-8}$ W/(m²·K⁴), feedback factor $g=0.60$ .

Calculate the equilibrium temperature of the Earth, then the equilibrium warming caused by a doubling of CO₂.

What the problem gives: the solar flux, the reflected share, the planet's emission and the feedback factor; the temperature and the warming are to be calculated.

Step 1

Mean absorbed flux: $(1-\alpha)\dfrac{S}{4}=0.70\times 340=238$ W/m².

Why

We first write what comes in, with its two factors: the 4 from the geometry and the (1 − α) from the reflection.

Step 2

$\sigma T_{e}^{4}=238$ , so $T_{e}=\left(\dfrac{238}{5.67\times 10^{-8}}\right)^{1/4}=254.6$ K.

Why

The equilibrium budget fixes the temperature at which the planet re-emits exactly what it absorbs.

Step 3

Forcing: $\Delta F=5.35\ln 2=3.71$ W/m².

Why

A doubling always gives the same forcing, because it is the ratio of concentrations that counts.

Step 4

Planck response: $\lambda_{P}=4\sigma T_{e}^{3}=3.74$ W/(m²·K), so $\Delta T_{0}=\dfrac{3.71}{3.74}=0.99$ K.

Why

The derivative of $\sigma T^{4}$ says how many extra W/m² the planet sends back per kelvin: this is what brings the budget back to equilibrium.

Step 5

Feedbacks: $\Delta T=\dfrac{0.99}{1-0.60}=2.5$ K.

Why

We divide by $1-g$ because each induced warming causes another; multiplying by $1+g$ would count only the first round.

The conclusion, written out

“The Earth with no atmosphere would be at $255$ K. A doubling of CO₂ would warm it by $0.99$ K without feedback, and by $2.5$ K with a feedback factor of $0.60$ .”

The classic mistake on this problem: Forgetting the factor 4 from the very first line: the equilibrium temperature comes out at 360 K, and everything that follows is calculated for an imaginary Earth.

Learn by heart

• $S=1361$ W/m²; mean flux $\dfrac{S}{4}=340$ ; absorbed $(1-\alpha)\dfrac{S}{4}=238$ W/m².
• $\sigma T_{e}^{4}=(1-\alpha)\dfrac{S}{4}$ : $T_{e}=255$ K; surface $288$ K; greenhouse effect $33$ K.
• Opaque layer: $T_{s}=2^{1/4}T_{e}$ ; absorptivity $f$ : $T_{s}=T_{e}\left(\dfrac{2}{2-f}\right)^{1/4}$ .
• CO₂: $\Delta F=5.35\ln\dfrac{C}{C_{0}}$ , $3.7$ W/m² per doubling.
• Planck: $\lambda_{P}=4\sigma T^{3}\approx 3.7$ W/(m²·K); feedbacks: divide by $1-g$ .
• Albedo: $\Delta F=-\dfrac{S}{4}\Delta\alpha$ ; one hundredth of albedo is worth $3.4$ W/m².
• Inertia: $\Delta T_{eq}\left(1-e^{-t/\tau}\right)$ , equilibrium is not instantaneous.

Frequently asked questions

Why do we divide the solar constant by four?

The Earth intercepts sunlight as a disc, whose area is pi times the radius squared, but it radiates from its whole spherical surface, which is four times larger. Averaged over the globe and over a day, each square metre therefore receives a quarter of the solar constant, about 340 watts per square metre.

What would the temperature of the Earth be without a greenhouse effect?

Setting the absorbed solar energy, with an albedo of 0.30, equal to the energy radiated by a blackbody gives about 255 kelvins, or minus 18 degrees Celsius. The real mean surface temperature is 15 degrees: the natural greenhouse effect therefore contributes about 33 degrees.

What is radiative forcing?

It is the change in the energy balance at the top of the atmosphere, in watts per square metre, caused by a disturbance such as the addition of carbon dioxide or aerosols, before the temperature has had time to respond. A positive forcing means more energy comes in than goes out, and the planet warms until balance is restored.

Why does the forcing of CO2 follow a logarithm?

Because the centre of the carbon dioxide absorption band, near 15 micrometres, is already saturated: the atmosphere absorbs almost everything there. The added gas only acts on the edges of the band, more and more weakly. That is why each doubling of the concentration adds the same forcing, about 3.7 watts per square metre.

Practise it

Corrected exercises: Greenhouse effect, albedo and radiative forcing, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys3-greenhouse-radiative-forcing. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: greenhouse effect, albedo and radiative forcing (203-SN3-RE)

The essentials

The budget of a planet

Two radiations that do not overlap

Forcing, response and feedbacks

The mistakes that cost marks

1. Forgetting the factor 4 between the solar constant and the mean flux

2. Using albedo as the absorbed fraction

3. Treating the forcing of CO₂ as proportional to the concentration

4. Multiplying instead of dividing by the feedback factor

5. Confusing the radiation emitted by the surface with what leaves for space

6. Assuming the equilibrium warming is reached immediately

Which method to choose

Which relation, depending on what the problem asks

How the answer is expected to be written

Calculating an equilibrium temperature, written up for the marker

Check before you hand in

The typical problem, taken apart

From the Sun to the warming of a doubling of CO₂

Learn by heart

Frequently asked questions

Why do we divide the solar constant by four?

What would the temperature of the Earth be without a greenhouse effect?

What is radiative forcing?

Why does the forcing of CO2 follow a logarithm?

Corrected exercises: Greenhouse effect, albedo and radiative forcing, 203-SN3-RE

See also

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