ATMS 501
Solutions Quiz #5
Nov. 10, 2009
Short and multiple choice questions (2 points each except where indicated otherwise; 40 points
total)
1. The (flux, flux density, intensity) of solar radiation varies inversely with the square of
distance from the sun. The other terms are independent of distance.
2. The inverse square law breaks down if the observer is (too far from, too close to) the star or
planet that is emitting the radiation. As the distance gets shorter, the arc of solid angle
subtended becomes smaller than (r/2d)2
3. Radiative equilibrium temperature increases (linearly, more or less exponentially, more or
less logarithmically) with geometric depth in the atmospheres of stars and planets. The
dominant factor is the exponential increase in the density of the absorbing gases with
increasing depth.
4. Monochromatic flux density has units of W m-2µm-1.
5. Over a horizontal surface the sky subtends an arc of solid angle of 2π steradians.
6. Downward isotropic radiation with intensity I incident on a horizontal surface is equivalent
to a flux density of πI W m-2.
7. A sphere with a radius of meter that emits isotropic radiation with intensity I emits a flux of
4π2I W. The flux density is πI and the flux is 4π (1)2 × πI
8. The Earth, which has an equivalent blackbody temperature of 255 K, emits peak
monochromatic intensity at a wavelength of around 12 µm. Based on the Wien Displacement
Law, Jupiter, which has an equivalent blackbody temperature of 125 K, should emit peak
monochromatic intensity at around 24 µm. The effective temperature is about half that of
Earth so, from the Wien Displacement Law, the peak emission should occur at a wavelength
twice as long.
9. The insolation incident on the North Pole at the time of the summer solstice is (greater than,
the same as, less than) that incident on the equator at the time of the equinox. 531 versus 435
W m-2 as per Exercises 4.17 and 4.18.
10. When subjected to solar radiation, the surface of a planet with an atmosphere consisting of
greenhouse gases that are nearly transparent in the visible part of the spectrum and strong
absorbers in the infrared will tend to be (warmer than, the same temperature as, cooler than)
the surface of a planet in the same orbit with the same albedo but having no atmosphere.
Greenhouse gases let solar radiation through with relatively little absorption and block
outgoing infrared radiation by absorbing it and reemitting it back toward the surface. To
counteract this downward emission from the atmosphere, the surface of the planet has to be
warmer than it would be in the absence of an atmosphere.
11. A cloud of greenhouse gases orbiting the sun would be (warmer, the same temperature,
colder) than a cloud of a hypothetical “gray” gas whose absorptivity kλ is independent of
wavelength. Greenhouse gases absorb relatively little solar radiation but are effective at
emitting infrared radiation. A “gray” gas emits the same fraction of blackbody radiation as
it absorbs of the incident solar radiation. In contrast, a greenhouse gas absorbs relatively
ATMS 501
Solutions Quiz #5
Nov. 10, 2009
little solar radiation and emits a larger fraction of blackbody radiation. Since for radiative
equilibrium, emission must equal absorption, it follows that a greenhouse gas is cooler than
a gray gas expose to the same solar radiation.
12. The equivalent blackbody temperature of Venus is (higher, the same as, lower than) that of
the Earth). Lower, even though it’s closer to the sun, the higher albedo of Venus more than
compensates for its being closer to the sun.
13. The albedo of the moon is about 1/3 as large as the albedo of the Earth and the radius of the
moon is about 1/4 as large as that of the Earth. The flux density of “earthlight” on the surface
of the moon is roughly equal to 48 times the flux density of moonlight on the surface of the
Earth. A factor of 16 for the square of the larger radius and 3 for the higher albedo of the
earth compared to the moon. It’s no wonder the astronauts were impressed when they viewed
the Earth from the moon.
14. Dirt on windows is more clearly evident when viewed looking (toward, away) from a light
source. .. because of the predominance of forward scattering in the Mie regime.
15. For 10 cm radar, raindrops with radii on the order of 1-2 mm are in the (Rayleigh scattering,
Mie scattering, geometric optics) regime in which the scattering is (mostly forward, isotropic,
mostly backward). The size parameter 2πr/λ is less than 0.1 so the scattering of radar waves
by raindrops is in the Rayleigh scattering regime in which the scattered radiation is isotropic.
16. Very low summertime insolation in high northern latitudes, a condition favorable for the
formation of continental ice sheets, occurs in the phase of the precession cycle when the
Earth is farthest from the sun during the northern summer, under conditions of (high
eccentricity and high obliquity, high eccentricity and low obliquity, low eccentricity and
high obliquity, low eccentricity and low obliquity) of the Earth’s orbit. High eccentricity so
as to maximize the Earth-Sun distance; low obliquity so as to maximize the zenith angle of
the sun in the summer sky.
2. A satellite orbiting the Earth and shaped like a cube and one of the six sides is always facing
directly toward the sun. The albedo of the satellite is 0.30, the same as that of the Earth and it
conducts heat strongly enough to keep all six of its sides are at the same temperature.
Calculate the equivalent blackbody temperature of the satellite under the assumption that it has
no internal heat source, it is far enough away from the Earth so that it is unaffected by the
Earth’s emitted radiation and it is not in the Earth’s shadow. (20 points)
The analysis is the same as for a spherical planet except instead of the factor 4, the ratio of the
area of intercepted solar radiation to the area emitting infrared radiation, we have the factor 6. So
the effective temperature of the satellite is equal to the effective temperature of the Earth times
the factor (4/6)1/4 = 255 × (0.667) 1/4 = 255 × 0.9037 = 230 K.
ATMS 501
Solutions Quiz #5
Nov. 10, 2009
3. Geothermal heating amounts to 0.06 W m-2 averaged over the entire surface of the Earth. How
much does the existence of this internal heat source raise the equivalent blackbody
temperature of the Earth?
We can use the relationship dTE/TE = 1/4 dF/F with TE = 255 K, F = 239.4 W m-2, and dF =
0.06 W m-2, which yields 0.016°C.
4. Suppose that stratospheric sulfate aerosols injected into the atmosphere from a volcanic
eruption instantaneously increase the Earth’s albedo from 0.30 to 0.35. Neglecting the heat
transfer between the atmosphere and the underlying surface, at what rate should the
atmosphere begin to cool in response to this radiative forcing? Express your answer in °C per
day. [Assume that the temperature of the atmosphere before the eruption is 250°K and that the
aerosol cloud is spatially uniform.]
It was a mistake on my part to give you the atmospheric temperature 250 K. You didn’t need it
and it let many of you off in the wrong direction. Accordingly, I am going to discard the grades
from this problem and grade the quiz on the basis of 80 points. Here is my solution...
The decrease in incoming radiation due to the increase in albedo, averaged over the entire
surface area of the Earth is (Fs / 4) W m-2 × δ(1 – A). Substituting numerical values, we obtain
(1368 / 4) –0.05 = –17.1 W m-2. This is the instantaneous radiative forcing F.
From the First Law of Thermodynamics
mcp dT/dt = F.
Substituting numerical values and solving for dT/dt we obtain
dT/dt = –17.1 W m-2 / (104 kg m-2 × 103 J kg-1 K-1)
= –17.1 × 10-7 K s-1
× 86,400 s day-1
= –0.198 K day-1
5. When the moon is partially full, the illumination fades gradually toward the terminator but the
the surface of the moon appears uniformly bright out to the limb, as shown in the figure below.
Explain. [To be added]