2. The temperature of Earth with an atmosphere—the greenhouse effect: In the pre

Question

2. The temperature of Earth with an atmosphere—the greenhouse effect: In the previous

problem, we imagined the Earth without an atmosphere. The absorbed photons heat the Earth’s surface which then emits new photons as a blackbody at the effective temperature, Te = 247K. Most of these photons are in the infrared part of the spectrum.

Recall that materials have different optical properties at different frequencies. The atmosphere is no exception—it is transparent to visible light but relatively opaque to infrared radiation. Thus, instead of the surface emitting the infrared blackbody photons directly into space, the infrared photons randomly walk their way through the troposphere before escaping. The randomly walking photons keep the atmosphere of the Earth warmer than the value we calculated in the last question (a good thing for life on Earth!).

A photon that randomly walks in a gas has a mean free path length, l, meaning that on average the photon “flies” about a distance of l before interacting or bumping into a gas particle. Imagine an infrared photon in the troposphere, but only in one dimension—it can move up (+l) or down (l) at each time step with equal probability. The average position is zero, the photon has no net drift pushing it up or down. However, the square of the displacements does have a net drift—even though on average photons are just as likely to have moved up as down, over a long period they will have had a chance to wander a long way from the starting point.

The mean square distance, D2, after N steps is D2 = Nl2. In three dimensions, the result becomes 3D2 = Nl2. Therefore, the number of steps that a photon needs to take on average to get to get to a height, h, is N = 3h2/l2. The random walk time is just the total distance covered divided by the speed of light, twalk = Nl/c = 3h2/(lc).

(a) Let’s estimate the size of the greenhouse effect. Above the troposphere, the Earth’s atmosphere becomes transparent to infrared photons, so infrared photons random walk only below the tropopause. Recall that the flux measures the energy per unit area per unit time. So, the flux emitted into space must be:

Frad = (height x radiation energy per unit volume) / random walk time to cover distance h

= h(aT4) / (3h2 / lc) = 4 T4l / 3h = 4 T4 / 3t

where = h/l denotes the optical depth of the troposphere, and where the radiation constant a = 4/c. But, we also know that Frad = Te4 , where Te =247K.

(b) Combine the results above to show that T4 = (3/4)tTe4 .

(c) A more refined analysis (the Eddington approximation in radiation transfer)

yields the result:

T4 = ¾ (t + 2/3) Te4

At the bottom of the troposphere (i.e., the Earth’s surface), t = 2. Using the Eddington approximation, calculate the temperature at the Earth’s surface.

(d) Suppose that the optical depth, t, at the bottom of the troposphere increases by

1%. Again, using the Eddington approximation, how much would this increase in t raise

the surface temperature of the Earth?

(e) Calculate the temperature at the tropopause, where t = 0. Notice that the temperature

at the top of the troposphere is cooler than the effective temperature, Te. What is the value of t when T = Te?

The height at which Te = T is the average location of the origin of the photons emerging from the atmosphere into space, not at the tropopause, where t = 0.

Explanation / Answer

answer- the effective temperatureat which photon emit Te = 247K

Frad = Te4

wherestephan constant = 5.670367×108 W m2 K4

so the flux emitted to the space will be =5.67*10-8 * 247*4 = 5601*10-8 = 5.6*10-5 W m2 K4

at the temperature at tropopause will be = 228.15K and it is cooler than the effective temperature that is 247k

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