1. The Greenhouse Effect Garver GEO 307

2. Take home points from Chapter 2  EMR carries energy through space  If an object can absorb energy, it can also emit energy.  Objects emit at a rate equal to T 4

3. Chapter 3: Layer Model  Algebraic calculation of the effect of an IR absorber (a pane of glass) on the equil. T of the Earth.  Not accurate or detailed  Not used for global forecasts Atmosphere Boundary to Space VIS IR

4. Bare Rock Model  T of Earth is controlled by the ways that energy comes from the Sun and is re-radiated to space as IR.  Sun’s T is high so its energy flux is high –See SB law

5. Simple model of Earth T  Layer model  Toy system to learn from  Assumption: energy in = energy out –F in = F out –Flux is in (W = J/S)  Incoming sunlight = I in = 1350 W/m 2  solar constant

6.  Albedo = reflected light =   Not absorbed and re-radiated as IR  Just ‘bounces’ back  Earth = 0.30 - clouds, snow, ice  Venus = 0.70 - sulfuric acid clouds –But carbon dioxide creates gh effect, T 758 deg K  Mars = 0.15 - no clouds –Some carbon dioxide, T 253 K (afternoon)

7. 9/7/2015 Temperature Scales  Kelvin  Celsius  Fahrenheit  Temperature Conversions: ºC = 5/9(ºF-32) K = ºC + 273 Absolute zero at 0 K is −273.15 °C (−459.67 °F)

8. Incoming solar energy not reflected  1350 W/m 2 (1 -  ) = 1000 W/m 2  Want flux for the whole planet (no m 2 )  F in (W) = I in (W/m 2 ) x Area(m 2 )

9. What area do we use?  Sun shines on half the Earth  Light is weaker/stronger  latitude, dawn/dusk NightDaySolar Constant

10. Sunlight hits Earth from same direction, makes a circular shadow, use the area of a circle, not a sphere. Earth receives influx of energy equal to the intensity of sunlight multiplied by the area of a circle =  r 2 earth Area (m 2 ) =  r 2 earth NightDaySolar Constant

11. Put them together, total incoming flux is:  F in =  r 2 earth (1 -  ) I in  Remember: F in = F out  I in = 1350 W/m 2  Reduce by albedo to 1000 W/m 2  Multiply by area of circle to get Flux (W)

12. First layer model has no atmosphere, just a bare rock!  We’re trying to find a single value for T e of Earth to go along with a single value of the heat fluxes F in and F out  Rate at which Earth radiates energy is given by SB law:  F out = Area x   T 4 earth   = emissivity, 0 to 1, unitless   = 1 would be a blackbody

13. Emissivity  From Wikipedia  The emissivity of a material (written ε or e) is the relative ability of its surface to emit energy by radiation. It is the ratio of energy radiated by a particular material to energy radiated by a black body at the same temperature. A true black body would have an ε = 1 while any real object would have ε < 1. Emissivity is a dimensionless quantity. radiatedblack bodyblack bodydimensionless quantityradiatedblack bodyblack bodydimensionless quantity  The effective emissivity of earth, about 0.612 emissivity

14.  As we did for solar energy, need to convert intensity, I, to flux, F, by multiplying by area.  What area do we use?  Energy leaves in all directions  So, we need the area of a sphere;  A sphere = 4  r 2 earth

15. Total Energy Flux from Earth  F out = 4  r 2 earth  T 4 earth  SB law x Area

16. F in = F out  r 2 earth (1 -  )I in = 4  r 2 earth  T 4 earth Flux in Flux out  T 4 earth (1 -  )I in

17. Can cancel out some factors 4  r 2 earth  T 4 earth =  r 2 earth (1 -  )I in  T 4 earth = (1 -  )I in 4 No atmosphere

18. We know everything here except T of earth Now, rearrange and put T earth alone:  T 4 earth = (1 -  ) I in T earth = 4 (1 -  ) I in 4 4 

19. T earth = 4 (1 -  ) I in 4  Now we have a model that shows the relationship between crucial climate quantities: 1.Solar intensity 2.Albedo But, if we calculate T earth we get 255 K (-15˚C). This is too cold, why?

20. Need a model with a single pane of glass as its atmosphere  Simple model lacks gh effect

21. Layer Model with GH Effect  Energy diagram for a planet with a single pane of glass for an atmosphere.  Glass is transparent to incoming VIS but a blackbody to outgoing IR VIS IR IR IR

22.  Incoming VIS passes thru atm, absorbed by surface  Surface radiates IR as  T 4 ground  I up,ground, which is entirely absorbed by atm  Atm has a top and bottom, so it radiates energy up and down:  I up,atm  I down,atm  T 4 ground

23.  The model assumes that the energy budget is in a steady state: energy in = energy out  This applies to individual pieces of the model as well.  So energy budget for atm: I up,atm + I down,atm = I up,ground (units are Watts/Area) Or,  T 4 atm  T 4 ground  T 4 ground 2  T 4 atm

24.  Budget for the ground is different now because we have heat flowing down from atm.  We still assume the energy budget is in a steady state: I in = I out  So the component fluxes are: I up,ground = I in,solar + I down,atm (units are Watts/Area) Or,  T 4 ground  I solar  T 4 atm  T 4 ground 2  T 4 atm 4

25. Finally, a budget for the Earth overall:  Draw a boundary above the atm and figure that if energy gets across the line in, it’s flowing out at the same rate.  I up,atm = I in,solar  The intensities are comprised of individual fluxes from the Sun and the atmosphere.  T 4 atm =  I solar  T 4 ground In = Out 2  T 4 atm 4 Page 25

26. Budget for the Earth overall:  T 4 atm =  I solar One unknown is T atm If we solve for T atm we get the same answer as solving for T earth in the bare planet model. This is an important point!  T 4 ground In = Out 2  T 4 atm 4

27. It tells us that the place in the Earth system where the T is most directly controlled by the rate of incoming solar energy is the T at the location that radiates to space. This is called the skin temperature of the Earth. (it’s equal to the outer most T atm ) Now that we know this we can plug that into the budget eqn for the atm. Page 25 - 26  T 4 ground In = Out 2  T 4 atm

28. And see that: (page 26) 2  T 4 atm =  T 4 ground Or, T ground = 2T atm This means that the T of the ground must be warmer than the skin T by a factor of the fourth root of 2, an irrational number that = 1.189. (~19%) Fourth root of 2 = ± 1.189207 4 Warmer by about 19% 2  T 4 atm  T 4 ground T atm

29. Bottom Line!  In our model where we slide in an atmosphere we have shown that:  The T of the ground must be warmer than the skin T (at top of atm) by roughly ~19%.

30. Final points for Chapter 3 Result from bare rock model, this T is more similar to skin T at top of atm Recorded T Data, not from a model Result from layer model (w/greenhouse) Too cold Too warm Pretty close 253 x 1.189 = 300

31. Atm is not an energy source like a giant heat lamp in the sky. So, how does it change the T of the ground?? Analogy: Equilibrium water level in a steadily filling/draining sink. Sink Drain Water Level 1. Water flows in, hangs out awhile, drain out. energy in = energy out 2. Drains faster as the level rises due to inc. pressure (weight). energy flows out faster as T rises 3. Eventually water level gets to a point where flow out equals flow in. This is the equlibrium T

32. Sink Drain Water Level Rises 4. Now, constrict the drain with a penny. Water flows out more slowly, The water level rises until the higher water level pushes more water down drain to balance the flow from the faucet again. Constrict flow Greenhouse gases are the ‘penny in the drain’. They make it more difficult for heat to escape the Earth, and as a result the equilibrium temperature has to go up until the fluxes balance each other again.

33. Take Home Points  The outflow of IR energy from a planet must balance heating from the Sun.  The planet accomplishes this balance by adjusting its temperature.  Absorption of outgoing IR by the atmosphere warms the surface of the planet, as the planet strives to balance its energy budget