Answers to questions from class on Friday, January 20, 2006

Note that the answers are more complete than I would have expected in class. If you have any questions, feel free to ask me (Clark) or Prof. Bitz.

1. Using your knowledge of pressure forces and gravity, describe why pressure in the atmosphere decreases with height (a diagram of forces may help you).

ANSWER: Let's assume that air in the atmosphere is not moving vertically. On the largest scales in the atmosphere, this is approximately true. This means that for a given air parcel, the sum of all forces pushing up must equal the sum of all forces pushing down. Thus, there is no net force acting on the parcel. Now the question remains: what are these forces? Well, they are the two given in the question: pressure force and gravity. Gravity acts toward the center of the earth, which is downward. If the parcel is not accelerating vertically, then there must be an equal net pressure force pushing up on the parcel. This net upward pressure force can only exist if the pressure force pushing up on the parcel is greater than the pressure force pushing down on the parcel. Restated, this means that the pressure below the parcel must be greater than pressure below the parcel. This explicitly tells us that pressure in the atmosphere must decrease with height.

2. If the sun's radiation were 10% more intense than today, using the climate sensitivity factor derived in class, what would be the new equilibrium temperature of the Earth? What distance from the sun would the Earth have to be from the sun to maintain our current temperature? (effective radiative temperature for this problem is fine)
   
   Fun note: The sun's intensity will be this value in about a billion years, so this problem isn't complete fantasy!

NOTE: This is more than I would expect an individual to do on their own in this class, which is why I gave it as a group question. It is also an example that ties together several different equations and is intended to stretch your thinking and understanding a bit (as well as be an interesting visual. How would you propose moving the Earth farther from the sun? :).

ANSWER:
First part:

i) Equation for determining climate sensitivity factor: lambda = dT / dF . Note that this includes the greenhouse effect on Earth.

ii) Teq = To + dT, Teq is the equilibrium T, To is the starting T, and dT the change in T.

iii) Solar radiation at the top of Earth's atmosphere: S * (1-A) / 4

Rewriting equation i): dT = lambda * dF.

Combining this with ii) equation yields: iv) Teq = To + lambda * dF

To solve for the future equilibrium temperature, we need to know To, lambda, and dF (the change in forcing).

From lecture: To (temperature today) = 288 K, lambda = .7

Now all we need is to know dF. dF, being the change in forcing, is the solar forcing on the climate in the future minus the solar forcing today. From iii):

v) dF = Ffuture – Fnow = [(1.1*S)*(1–A)/4] – [S*(1–A)/4] = (.1*S)*(1–A)/4

Now the average albedo of the earth is A = .3 and the value of S at Earth's distance from the sun is S = 1368 W/m^2.

Plugging in: dF = (.1 * 1368) * (1 - .3) / 4 = 136.8 * .7 / 4 = 23.94 W/m^2

Inserting this into iv) we get the new equilibrium temperature:

Teq = To + lambda * dF = 288K + .7 (K/(W/m^2)) * 23.94 W/m^2 = 304.8K.

Second part:

vi) Steffan-Boltzman Law (emitted radiation): F = sigma * Te^4

vii) Inverse-Square Law: S = So * (ro/r)^2

Assume that Earth's albedo doesn't change in the future and the sun doesn't expand.

The goal of the second part of this question is to move the earth farther from the sun to perfectly balance the increase in solar radiation. I also hinted that using the effective radiating temperature of the Earth is the way to go (thus we are ignoring the greenhouse effect, which would complicate the calculation further). If the Earth's Te (effective radiating temperature) doesn't change, then the outgoing radiation from Earth is shown in vi) with Te from today. Since we're interested in equilibrium in this question, then what goes out must equal what comes in, so from iii) and vi):

viii) Sf * (1-A)/4 = sigma * Te^4 , where Sf = S in the future.

Now, we know sigma, Te, and A. All we need is Sf. From vii) (using Sf notation):

ix) Sf = (1.1 * So) * (ro/r)^2.

In the future, the suns intensity is 10% greater than today (So in ix) is today's value), we assumed ro is unchanged, but r is unknown. This is the new sun-Earth distance. If we combine viii) and ix), then:

x) (1.1 * So * (ro/r)^2) * (1-A) / 4 = sigma * Te^4

Now we have an equation where we know everything except r, the sun-Earth distance, which is the goal of this question. Now we have to solve for r:

(1.1 * So * (ro/r)^2) * (1-A) / 4 = sigma * Te^4 multiply by 4 / (1.1 * So * (1-A))

(ro/r)^2 = sigma * Te^4 * 4 / (1.1 * So * (1-A)) square root both sides

ro/r = (sigma * Te^4 * 4 / (1.1 * So * (1-A)))^.5 divide by ro

1/r = (1/ro) * (sigma * Te^4 * 4 / (1.1 * So * (1-A)))^.5 invert both sides (-1 power)

xi) r = ro * (sigma * Te^4 * 4 / (1.1 * So * (1-A)))^(-.5)

plugging in the following values into xi):

So = 63,000,000 W/m^2, ro = 696,000 km, A = .3,

sigma = 5.67 * 10^-8, Te = 255K (page 41)

r = 696,000km * (5.67*10^-8 * 255^4 * 4 / (1.1 * 63000000 * (1-.3)))^(-.5) = 156.5 million km, or about 4.6% farther than the current Earth-sun distance.

3. What is the most important property that the atmospheric meridional circulation cells (like the Hadley cell) transport? In what direction is this property transported, on average?

ANSWER: Heat is the most important property transported by the atmospheric meridional circulation cells. Heat is transported up and poleward on average. For the Hadley cell, there is an upward transport due to the release of latent heat by the warm, moist air that is lifted on the equatorward edge. The poleward transport of heat can be seen by looking at the poleward and equatorward branches of the cell. Since the temperature in the atmosphere is generally warmer toward the equator no matter what your height is, the upper-level, poleward branch of the Hadley cell is transporting warm air to colder regions (a transport of heat toward the pole). The near-surface equatorward branch of the Hadley cell is transporting cool air to warmer regions. This means that it is transporting 'cold' toward the equator, which is another way of saying that it is transporting heat toward the pole. Figure 4-3 on page 60 or figure 4-7 on page 62, along with knowing that colder air is toward the pole in general should help in visualizing this.

4. Name the type(s) of heating/cooling that occur on the vertical components of the Hadley cell.

ANSWER: On the upward branch of the Hadley cell, there is adiabatic cooling due to the lifting of air from higher pressure near the surface to lower pressure aloft (remember, atmospheric pressure is analogous to the weight of the air above you). On the downward branch, there is adiabatic warming due to the descent of air from lower pressure to higher pressure. There is also latent heating taking place in the upward branch of the Hadley cell. This release of latent heat in the atmosphere is due to the fact that the adiabatically cooled air here cannot hold all of its the water vapor anymore (figure 4-25, page 78). When water transitions from a gas to a liquid (condenses), it releases large amounts of heat.

5. Would you expect the regions of the Earth under the descending branch of the Hadley cell to be cloudy or not? Why?

ANSWER: Regions of the Earth under the descending branches of the Hadley cells tend not to be cloudy. The reason is that this descending air is adiabatically warming since it is moving from low pressure higher in the atmosphere to higher pressure near the surface. Since the amount of water vapor that air can hold increases with when temperature goes up (figure 4-25, page 78), even if the air was saturated with water before descending, it would be unsaturated (meaning it could hold more water vapor than it does). Clouds only form when air is saturated, allowing the condensation of water vapor into cloud droplets. Thus, we would not expect clouds to form in generally descending regions.

6. What type of clouds are formed by the deep convection of the Hadley cell? Would you expect these clouds to affect the temperature of the region in which they form? Why?

ANSWER: The type of cloud formed by deep convection is cumulonimbus (aka – thunderheads). These are extremely thick clouds that can reach all the way up to the tropopause. This means that they have a high albedo, reflecting much solar radiation as well as having cold tops, which radiate relatively small amounts of radiation to space (strong warming effect). Since these clouds have strong warming and cooling effects, the last part of this question was a trick on my part. Without more quantitative information, it would be difficult at best to determine the net effect of these tropical cumulonimbus clouds. It turns out, however, that the net effect is a cooling effect in the tropics.

7. Would you expect the Intertropical Convergence Zone (ITCZ) to be farther north or south in the southern hemisphere summer relative to the southern hemisphere winter? Why?

ANSWER: Let's start with a question: what is the ITCZ? It is the location of the deep convection associated with the Hadley cells (figure 4-3, page 60). This deep convection starts due to strong surface warming by solar radiation absorption. Thus, if the ITCZ moves, it should locate itself close to where the maximum solar radiation absorption is taking place. Solar radiation is a maximum when the sun is directly overhead, and this point (latitude, actually) changes throughout the seasons due to the tilt of the earth (figure 4-16, page 68). During the southern hemisphere winter, the latitude where solar radiation absorption is a maximum is between 0 and 23.5 degrees north, or in the northern hemisphere. Thus, we would expect that the ITCZ would migrate north into the northern hemisphere during the southern hemisphere winter, and south of the equator during the southern hemisphere summer.

8. Explain why the trade winds blow from the east.

ANSWER: Looking at figure 4-3 on page 60, we see that the Hadley circulation near the surface is toward the equator. If we now mentally apply the coriolis force to this equatorward motion in the northern hemisphere, it would deflect this equatorward (southward) motion to the right (west). Since winds are named by the direction they come from, a wind going to the west is from the east and called an easterly. Now do the same exercise for the southern hemisphere. The differences are that now the equatorward surface motion is to the north and that the coriolis force acts to the left. So if we deflect air moving north to the left, it goes west, is from the east, and (surprise, surprise) is another easterly. Historically, these easterly winds were called the trade winds due to their importance for sailing TRADE ships.