Solar Radiation and the Seasons. LAST TIME in GEOG 1001! Chapter 1. 8/29/2010. Objectives

Transcription

1 Solar Radiation and the Seasons GE G 1001 Professor Holly Barnard August 26, 2010 LAST TIME in GEOG 1001! Introductions Social contract No audible ringing or answering of cell phones Please text responsibly (do not break the golden rule) Computers to the sides and back of the room, please Please raise your hand My name: Dr. Barnard, Prof. Barnard, Holly, or Dr. B! Add-drop Objectives Chapter 1 1) Find your way around the universe 2) Understand Earth s orbit 3) Find your way around Earth 4) What does Jimmy Buffet have to do with Geography? M27: Not a Comet Credit & Copyright: Matthew T. Russell 1

2 Our Solar System Figure 2.1 Dimensions and Distances Earth s orbit Average distance from Earth to the Sun is 150,000,000 km (93,000,000 mi) Perihelion closest at January 3 147,255,000 km (91,500,000 mi) Aphelion farthest at July 4 152,083,000 km (94,500,000 mi) Earth is 8 minutes 20 seconds from the Sun Plane of Earth s orbit is the plane of the ecliptic Chapter 2: Solar energy to Earth: Where are you on Earth? What powers Earth and how? What causes the seasons? What is the relationship between energy and temperature? 2

3 1) Solar wind What powers Earth and how? Energy from Sun Energy from Sun to Earth 2) Radiant or Solar energy (in portions of the electromagnetic spectrum 3) Uneven intercepted energy at the top of the atmosphere 1) Solar wind Effects on: the Auroras (Australis and Borealis) Climate variations (~11 year cycle?) Energy from Sun to Earth 3. Uneven distribution of INtercepted SOLAr radiation or also called? IN- SOLA- TION 2) Radiant or Solar energy in portions of the electromagnetic spectrum i.e. Electromagnetic Energy Solar Radiation Figure 2.8 Why is it uneven? Where is the place receiving the max insolation? What is it called? 3

4 Lambert s Cosine Law Distribution of Insolation The way energy changes with angle is described by Lambert s Cosine Law: E = E o X cosθ Increasing angle increases area illuminated at the surface and therefore per unit area decreases. θ Tropics receive more concentrated insolation due to Earth s curvature Tropics receive 2.5 more than poles θ 13 Daily Net Radiation The Seasons Seasonality Reasons for seasons Figure

5 Seasonality Seasonal changes Sun s altitude angle above horizon Declination location of the subsolar point Daylength Revolution and Rotation Figure 2.13 Axial Tilt & Parallelism (Earth) Sphericity results in uneven insolation Deficit Surplus of energy Figure 2.14 What weather phenomena occur to even the difference between areas with surplus and areas with deficit? 5

6 Daily Net Radiation Reasons for Seasons Revolution Earth revolves around the Sun Voyage takes one year Earth s speed is 107,280 kmph (66,660 mph) Rotation Earth rotates on its axis once every 24 hours Rotational velocity at equator is 1674 kmph (1041 mph) Figure 2.11 Reasons for Seasons Tilt of Earth s axis Axis is tilted 23.5 from plane of ecliptic Axial parallelism Axis maintains alignment during orbit around the Sun Sphericity Solar Energy: From Sun to Earth Intercepted energy at the top of the atmosphere Electromagnetic spectrum of radiant energy 6

7 Radiation, Distance, Energy at Top of Atmosphere For earth system science we are interested in amount of energy incident at earth. Energy will decrease as we move away from the sun (or light bulb) because light is spread out over larger surface. Here E (energy) is: E top of atmophere = Energy sun / Area of Sphere E top of atmophere = Energy sun / 4 π R 2 Note that R is the radius of a sphere defined by our distance from the sun. 25 Radiation, Distance, Energy at Top of Atmosphere Question: What is the incoming energy of a 100 W light bulb at a distance of 1 meter? Remember energy is: E 1-meter away = E bulb / area of a sphere Area of sphere: Area = 4πr 2 = 4 X 3.14 X (1m) 2 = 12.6 m 2 Energy (E) = 100 W / 12.6 m 2 = 8 W m -2 1 m 26 Radiation, Distance, Energy at Top of Atmosphere UNITS What if we are now 2 meters away from the 100 W light bulb? E 1-meter away = E bulb / area of a sphere Area of sphere: Area = 4πr 2 = 4 X 3.14 X (2m) 2 = 50 m 2 Energy (E) = 100 W / 50 m 2 = 2 W m -2 2 m 27 How did we end up with units of Watts per meter squared (W m -2 )? Energy (E) = 100 W / 25 m 2 = 4 W m -2 Energy (E) = W / m 2 = W m -2 2 m 28 by doubling our distance from 1 m to 2 m we reduced incident radiation by 75% 7

8 Radiation and Distance Now for earth we do the same thing except we replace sun s energy for the light bulb and we replace the radius with Earth s orbital radius: E sun Stefan-Boltzmann Law the warmer the object, the greater the energy emitted by that object E toa = 3.92x10 26 W / (4π X r eo2 ) toa = top of earth atmosphere E TOA Assume earth orbital radius (r eo ) = 1.5 x m E toa = 3.92x10 26 W / (4π X (1.5x10 11 m) 2 ) E toa = 1404 W m -2 8