Take away concepts
Solar Radiation
Emission and Absorption
1. 
2. 
3. 
4. 
5. 
6. 
Conservation of energy.
Black body radiation principle
Emission wavelength and temperature (Wein’s Law).
Radiation vs. distance relation
Black body energy flux (Stefan-Boltzmann Law)
Effective temperature calculation, differences from
actual temperature.
V1003 - Climate and Society
What is Energy?
Energy: “The ability to do work”.
Energy measured in Joules (1 J = 0.24
calories).
Power measured in Watts (1 J/s)
Energy is always conserved (1st law of TD).
Energy can be changed from one form to another,
but it cannot be created or destroyed.
EM Radiation
Solar Energy
Nuclear fusion: H to He
Emits Electromagnetic
radiation (radiant E)
EM waves behave like
particles and waves
EM travels at c
(3 x 108 m/s)
Properties of waves
Amplitude (A)
Wavelength (µm)
Period (sec)
Frequency (1/sec)
c is constant
Since c is constant, frequency of EM wave emission related to electron vibration
Warm things have more energy than cold things, so ….?
1
Wein’s Law
Blackbody Radiation
λmax = a / T
A “blackbody” absorbs and
emits radiation at 100%
efficiency (experimentally,
they use graphite, or
carbon nanotubes)
Where:
λmax
a
T
= 2898, constant
emitter temperature (in K)
Recall that K = T°C + 273.15
Across all wavelengths
Sun’s temperature is 5800K
What’s its wavelength?
The Sun’s temperature is 5800 K, that is the
wavelength of its radiation?
λmax = a / T
5000 µm
50 µm
0.5 µm
2 µm
20µm
is wavelength of emitted
radiation (in µm)
energy in = energy out
a. 
b. 
c. 
d. 
e. 
emission wavelength and temperature
λmax
is wavelength of emitted
radiation (in µm)
a
T
= 2898, constant
emitter temperature (in K)
What’s your wavelength?
λmax = a / T
(a = 2898)
Your body is 37°C
or 37+273 = 310K
Recall that K = T°C + 273.15
λmax = ?
9.4 µm (far infrared)
Earth as we see it (visible)
Earth’s Infrared “Glow”: 15µm
2
Electromagnetic spectrum
Visualizing emission temperatures
9 µm
“hot”
“cold”
Sunny day: 6000K
Sunset: 3200K
Candlelight: 1500K
0.5 µm
Blackbody applet: http://qsad.bu.edu/applets/blackbody/applet.html
1 µm = 1000 nm
The effect of distance on radiation
“the 1 /
r2
Mars is 1.52 AU
rule”
(1 AU = earth-sun distance =
1.5 x 1011 m)
Using 1/ r2 rule…
1 / (1.5*1.5) = 0.44
Mars receives ~44% of the
Earth’s solar radiation.
Sun emission decreases in proportion to
1 / r2 of the Sun-Planet distance
Jupiter is roughly 5 AU from the Sun, what
fraction of Earth’s solar radiation does it get?
Summary so far…
Wein’s Law (emission freq. and temperature)
a.  1/2
b.  1/5
c.  1/10
d.  1/25
e.  1/125
The “1 / r2” law (radiation amt and distance)
Now let’s calculate the total radiative energy flux
into or out of a planet using the:
Stefan - Boltzmann Law
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Stefan - Boltzmann Law
Energy emitted by a black body is greatly
dependent on its temperature:
I = (1-α) σ
Calculating the Earth’s
“Effective Temperature”
Easy as 1-2-3…
T4
Where:
I = Black body energy radiation
σ = (Constant) 5.67x10-8 Watts/m2/K4
T = temperature in Kelvin
1.  Calculate solar output.
2.  Calculate solar energy reaching the Earth.
3.  Calculate the temperature the Earth should
be with this energy receipt.
α = albedo (“reflectivity”)
Example: Sun surface is 5800K, so I = 6.4 x 107 W/m2
1. Calculate solar output.
Calculate Sun temperature assuming it behaves
as a blackbody (knowing that λsun= 0.5µm).
2. Calculate solar energy reaching the Earth.
Simple Geometry.
(recall the inverse square law..)
From S-B law: Isun = 6.4 x 10 7 W/m2
Earth-Sun distance (D): 1.5 x 1011 m
Area of sphere = 4π r 2
We need surface area of sun:
Area = 4πr2 = 4π (6.96x108 m) = 6.2 x 10 18 m2
So, 3.86 x 10 26 Watts / (4π (1.5 x 1011 m)2 )
Total Sun emission: 3.86 x 1026 Watts (!)
Earth’s incoming solar radiation: 1365 W/m2
Solar
Emission
Power
3. Earth energy in = energy out
You have Iearth, solve for Tearth
Stefan - Boltzmann law: Iearth = (1-α) σ Tearth4
Incoming solar radiation: 1365 W/m2
About 30% is reflected away by ice, clouds, etc.: reduced
to 955 W/m2
Incoming on dayside only (DISK), but outgoing
everywhere (SPHERE), so outgoing is 1/4 of
incoming, or 239 W/m2
Solve for Teffective = 255K
Earth Effective temp: 255 K, or -18°C
Earth Actual temp: 288K, or +15°C
… the difference of +33°C is due to the natural
greenhouse effect.
that is: (0.7)*(0.25)*1365 = energy that reaches Earth surface
Energy in = 239 W/m2 = σ T4
4
So what Earth’s radiation
wavelength?
Emission Spectra: Sun and Earth
λmax = a / T
Where:
λmax
is wavelength of emitted
radiation (in µm)
a
T
= 2898, constant
0.5 µm
9 µm 15 µm
emitter temperature (in K)
If Earth effective temperature is 255K
What’s the wavelength?
Radiation and Matter
Emission Spectra: Sun and Earth
Also dependent upon the frequency of radiation!
(next lecture)
Blackbody emission curves and
absorption bands
Why is the Sky blue?
Rayleigh scattering of incoming,
short wavelength radiation
(photons with specific energy)
Radiation scattered by O3, O2 in
stratosphere (10-50 km)
5
Why are sunsets red?
Blue wavelengths are scattered/
absorbed
Red and orange pass through
to surface
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