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Stefan-Boltzmann's law, atmosphere

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fluidistic
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Homework Statement


To understand why it's hotter inside a car than its surrounding when exposed to sunlight, let's try a simplificate model that is equivalent to what happens with the atomosphere. To start, let's suppose that the Earth's surface is flat and that it behaves like a black body for the sunlight (forget about the atmosphere for a moment). There will be a radiation flow over the surface of the Earth coming from the Sun (1350W/m²) that will produce a heating that will eventually reach an equilibrium temperature given by Stefan-Boltzmann's law. Let's now introduce in our model the atmosphere, under the form of a glass a few meters over the Earth's surface that let pass all the radiation from the Sun but that behaves like a black body for the Earth's radiation. Determine the equilibrium temperature of this situation.


Homework Equations


[itex]P/A=\varepsilon \sigma T^4[/itex].


The Attempt at a Solution


I calculated the temperature without considering the atmosphere with the given equation. This gave me about 392.81K which seems somehow big to me but possible after all.
Now I consider the atmosphere. I've made a sketch but I get confused on how to solve the problem. Basically I consider a "ray" of sunlight passing through the atmosphere, reaching the Earth. The Earth will then emit as if it was a body at 392.81 K. The atmosphere will aborb this radiation and re-emit it since it behaves as a black body. I can continue this process ad infinitum and don't reach anything.
If I remember well, the solution contained infinite series. Can someone help me a bit?
 

Answers and Replies

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D H
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I calculated the temperature without considering the atmosphere with the given equation. This gave me about 392.81K which seems somehow big to me but possible after all.
This value is high for two reasons.
  1. Not all of the incoming solar radiation is absorbed by the surface. Some is absorbed by the atmosphere, some is reflected, mostly by clouds. The Earth's albedo is about 0.3.
  2. The Earth's cross section to the incoming radiation is the area of a circle with radius RE while the outgoing radiation is emitted from the surface of a sphere with radius RE.

Now I consider the atmosphere. I've made a sketch but I get confused on how to solve the problem. Basically I consider a "ray" of sunlight passing through the atmosphere, reaching the Earth. The Earth will then emit as if it was a body at 392.81 K. The atmosphere will aborb this radiation and re-emit it since it behaves as a black body. I can continue this process ad infinitum and don't reach anything.
Assuming an atmospheric emissivity of 1, the upper atmosphere emits as if was a black body at the equilibrium temperature -- inward and outward. The surface receives the incoming solar radiation plus the thermal radiation from the atmosphere. To be in balance, the energy from the outgoing thermal radiation from the surface must equal this incoming energy to the surface. It gets a little hairier (but not much) if the emissivity of the atmosphere is less than 1.
 
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fluidistic
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This value is high for two reasons.
  1. Not all of the incoming solar radiation is absorbed by the surface. Some is absorbed by the atmosphere, some is reflected, mostly by clouds. The Earth's albedo is about 0.3.
  2. The Earth's cross section to the incoming radiation is the area of a circle with radius RE while the outgoing radiation is emitted from the surface of a sphere with radius RE.
Ah I see, thanks.

Assuming an atmospheric emissivity of 1, the upper atmosphere emits as if was a black body at the equilibrium temperature -- inward and outward. The surface receives the incoming solar radiation plus the thermal radiation from the atmosphere. To be in balance, the energy from the outgoing thermal radiation from the surface must equal this incoming energy to the surface. It gets a little hairier (but not much) if the emissivity of the atmosphere is less than 1.
Ah ok, I totally missed the part that the atmosphere will also emit in space. In fact in my draft I considered it as a mirror. I just corrected it now thanks to your post.
Does this make sense if I say that the Earth absorbs (and also emits) in total [itex]\sum _{n=0}^{\infty } \frac{P}{2^n}=2P[/itex]? If so, I calculated the new equilibrium temperature to be around 467.13K.
 
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D H
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That 467.13 looks good. That is of course ignoring albedo, the difference between cross section and surface area, and the emissivity of that atmosphere.
 
  • #5
fluidistic
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That 467.13 looks good. That is of course ignoring albedo, the difference between cross section and surface area, and the emissivity of that atmosphere.
Ok thank you very much!:biggrin:
 

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