1. Use pressure based on the US standard atmosphere to estimate the proportions of the atmospheric mass in the troposphere, the stratosphere, and the mesosphere. How do the relative proportions in the tropospheric and stratospheric parts of the column change from the tropics to the polar latitudes?
If you use an altitude of 11 km, as the book does,
to define the tropopause, the pressure there is about 210-250 mbar. Since
pressure is directly related to the mass of air in the column above, this
means that above 11 km there is 21-25% of the earth's atmosphere. The mesosphere
is at
about 50 km, where the pressure is 1 mbar or
less, so less than 0.1% of the earth's atmosphere is in the mesosphere.
This leaves, by difference, 75-79% of the earth's atmosphere (by mass)
in the troposphere.
The tropopause at midlatitudes is typically at 100 mbar (16 km or so) so that 10% of the mass of the atmosphere is in the stratosphere and 90% in the troposphere there. In the tropics, the tropopause is a bit higher (17-18 km), where the pressure is about 85 mbar, and in the polar regions the tropopause is much lower...near 280 mbar, and sometimes even lower in altitude. So there is proportionately more air in the stratosphere over the polar regions than midlatitudes and tropics. The reason for this is not that the atmosphere bulges at the equator, but rather because air is rising at the equator where it is warmest and sinking at the poles where it is coldest, so the tropopause is pushed at at the equator and down at the poles.
You could also have done this problem by calculating the mass of the total atmosphere using pressure, then similarly the mass of the atmosphere above the tropopause and stratopause. When you take ratios, you find all the terms drop out but the ratio of pressures (or densities)...which makes it much easier to do the problem using pressures alone.
2. Suppose that the maximum mass mixing ratio of ozone is 1.3x 10-5 at a height of 35 km above sea level. Using a reasonable value for air density there, find the density and volume mixing ratio (ppmv) of ozone.
You can estimate the density of air at 35 km using a scale height of 7.5 km or so, and a value at the surface of about 1.2 kg/m^3. You get around 1x10^-2 kg/m^3. Multiply this by the mass mixing ratio (defined simply as the ratio of the mass of ozone to the total mass of air), and you get the density of ozone, or 1.3x10^-7 kg/m^3. To calculate the volume mixing ratio (or number mixing ratio, since gases obey the ideal gas law that says for a given temperature and pressure the number is proportional to the volume) you simply divide the mass mixing ratio by the molecular weight of ozone (48 g/mole) to get number and multiply by the molecular weight of air (28.8 g/mole) to convert the denominator to number of air molecules, and you end up with number of ozones per number of air, or about 8x10^-6 (8 parts per million, or ppm).
3. Use the PAOS Weather Center Web site (Archives section, general sounding archive) for this question. Output the text format of the sounding for Denver on January 6, 1999 at 12Z (99010612).
Note the reversal of the direction of the equatorial surface meridional winds from northerly in July (bringing warm gulf air up to the subtropics in northern hemisphere summer (the "monsoon" pattern) to southerly in January (allowing dry, colder air southward in the winter). Also note that there is a strong counter cell (Ferrel cell) in the winter hemisphere, not observed strongly in the summer hemisphere. This is because the equator-to-pole temperature gradients are strongest in winter. Also note the asymmetry in the two hemispheres. There is a strong Ferrel cell in the southern hemisphere even in summer...this is due to the cold circumpolar current that flows at high mid-latitudes in the S.H. that keeps this part of the globe cold, even in summer.
5. Radiosonde data are the primary input for numerical weather prediction models even though there are many fewer radiosonde stations than surface stations. Explain why surface stations alone can't be used. What regions of the globe are likely to have poorer weather forecasts? Why?
The lack of vertical information makes forecasting tough...since steering currents are important and because of water condensation as air rises. Also, regions of globe that are data poor (oceans, southern hemisphere, poor countries, mountain regions) are generally areas where forecasts are suspect.
6. Contrast the appearance on visible and infrared satellite imagery of a tropical thunderstorm (with dense clouds rising from low levels to the tropopause) and low level overcast stratocumulus clouds over a subtropical ocean. Estimate the infrared brightness temperature in each case.
First, you need to make sure that the visible image is obtained in sunlight! Many of you worked at night, picked the US sector, and didn't see any clouds. No wonder! Note that you always see clouds in the infrared, just that those near the surface are hard to see because they are warm (~285-290 if over ocean) and there isn't much contrast against the warm surface. So in the IR, the 'brightest' clouds are the highest (i.e. coldest) in the field of view. So, the proper way to do this is to look for those clouds that appear in both visible (in daylight!) and IR imagery, and those are probably the coldest...~tropopause temps (210 K or so). The ones over the ocean that appear in the visible (in daylight!) and NOT in the IR are the low level, warm (i.e. ~280-290 K) clouds.