ATMS 211 - W08
Problem Set #2 - Solutions
Due 1 February 2008
1.
Using the Earth’s emission temperature of 256 K, calculate its peak wavelength
of emission. Also, what is the peak wavelength of emission for a human of
average body temperature (37◦ C)?
Recall: TK = T◦ C + 273, Wien’s Law: λmax =
2898 µm K
T K
Peak wavelength of emission for the Earth:
λmax =
2898 µm K
= 11.3 µm.
256 K
Convert average body temperature to Kelvin:
TK = 37 + 273 = 310 K.
Peak wavelength of emission for a human:
λmax =
2898 µm K
= 9.3 µm.
310 K
The Earth’s wavelength of peak emission is used by meteorological satellites
to sense clouds at various levels of the atmosphere and at night (see Figure
1). The sensor on the satellite is sensitive to wavelengths near 11 µm. Colder
temperatures appear lighter than warmer temperatures. Thus, clouds higher
in the atmosphere are white, whereas clouds closer to the surface appear grey.
2.
Infrared (IR) light has a wavelength on the order of 10 micrometers (microns).
What is its [wave number (cm−1 ) and] energy? Visible light has a wavelength
of about 0.5 microns. What is its frequency (Hz) and energy? KUOW operates its radio station with a frequency of 94.9 MHz, what is the wavelength of
the radio waves emanating from KUOW’s transmitter?
Recall: c = λ ν, E = h ν = h λc , 1 micron (µm) = 1 × 10−6 m,
1 MHz = 1 × 106 Hz, 1 Hz = 1s
Energy of a photon of infrared light
E=
(6.6 × 10−34 J s) (3 × 108 m/s)
= 1.98 × 10−20 J
10 × 10−6 m
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Figure 1: GOES West IR Imagery
wavenumber =
1
= 1000 cm−1
−4
10 × 10 cm
Frequency of visible light
ν=
c
3 × 108 m/s
=
= 6 × 1014 Hz
λ
0.5 × 10−6 m
Energy of a photon of visible light
E=
(6.6 × 10−34 J s) (3 × 108 m/s)
= 39.6 × 10−20 J
0.5 × 10−6 m
Wavelength of radio waves
3 × 108 m/s
c
λ= =
= 3.16 m.
ν
94.9 × 106 1/s
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3.
Consider Mars. It receives ∼ 600 W/m2 from the sun at the top of its atmosphere. About 15% of this radiation is directly reflected back to space (Mars
Albedo = 0.15). Assume that Mars’s atmosphere is unimportant in its energy
balance (as we did for Earth). Calculate the average surface temperature of
Mars in Kelvin. How does your prediction compare with the observed value
∼ 216 K?
You can use the same equation we ended up with on Thursday for the temperature of Earth, just use the correct numbers for Mars (Iin = 600 W/m2
and A = 0.15). Note that Mars is about 1.511 times farther from the sun
than Earth, so, from the inverse square law,
Iin (Mars) =
1
1 2
(1370 W/m2 ) = 600 W/m2 .
Iin (Earth) =
1.511
2.283
Now put these numbers into the equation for the equilibrium temperature of
a planet:
TM =
I
in
1/4
600 W (1 − A) 1/4
m2 K4
=
(1
−
0.15)
4σ
m2
4 · 5.67 × 10−8 W
TM = 217.8 K = −55.4◦ C.
Indeed, the calculated value for the average surface temperature of Mars is
approximate to the observed value of 216 K. Combined with our physical
understanding of Mars’ thin atmosphere, this result lends credence to our assumption that the atmosphere is relatively unimportant in its energy balance.
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4.
Give two assumptions we made in calculating the surface temperature of Earth
without an atmosphere (in addition to the assumption that the atmosphere isn’t
important). Briefly discuss the validity of each assumption.
Any of the following:
Equilibrium (energy in = energy out) – Approximately correct on time
scales longer than one year. Differences between energy coming in and out on
these longer time scales lead to climate change. Currently energy in is slightly
greater than energy out because of human greenhouse gas emissions.
Solar flux constant in time – Nearly correct for time scales longer than a
year and shorter than a thousand years. Over the last 30 years the annualmean solar flux at the top of the atmosphere has varied by about ±0.1%.
During a year the solar flux at the top of the atmosphere varies by about
±3.4% due to variations in the Earth-Sun distance. Over a day the flux varies
by 100% most places.
Everywhere on Earth receives the same average solar flux – Very bad assumption; solar flux varies greatly by latitude. The reason we can make the
assumption is that we are interested in temperature averaged globally and
annually, so the regional and temporal variations get averaged out.
Earth is a blackbody – This isn’t exactly what we assumed We assumed
Earth emits as a blackbody, but we allowed it to scatter some light (28%)
back to space. This assumption is quite accurate. The reason it works is
that Earth receives its energy from the sun mostly in the form of ultraviolet,
visible, and shortwave infrared light (less than 4 µm), but emits mostly in
the form of longer-wave infrared light (greater than 4 µm). At the shorter
wavelengths, most Earth surfaces are not like blackbodies (they don’t absorb
all of the incident light), but at the longer wavelengths most Earth surfaces
are very similar to blackbodies.
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