Introduction to Atmospheric Science Fall 2023 Lecture notes 2 - PAT

LE/ESSE 1011 3.0A

Lecture notes mostly based on material from Stull, R., 2017: "Practical Meteorology: An Algebra-based
Survey of Atmospheric Science" -version 1.02b. Univ. of British Columbia. 940 pages. isbn 978-0-88865-283-6 .
Available at https://www.eoas.ubc.ca/books/Practical_Meteorology/

Copyright © 2017 by Roland Stull. Practical Meteorology: An Algebra-based Survey of Atmospheric Science. v1.02

2 SOLAR & INFRARED RADIATION

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Stull puts in a lot of material on Orbital Factors - interesting but no time, we will
just cover some essentials.

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Note - Not to scale. Earth radius - 6371 km, a = 149.598 Gm [G = x109] Because
the Earth is rotating around the Earth-moon barycenter while this barycenter is
revolving around the sun, the location of the center of the Earth traces a slightly
wiggly path as it orbits the sun. This path is exaggerated in Fig. 2.2 (thick line).
The closest distance (perihelion) along the major axis between the Earth and sun is
a – c = 146.96 Gm and occurs at about dp ≈ 4 January. The farthest distance
(aphelion) is a + c = 151.96 Gm and occurs at about 5 July. The dates for the
perihelion and aphelion jump a day or two from year to year because the orbital
period is not exactly 365 days.
So in N hemisphere sun is closest in ????
Other planets?
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Figure 2.1 Planetary orbital periods versus distance from sun. Eris (136199) and
Pluto (134340) are dwarf planets. Eris, larger than Pluto, has a very elliptical orbit.
Log-log graph.

Some Angle issues:

The tilt of the Earth’s axis relative to a line perpendicular to the ecliptic (i.e., the
orbital plane of the Earth around the sun) is presently Φr = 23.44° = 0.409 radians.
The direction of tilt of the Earth’s axis is not fixed relative to the stars, but wobbles
or pre-cesses like a top with a period of about 25,781 years.

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 The solar declination angle δs
is defined as the angle
between the ecliptic and the
plane of the Earth’s equator
(Fig. 2.4). Assuming a fixed
orientation (tilt) of the Earth’s
axis as the Earth orbits the
sun, the solar declination
angle varies smoothly as the
year progresses. The north
pole points partially toward
the sun in summer, and
gradually changes to point
partially away during winter
(Fig. 2.5).

? Suppose that there were no tilt? What would the consequences be?
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Some fluxes are associated with fluid movement or "advection", sometimes called
"convection" but in atmospheric science, convection is usually upwards and
advection is used for horizontal fluxes. Here we focus on radiative fluxes -
travelling at speed of light along a beam of EM radiation.
https://en.wikipedia.org/wiki/Advection https://dictionary.cambridge.org › dictionary › english › convection
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We will work primarily in terms of wavelength to distinguish the type of EM radiation.

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 Planck's Law tells us about
the spectrum of radiation
from a black body. It has a
complex variation with
wavelength (λ) and T (in K).
If we integrate (sum) over
contibutions from all
wavelengths we get the
Stefan-Boltzmann Law.

for total "Emittance". Note

is the Stefan-Boltzmann
constant.
CHECK the units. W = Js-1.

Do the calculations suggested for

todays current T.
Exitance per m2. Compare with
weather station values.
https://www.yorku.ca/pat/weatherStation
Upwelling longwave.

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Be careful interpreting plots with log scales! Especially with pdfs (probability density
functions) like this.
Stull says " The area under the solar-radiation curve in Fig. 2.10 is the total (all wavelengths)
solar irradiance (TSI), So, reaching the Earth’s orbit. We call it an irradiance here, instead of an
emittance, because relative to the Earth it is an incoming radiant flux. This quantity was
formerly called the solar constant but we now know that it varies slightly. The average value
of solar irradiance measured at the Earth’s orbit by satellites for a quiet sun (during sunspot
minima) is about So = 1361 W·m–2.
Many factors but should incoming solar radiation ≈ outgoing long wave radiation.? Global
average. Annual average. Why are "areas undr the curves" so different?
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Use distances and r2 to compare solar "constants" on
Earth and Mars.
MARS values
Closest: 205 million km / 127 million miles (1.38 AU)
Furthest: 249 million km / 155 million miles (1.66 AU)
Average: 228 million km / 142 million miles (1.52 AU)
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 Earth orbit is slightly
elliptical, so distance
to Sun's centre
varies with time of
year, and day
because of moon but
just using barycentric
orbit values and

Stull shows results

on left.

So far we have assumed surfaces normal (perpendicular) to radiation direction but

for surfaces on Earth surfaces that are horizontal, or at other orientations we need
to allow for angles between normal and radiation direction.
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Stull uses elevation angle to make
adjustments, I prefer zenith angle =
π/2 - ψ.

The solar zenith angle is the zenith

angle of the sun, i.e., the angle
between the sun’s rays and the vertical
direction. It is the complement to the
solar altitude or solar elevation, which
is the altitude angle or elevation angle between the sun’s rays and a horizontal
plane. An azimuth is the angular measurement in a spherical coordinate system
which represents the horizontal angle from a cardinal direction, most commonly
north.

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NH mid summer at poles? Elevation angle 22.5°, and constant all day, so
I0 sin(22.5°) = 1327 * 0.38 = 508 Wm-2.

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These are properties of surfaces and their
sum is 1.0

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In Lab 2 you will measure
surface albedo for solar
radiation, but can also
determine from EMOS
measurements, over grass.
Approx values at solar max
give
A = 150/780 = 0.19
See other values in Stull.
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Can also see
upwelling Long
Wave radiation. If
emissivity = 1 and
we note
temperature
variations. These
though are 2m air T
and temperature 5
cm below surface.
The actual surface
will probably be different, maybe higher than both. Sit in the sun for a while and
your skin temperature can get high. We could use the radiation measurements to
estimate the skin or surface radiative temperature, from S-B law. Peak outgoing
long wave is about 460 Wm-2. From
https://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html,
a typical emissivity for a grass surface is ε = 0.98. The Stefan-Boltzmann Constant
is σ = 5.6703 10-8 Wm-2K-4. Then we have 460 = 0.98 *5.6703 10-8 * T4 and
calculation gives T = 301.6K or 28.5°C. Warmer that the air T (max 21°C). We can
also note that T1 - T9.5 was measured as about 1.5 °C and the surface radiative T
could well have been 28.5°C. If we assumed a higher emissivity (1.0) we would get
T = 300.1K or 27.0 °C so a slight difference only.

At night the surface would be colder that the air. Let's look at 0400 EDT, 0800 UTC
with outgoing LW = 350 Wm-2. S-B law then gives T = 281.7K = 8.6°C while air
temperature was about 11°C, warmer that the surface radiative T.

How about the downwelling LW? Where does that come from?

Stull: Within the air, however, the process is a bit more complicated. One approach is to treat
the whole atmospheric thickness as a single object. Namely, you can compare the radiation at
the top versus bottom of the atmosphere to examine the total emissivity, absorptivity, and
reflectivity of the whole atmosphere.
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Transmission through the atmosphere

The surface of the Earth (land and sea) is a very strong absorber and emitter of
radiation. Within the air, however, the process is a bit more complicated. One
approach is to treat the whole atmospheric thickness as a single object. Namely,
you can compare the radiation at the top versus bottom of the atmosphere to
examine the total emissivity, absorptivity, and reflectivity of the whole
atmosphere. Over some wavelengths called windows there is little absorption,
allowing the radiation to “shine” through. In other wavelength ranges there is
partial or total absorption. Thus, the atmosphere acts as a filter. Atmospheric
windows and transmissivity are discussed in detail in the Satellites & Radar and
Climate chapters.

2.3.6. Beer’s Law

Sometimes you must
examine radiative
extinction (reduction of
radiative flux) of radiation
in a direct beam across a
short path length Δs within
the atmosphere (Fig. 2.12).
Let n be the number
density of radiatively
important particles in the
air (particles m–3), and b be
the extinction cross section
of each particle (m2
particle–1), where this latter quantity gives the area of the shadow cast by each
particle. Extinction can be caused by absorption and scattering of radiation. If the
change in radiation is due only to absorption, then the absorptivity across this
layer for a narrow beam of photons is

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Comes from dE/ds = - γE, where the extinction coefficient has units of m-1.
What are n and b, and their
units?

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https://www.yorku.ca/pat/weatherStation/index.php. (all + irradiances)

Final notes from Stull, and suggested questions.

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