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# GEOL 1302, Week 4 Notes GEOL 1302

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This 4 page Class Notes was uploaded by Theresa Nguyen on Sunday September 18, 2016. The Class Notes belongs to GEOL 1302 at University of Houston taught by yunsoo choi in Fall 2016. Since its upload, it has received 25 views. For similar materials see Intro To Global Climate Change in Geology at University of Houston.

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Date Created: 09/18/16

Chapter 4: A “simple” climate model Power of the Sun P = sT^4a = sT^4(4pr^2) σ = 5.670373 * 10−8 (W/m2)/K4 T = 5778 K Radius of the Sun (r) = 6.9634 * 108 m 3.8510 * 1026 W ≈ 3.8 trillion trillion Watts Earth’s solar constant (S) Earth-Sun Distance =150,000,000 km = 150*109 m Surface area of a sphere = 4*pi*r2 SA = 4*pi*(150*109 m)2 =2.83 * 1023 m2 Earth’s Solar Constant (S) = Power of Sun / Surface area sphere = = 3.8*1026 W / 2.8*1023 m^2 = 1,360 W/m^2 Solar constant on Venus (Sv) Venus-Sun Distance = 108,000,000 km = 108*109 m Surface area of a sphere = 4*pi*r^2 SA = 4*pi*(108*109 m)^2 = 1.47 * 10^23 m^2 Solar Constant Venus (Sv) = Power of Sun / Surface area sphere = 3.8*1026 W / 1.47*1023 m^2 - Sv = 2,600 W/m^2 - What is S? 1,360 W/m^2 Total power of Solar energy shining on Earth Solar constant = 1,360 W/m^2 What is Sunlight area of Earth? Radius of Earth (R) = 6,400 km = 6.4*10^6 m Area of the Earth’s shadow ( AEs ) = pi*R2 = 1.3*10^14 m^2 Total power of Solar energy shining on Earth = A Es - = Solar constant (S) * Area of Earth’s shadow ( ) - = (1360 W/m^2) * (1.3*10^14 m^2) = 1.8*10^17 W - = 180,000 trillion watts = 180,000 terawatts (TW) Each day humans consume ≈ 15 TW. So if we could capture 0.01% of solar energy shining on Earth, we could satisfy all of the world’s current energy needs. Earth’s Albedo (α) Some of the sunlight that shines on Earth is absorbed and the rest is reflected back into Space. The reflectivity of a planet, or the fraction of reflected photons, is called its’ “albedo” or “α”. The global average albedo of Earth is approximately 0.3, meaning that 30% of photons are reflected. - What percent of the photons are absorbed? Fraction of photons absorbed = 1 – α = 0.7 or 70%. Thus the total rate of energy absorbed by Earth ( E ¿ ) = E ¿ = S(1 – a)pR^2 ( E ¿ ) = 180,000 TW * 0.7 = 126,000 TW Energy Input ( E ¿ ) aka Power Input ( E ¿ ) = 180,000 TW * 0.7 = 126,000 TW More useful to describe E¿ in units of Watts per square meter. 2 E¿ S 1−a )pR S(1−a ) area= 4 pR 2 = 4 E ¿ / area = (1,360 W/m^2 * 0.7) / 4 = 238 W/m^2 How much is reflected back into space? - Ereflect= (1,360 W/m^2 * 0.3) / 4 = 102 W/m^2 How much is average total E reaching top of our atmosphere? - 238 W/m^2 + 102 W/m^2 = 340 W/m^2 Energy Input ( E ¿ ) varies spatially So the Earth absorbs an average of 238 W/m^2 from the Sun, but that does not mean that every square meter absorbs this amount. - Where is it absorbing less? - Where is it absorbing more? Energy loss to Space In the early 1800s, Joseph Fourier, on of history’s greatest mathematicians asked the following simple question: - “Because energy is always falling on the Earth from the Sun, why doesn’t the Earth heat up until it is the same temperature as the Sun?” The answer he determined is that the Earth is losing energy at the same rate as it is receiving it from the Sun. - Otherwise it would heat up. - Therefore, the global average E ¿ = 238 W/m^2 = E out. If E¿ = Eout, what is the temperature of Earth E¿ = 238 W/m^2 = E outfrom Earth’s surface. P 4 E¿ S 1−a ) S 1−a ) a =sT area = 4 sT^4 = 4 T = 4 s1−α ) √ 4s S = 1,360 W/m2 α = 0.3 σ = 5.67 * 10-8 T Earth sSur= 255K (-18°C) Is this correct? - Why? The greenhouse effect – Simple Model Start off with a few assumptions: - 1) The Earth’s atmosphere is transparent to visible photons emitted by the Sun (wavelengths from 0.3 –0.8 microns). - 2) The Earth’s atmosphere is opaque to infrared photons emitted by the Earth’s surface (4 – 20 microns), so all of these photons are absorbed by the atmosphere. - 3) The atmosphere behaves like a blackbody (i.e., it emits photons based on it temperature). - 4) The atmosphere emits ½ of these photons upwards to Space (carrying energy away from Earth) and ½ of these photons back down to Earth’s surface (absorbed by Earth’s surface). One-Layer Model Also assume that the temperature on Earth is not changing. Is Earth’s surface in Energy balance? Is the atmosphere in balance? Calculate surface T with one-layer Model E P/a = sT^4 = 476 W/m^2 = out T = 4 476 onlay√5.67x10−8 = 302.7 K Is this temperature right? - What assumption is incorrect? Also need to consider all planet is at one temperature, which is equivalent to assuming that energy is transported very rapidly around the planet. Thus, addition of an atmosphere that completely absorbs infrared (IR) radiation have significantly warmed the planet’s surface. This occurs because the surface of the plant with an atmosphere is heated by the Sun and also by the atmosphere. You cannot see the atmosphere heating the Earth because IR photons are not visible to humans, but they still carry energy. When scientists talk about the greenhouse effect, it is this heating of the surface by the atmosphere to which they are referring. Another way to think about this: - The atmosphere makings it harder for the surface to lose energy to Space. - Without an atmosphere, the surface emits only 238 W/m^2 to stay in energy balance. - With a one-layer atmosphere, only ½ the energy emitted from the surface escapes to space, the other half is returned to the surface. - This means that the surface must emit 2X as much (476 W/m^2) in order for 238 W/m^2 to be emitted to Space. - Which means that a planet with an atmosphere must have a warmer surface than a planet without an atmosphere. Calculate surface T with two-layer Model 4 714 Tonlayer5.67x10−8 = 335 K = 62 degrees C √ Summary Created a very simple climate model based on fact that solar energy received by a planet ( E¿ ) must be balanced by energy that is radiating to space ( E out) For a planet, E¿ = S(1-α)/4. S is solar constant, which is the amount of energy per square meter that sunlight is delivering at the planet’s orbit. - And α is planet’s albedo. Energy out for a planet is due to blackbody radiation; Eout = σT4

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