1. Assume the surface emits like a blackbody at 290 oK.
(a) Calculate the total flux across the spectrum emitted by the surface.
E = sT4 = 400 W m-2
(b) What is the wavelength of peak emission?
l = 2897 mm K/290 K = 10 mm
(c) Calculate the monochromatic flux at the wavelength of peak emission. Over how wide a spectral interval would this monochromatic flux have to be emitted to equal the flux from part (a)?
Using equation 6.5, E = 2.63x107
W m-3
400 W m-2 divided by 2.63x107
W m-3 = 1.5x10-5 m (or 15 mm)
2. Calculate the solar constant S and radiative equilibrium temperature Te for Venus and Mars. Use a solar constant of S=1380 W m-2 for the Earth.
Planet Distance from Sun
Albedo Avg. Sfc.
Avg. Sfc.
(km)
Temperature (K) Pressure (mb)
Venus 108x106
0.77
730
92000
Earth
150x106
0.30
288
1013
Mars
228x106
0.24
218
7
S(venus) = 1380 W m-2 x (150x106)2/(108x106)2
= 2662 W m-2
S(mars) = 1380 W m-2 x (150x106)2/(228x106)2
= 597 W m-2
Te(venus) = 228 K
Te (mars) = 211 K
3. The optical depth of sunlight at sea level due to Rayleigh scattering from air molecules is about
d R = 0.0090 l-4
where the wavelength l is in m
m.
(a) Compute the clear sky transmissivity of collimated solar flux in the blue (l =0.47 mm) and the red (l =0.64 mm) at solar zenith angles of j =30o and j =85o (four cases). Ignore absorption by ozone and water vapor (which is small at these wavelengths) and the effects of aerosols. Assume the atmosphere is a planar slab (i.e. ignore the curvature of the Earth).
j =
30o
0.81 for blue, 0.94 for red
j
= 85o
0.12 for blue, 0.54 for red
(b) Use these results to explain why the sky is blue and why the setting sun is red.
Sky is blue because blue light is scattered more efficiently than red...setting sun is red for the same reason, scattering removes more blue from the direct beam, hence, more red is left.
4. (a) Calculate the visible optical depth of a cloud of thickness D Z=250 m, liquid water content of W=0.2 g m-3, and droplet number concentration of 50 cm-3 droplets. Assume the droplets all have the same size. Use a scattering efficiency (scattering area coefficient) of K=2. Why is the scattering efficiency equal to 2 for these cloud droplets?
Using 4/3 p r3,
you can calculate the radii of the particles to be 10 mm
(9.85, to be exact) from the total mass of water and the number concentration.
Then, from area = p
r2 = 1219 mm2, and 50x106
particles m-3, you can calculate the optical depth to
be about 7.6!
That is, if you were in this cloud you would
only see about 10 meters.
Note that K = 2 because a = 125.
(b) Calculate the visible albedo (reflectivity) of this cloud for a solar zenith angle with cosj 0=2/3. Use an asymmetry parameter of g=0.85.
For this solar zenith angle, A = (1-g)t/[4/3 + (1-g)t]
t = 7.6, so A = 0.46
(c) Emissions of sulfate aerosols by industry have the potential to change cloud properties. This is called the indirect effect of aerosols on climate. Calculate the albedo of this cloud if the number of cloud condensation nuclei, and hence cloud droplets, doubled, and assuming the liquid water content remains constant.
didn't assign, but you can guess that a larger
number of smaller particles might have a larger albedo, as particles smaller
than 10 mm will be more effective scatterers
of visible light individually, and there will be more of them.