MATH 412: Homework # 1 Tim Ahn June 15, 2016

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1 MATH 412: Homework # 1 Tim Ahn June 15, 2016 Chapter 1 Problem 1 (# 2 & 3) (a) Arctic sea ice extent averaged over January 2011 was million square km. This was the lowest January ice extent recorded since satellite records began in It was 50,000 square km below the record low of million square km set in 2006, and 1.27 million square km below the 1979 to 2000 average. What do these data suggest about the Earth s climate? These data could suggest that the oceans and land masses are absorbing more energy than in the previous 31 years, which would also indicate higher temperatures. The net effect is Earth radiating out more energy than what is incoming from the Sun. It could also be used as supporting evidence for the ongoing occurrence of global warming. (b) The extent of the Northern Hemisphere snow cover during January 2011 was 1.76 million square km above the long-term average of 46.7 million square km, marking the sixth largest January snow cover extent and the tenth largest monthly snow cover extent on record for the hemisphere. It also marked the fourth consecutive January with above-average snow cover extent for the Northern Hemisphere. What do these data suggest about the Earth s climate? Snow cover extent, even moreso than ice extent, would intuitively seem to be more of an indicator of climate variability than climate change in the presence of sparse data. However, drawing broad conclusions based strictly on these data would indicate a trend that Earth is radiating less energy in the Northern Hemisphere due to lower temperatures. With greater snow cover, Earth s albedo is increased thereby lowering the incoming energy and perpetuating the trend. Problem 2 (# 6) Figure 1.8 shows many possible dipole connections. Choose two such connections which are identified in the figure and not discussed in the text and write a brief summary for each, using publicly available resources. As per the NOAA website, the Arctic Oscillation (AO) is a climate pattern characterized by winds circulating counterclockwise around the Arctic at around 55 N latitude. Strong winds occur when the AO index is positive which confines the cold air to the polar regions. When the AO index is negative, the winds are weaker and allow the cold winds to travel further south. The Pacific-North American (PNA) teleconnection pattern involves the air pressure at four locations over the Pacific Ocean and North America. The positive phase of PNA occurs with above average air pressure in the Pacific near Hawaii and over the mountainous region of North America, with a corresponding below average air pressure pattern south of Alaska and over the southeastern United States. This positive phase generally results in above average temperatures in western Canada and US, and below average temperatures in the southern region of the US. It is also associated with lower precipitation in the Pacific Northwest and eastern half of the US. The negative phase of PNA generally produces opposite results of the positive phase. 1

2 Chapter 2 Problem 3 (# 4) Assume that the temperature of the Sun is T and that the Sun radiates as a perfect black body, so the total energy radiated by the Sun per unit surface area and per unit time is F S = σt 4. Assume, furthermore, that this energy is radiated uniformly in all directions. The amount of energy passing per unit time through a unit area at the Earth s surface is found by multiplying F S by the ratio of the surface area of the Sun (πrs 2 ) and the surface area of a sphere at the mean distance from the Sun to the Earth (πrse 2 ), S = (R S/R SE ) 2 F S. Taking R S = m and R SE = m, and assuming T = 5, 780K, find the value of S. How close is this value to the solar constant? What could explain the difference? F S = σt 4 σ = W m 2 K 4 πr 2 S = Surface area of the Sun πr 2 SE = Surface area of a sphere at the mean distance from the Sun to Earth R S = m R S E = m T = 5, 780K S = (R S /R SE ) 2 F S (1) ( ) = , (2) = 1, (695.5/(149.6*10^3))^2 * 5780^4 * 5.67*10^(-8) (3) ## [1] The measured value of the solar constant is 1,368 Wm 2 (pp 16) which is nearly equal to the calculated value of 1, The difference can be explained by myriad reasons. The radii may not be precise enough due to significant digits, the temperature of the Sun may not be exactly 5,780K, or the difference could be explained by a variation in the sunspot cycle. 2

3 Problem 4 (# 8 i,ii) Satellite data of the Sun s radiation indicate that the solar constant S 0 varies approximately between 1, Wm 2 and 1, 367 Wm 2, with a period of about 11 years. (i) Use the balance equation (2.8) with α = 0.3 and ε = 0.6 to estimate the resulting variation (difference of max and min) in the Earth s global mean surface temperature T. Use a suitable linearization if possible. σ = W m 2 K 4 Q = 1 4 S 0 Q min = = W m 2 4 Q max = = W m 2 4 alpha <-.3 epsilon <-.6 sigma <- 5.67*10^(-8) solcon <- seq(1360, 1370, by=.01) baleq <- function(s0){ (((1-alpha)*(S0/4))/(epsilon*sigma))^(1/4) } (1 α)q = εσt 4 (4) ( ) 1/4 (1 α)q T = (5) εσ ( ) 1/4 (1 0.3) Tmin = (6) = K (7) ( ) 1/4 (1 0.3) Tmax = (8) = K (9) T max T min = = K (10) plot(solcon, baleq(solcon), type='l', asp=t, xlab='s_0', ylab='temp (K)', main='energy Balance') points(1365.5, baleq(1365.5), pch=8, cex=1.5, lwd=3, col='gold') points(1367, baleq(1367), pch=8, cex=1.5, lwd=3, col='gold') 3

4 Energy Balance Temp (K) S_0 cat('t_max - T_min = ',baleq(1367)-baleq(1365.5),'k') ## T_max - T_min = K (ii) Use the balance equation (2.14), (1 α)q = A + BT, with A = and B = 2.09 to estimate the resulting variation (difference of max and min) in the surface temperature T. Use α = 0.3. (1 α)q = A + BT (11) (1 α)q A T = B (12) Tmin = 2.09 (13) = C = K (14) Tmax = 2.09 (15) = C = K (16) Tmax Tmin = = 0.13 K (17) 4

5 A < B < budyko <- function(s0){ (((1-alpha)*(S0/4)-A)/B) } plot(solcon, budyko(solcon), type='l', asp=t, xlab='s_0', ylab='temp (K)', main='budyko\'s Model') points(1365.5, budyko(1365.5), pch=8, cex=1.5, lwd=3, col='gold') points(1367, budyko(1367), pch=8, cex=1.5, lwd=3, col='gold') Budyko's Model Temp (K) S_0 cat('t_max - T_min = ',budyko(1367)-budyko(1365.5),'k') ## T_max - T_min = K 5

6 Problem 5 (Supp #4) The albedo of Mars is approximately α = Mars is approximately 1.5 times as far away from the Sun as Earth. The solar constant of Earth is approximately 1368 W/m 2. Use this information and the Stefan-Boltzmann law to compute the black body temperature of Mars. We know that the solar constant S 0 of Earth is approximately 1,368 Wm 2 and can be calculated by ( rsol ) 2 R (Energy Output of Sun), where rsol is the radius of the Sun (approximately 432,288 miles), and R is the distance from the Sun to Earth (approximately 92,960,000 miles). We can calculate the approximate energy output of the Sun by using the given values: Energy output of Sun = 1368 ) 2 63, 260, 405 W m 2 We can then use the Stefan-Boltzmann law to derive an equation to solve for temperature T by: ( rsol R ( (1 α)q T = σ where Q = Energy flowing into a square meter of a planet = 1 4 S 0. By multiplying R by 1.5 in the solar constant equation, we find the solar constant of Mars to be approximately 608 Wm 2, so Q mars = = 152. Therefore, the black body temperature of Mars can be calculated by: ) 1/4 T mars = ( ) 1/ (18) = K (19) r_sol < # radius of Sun in miles R < # distance from the Sun to Earth in miles S0 < # solar constant of Earth sun <- S0 / (r_sol/r)^2 # Watts per square meter Q.mars <- (r_sol/(1.5*r))^2 * (1/4) * sun alpha <-.25 sigma < * 10^(-8) # Stefan's constant Tstar <- (((1-alpha)*Q.mars)/sigma)^(1/4) cat("the black body temperature of Mars is approximately", Tstar, "K") ## The black body temperature of Mars is approximately K 6