Earth and Planetary Sciences 5 Midterm Exam March 10, 2010

Earth and Planetary Sciences 5 Midterm Exam March 10, 2010 Name: Teaching Fellow: INSTRUCTIONS PUT YOUR NAME ON EACH PAGE. The exam will last 80 minutes. Complete the problems directly on the exam. Extra paper is available if needed. Please show ALL your work so partial credit can be given! Neatness is appreciated. Scoring: Problem Score Multiple Choice / 12 Avatar / 15 Jekyll Island / 15 TOTAL SCORE / 42

Section I: Multiple choice questions. Circle the best answer. (1 point each) 1. The following figure shows the energy balance of the Earth. If the Earth is in radiative equilibrium, and as shown has 100 units of incoming solar radiation, how many units of outgoing infrared radiation at the top of the atmosphere will it have? a. 100 b. 95 c. 75 d. 70 2. The average albedo of the moon is 0.07. This means that, on average, a. 7% of the sunlight that strikes the moon is absorbed by the lunar surface. b. 7% of the sunlight that strikes the moon is reflected by the lunar surface. c. 7% of the radiation emitted by the moon is absorbed by the lunar atmosphere. d. In the N-layer model for the moon, N = 0.07. 3. Which of the following is a NEGATIVE feedback for a change in Surface Temperature? a. Temperature increases polar ice melts albedo decreases? b. Temperature increases atmospheric water vapor increases atmospheric IR absorption increases? c. Temperature increases forest fires increase more dark soot is emitted? d. Temperature increases deserts expand albedo increases? 4. Which of the following will certainly result in an increase in pressure of a gas? a. Decrease volume & increase temperature b. Decrease volume & decrease temperature c. Increase volume & increase temperature d. Increase volume & decrease temperature

5. The following figure shows the infrared emission spectrum of the Earth measured by a satellite over the Mediterranean Sea. Based on this figure, what is the approximate temperature at the surface of the Mediterranean? a. 220 K b. 260 K c. 280 K d. We can t tell from this figure. 6. An object with POSITIVE buoyancy (net force on it is up): I. Has a smaller volume than the volume of fluid it displaces (V OBJECT < V DISPLACED ) II. Has a lower density than the density of fluid it displaces (ρ OBJECT < ρ DISPLACED ) III. Has a smaller weight than the weight of fluid it displaces (ρ OBJECT * V OBJECT *g < ρ DISPLACED *V DISPLACED *g) a. II only b. I and III only c. II and III only d. I, II, and III 7. Which statement about the Hadley circulation is NOT true? a. The lower level winds are directed from west to east in both hemispheres b. The circulation is thermally driven c. The Earth s rotation restricts the Hadley cells to the tropics d. The circulation leads to subsidence and an arid climate around 30 degrees latitude 8. Which of the following describes circulation around low pressure systems? a. Clockwise flow in the Northern Hemisphere & clockwise flow in the Southern Hemisphere b. Clockwise flow in the Northern Hemisphere & counterclockwise flow in the Southern Hemisphere c. Counterclockwise flow in the Northern Hemisphere & clockwise flow in the Southern Hemisphere d. Counterclockwise flow in the Northern Hemisphere & counterclockwise flow in the Southern Hemisphere

9. Where is the atmosphere stable? a. Region I only b. Region II only c. Region III only d. Region I & II e. Region I & III 10. Which of the following is a potential final resting point for a block with density 500 kg m -3 suspended in three fluids with the following different densities. a. b. c. d. 11. Which gas is the most dense for a given pressure and temperature? The atomic weight of N is 14 g mol -1 and of O is 16 g mol -1. a. NO b. NO 2 c. N 2 O d. N 2 e. O 2 12. Which of the following does not represent a partial pressure of water vapor of 1 hpa? a. 50% RH and 2 hpa saturation vapor pressure b. 25% RH and 4 hpa saturation vapor pressure c. 0.1% mixing ratio and 1000 hpa atmospheric pressure d. 0.5% mixing ratio and 500 hpa atmospheric pressure

Section II: Problems Show ALL your work so partial credit can be given! I. Avatar (15 points) In the film Avatar, James Cameron created an entire world on a fictional moon called Pandora. Let s explore how realistic some elements of Pandora were. a.) Pandora is supposedly about 7.41 10 8 km from its star, Alpha Centauri A. Alpha Centauri A has a radius of 8.53 10 5 km and a radiating temperature of 5790 K. Calculate the solar constant (F, in W.m -2 ) at Pandora. (3 points) We use the equation for the solar constant with the parameters defined in the problem: F = σt 4 2 α 4πR α = (5.67 10 8 Wm 2 K 4 )(5790K) 4 (8.53 10 5 km) 2 2 4πd P (7.41 10 8 km) 2 = 84.4Wm 2 b.) If we use the solar constant from part (a) and make some assumptions about the albedo of Pandora, we can calculate Pandora s effective temperature, a chilly -146 C! But the surface temperature must be much warmer to sustain liquid water and the planet s biosphere. Assume the surface temperature of Pandora is actually 20 C. Using the N-layer model, calculate the number of atmospheric layers (N) that Pandora atmosphere must have. (3 points) We use the equation for the N-layer model of the atmosphere with the given temperatures: T surface = (N +1) 1/ 4 T eff (N +1) 1/ 4 = T surface N +1= T 4 surface T eff T N = T 4 surface eff T 1 eff 4 273.15 + 20 So, N = 1 = 27.3 273.15 146 c.) Regardless of your answer to part (b), describe what it means to have an N-layer atmosphere and explain how having an atmosphere can increase the surface temperature of a planet or moon like Pandora. (3 points) The N-layer model describes the atmosphere as a series of layers that each absorb outgoing longwave radiation. Each layer absorbs radiation from the layer above and below and re-radiates that energy both upwards and downwards. As a result, each layer warms the layers below (and as a result, the surface below the lowest layer) with energy that would otherwise have been lost to space if the layer didn t exist.

d.) One of the most spectacular features of Pandora is the floating Hallelujah Mountains. How they float is never explained one possible explanation might be buoyancy. Let s assume: the mass of a mountain is 1.0 10 12 kg the mountains are located at 1 km altitude each mountain has volume of 0.40 km 3 the atmospheric density at 1 km is 1.05 kg/m 3 the gravitational acceleration (g) is the same on Pandora as on Earth (9.81 m/s 2 ) What is the buoyant force on the mountains due to the atmosphere? Is the buoyancy sufficient to make the mountains float? (3 points) The mountains will float if they experience no net force at 1 km. We can therefore compare the force due to gravity (F g ) and the buoyant force (F b ): F g = m mountain g = (1.0 10 12 kg)(9.81 m/s 2 ) = 9.8 10 12 N F b = ρ displaced V displaced g = ρ air V mountain g = (1.05 kg/m 3 )(0.4 km 3 )(1000 m / 1 km) 3 (9.81 m/s 2 ) = 4.1 10 9 N The gravitational force is much larger than the buoyant force (3 orders of magnitude)! So buoyancy is not enough to support the floating mountains, and there must be another force acting on them. In fact, fans of the movie suggest some sort of superconductivity effect is at play (of course, no need to know that to get this question right!). e.) In the movie, the Hallelujah Mountains are often shown enshrouded in clouds. We can calculate the likely height of cloud formation on Pandora. If the temperature at sea level on a given day is 20 C and the dew point temperature is 9 C, at what altitude will clouds begin to form? Is the base of the Hallelujah Mountains (at 1 km above sea level) above the level where the clouds form? You can assume the lapse rates are the same as on Earth: Γ dry = 9.8 K km -1, Γ moist = 6.0 K km -1, Γ dew = 2.0 K km -1. (3 points) We want to know the altitude of cloud formation; i.e. when the temperature of an air parcel and the dew point temperature are equal. Because the air parcel rises below the cloud base, its motion is governed by the dry adiabatic lapse rate: T air = T surface - Γ dry (z z surface ) T air = 20 C (9.8 C/km)z (since z surface = 0 km) We can similarly describe the dew point temperature at an arbitrary altitude: T dew = T dew,urface - Γ dew (z z surface ) T dew = 9 C (2.0 C/km)z When these are equal: T air = T dew 20 C (9.8 C/km)z = 9 C (2.0 C/km)z 11 C = (7.8 C/km)z z = (11/7.8) km z = 1.4 km At the altitude of the mountains (z = 1 km), cloud formation will not yet have begun. So the mountaintops may be in the clouds but the bases won t at least in these weather conditions!

II. Jekyll Island (15 points) Jekyll Island, Georgia is vacation destination located at 31 N, 81 W. In this problem, we ll think about what s going on in the atmosphere when you go to Jekyll Island for vacation. a.) It's a warm summer day with temperature 30 C on your beach vacation. The relative humidity is 70%. What is the saturation partial pressure of water vapor? To what temperature would the air need to cool to in order for condensation to occur? Use the figure below. (3 points) We first read P H2O,sat from the graph, then calculate P H2O RH = (P H2O / P H2O, sat ) * 100% P H2O = (RH/100) * P H2O, sat P H2O = (70/100) * 40hPa = 28hPa We then read T dew from the graph: T dew = 23 o C b.) The air above the ocean just offshore is a slightly cooler 25 C. What is the scale height over the island? What is the scale height over the ocean? The average molar mass of dry air is 29 g mol -1. (3 points) H = (RT)/(M a g) or (kt)/(mg) Island: H = (8.314 J mol -1 K -1 * 303 K )/(0.029 kg mol -1 * 9.8 m s -2 ) = 8.86 km Ocean: H = (8.314 J mol -1 K -1 * 298 K )/(0.029 kg mol -1 * 9.8 m s -2 ) = 8.72 km

c.) In which direction (toward land or sea) does the wind blow at the surface as you lay on the beach in the afternoon? Why? (3 points) The breeze blows from sea to land. This is a land-sea breeze problem during daytime. The scale height is greater over the island, so the pressure falls off less quickly with altitude relative to the ocean. This sets up a pressure gradient at altitude in which air blows from the higher pressure over land to the lower pressure over the sea. There is rising air over the warm land and sinking air over the cool sea. There is a return flow at the surface from ocean toward land. d.) You re worried that your beach vacation may be under threat. While watching the local news, you saw the following satellite image of Tropical Storm Fay headed your way (left). Given all the clouds in the image, is the pressure at the center of the storm higher or lower than the surrounding pressure? The map on the right shows the storm track for Fay. When Fay is located at position A, which direction will you feel the wind blowing at Jekyll Island (assuming geostrophic balance applies)? What about when Fay is at position B? (3 points) A tropical storm (or any storm) is a low pressure system (and you can tell that from all the cloud formation!). In the Northern Hemisphere, flow around a low is counterclockwise. At A, the storm is to the south of the island, and wind is from east to west. At B, the storm is to the west of the island, and wind is from south to north. e.) Fay was associated with sustained winds of 113 km/hr, much stronger than those in typical mid-latitude storms. Using your knowledge of geostrophy, do you expect the pressure gradient between Fay and the surrounding area to be larger or smaller than the pressure gradient between a typical storm and its surrounding area? Justify your answer. (3 points) The pressure gradient is much stronger. Wind speed is directly related to the magnitude of the pressure gradient force a stronger pressure gradient leads to a stronger force on the air, causing increased wind speeds. Hurricanes and other tropical storms typically have very low pressures at their centers.

FORMULAS Ideal gas law: pv=nkt p=ρkt/m a p=nkt Stefan-Boltzmann Law: Energy flux = σt 4 Solar Constant: (blackbody, energy at all wavelengths) F S = σt 4 2 SUN 4πR SUN 2 4πd EARTH SUN 1/ 4 F(1 A) Effective temperature: T eff = 4σ Surface Temperature: T g = (N +1) 1/ 4 T eff Barometric Law: Scale height: Dry Adiabatic Lapse Rate: P(z) = P 0 e z H H = RT g = kt mg Γ dry = ΔT Δz = g C p Angular velocity: Ω = 2π t Coriolis Force: F C = 2mvΩsin(λ) Geostrophic Velocity: 1 ΔP v = 2Ωρsinλ Δx UNITS AND CONVERSIONS 1 m = 100 cm = 1000 mm = 0.001 km 1 kg = 1000 g K = C + 273.15 1 mole = 6.02 10 23 molecules 1 mb = 1000 N m -2 1 Pa = 1 N m -2 1 cm 3 = 10-3 L 1 J = 1 N m 1 Watt = 1 J s -1 CONSTANTS Stefan-Boltzmann constant σ 5.67 10-8 W m -2 K -4 Earth s radius R Earth 6.38 10 6 m Radius of the Earth s orbit around the Sun R Sun-Earth 1.5x10 11 m Solar constant for the Earth: F S 1380 W/m 2 Acceleration due to gravity g 9.81 m s -2 Pressure at sea level P surf 1.013 10 5 Pa Angular velocity for the Earth Ω 7.3 10-5 s -1 Boltzmann s constant k 1.381 10-23 J K -1 Universal gas constant R 8.314 J mol -1 K -1 Avogadro s number A v 6.02 10 23 molecules mol -1 Dry adiabatic lapse rate for the Earth Γ dry 9.8 K km -1