# Earth black body temperature wrong?

1. May 30, 2007

### Andre

The blackbody temperature of Earth is supposed to be -18C or 255K. This can be derived by reworking Stefan Boltzmann law to:

TK = (S*(1-a)/4*rho) ^0,25

http://www.lwr.kth.se/Grundutbildning/1B1292/Compendium_online/ch05s02s01.html

in which S is solar influx, we use 1367 w/m2
a is albedo: 0.3
rho is the Stefan Bolzmann constant: 5.667E-8
The factor 1/4 is caused by the difference in influx surface. The cross section of Earth (pi*r2) is intercepting the solar radiation but it has been equally divided over the surface of the Earth (4*pi*r2).

I think the latter reasoning is wrong. You can average out the influx but that doesn not average out the temperatures because the relationship is not lineair.

I may be wrong of course but If I assume the earth to be 180 slices of a sphere and calculate the radiation temperature for every degree of lattitude seperately, then I get an average blackbody temperature of about -21.1C or 251.9K.

Before I show the (simple) calculations, I like to invite everybody to attempt the calculation likewise since independent duplication is the best confirmation.

What would be the consequences on all climate models etc if the blackbody temperature is three degrees lower than always assumed?

Last edited by a moderator: Apr 22, 2017 at 5:52 PM
2. May 30, 2007

### billiards

You know, I'm pretty sure that the 'latter reasoning' is correct, and is a simple application of Gauss' Law.

3. May 30, 2007

### sneez

Andre nowhere in the derivation it is assumed to be linear and it poses no restriction on average.

Andre, you must do something wrong. If you use the same values it does not matter if you divide the earth in 180 slices or 1000 slices. When you add them up it should give you the average or 255K since thats how what you are doing anyway you are adding up averages.
[It looks to me as if you divided the earth into 179 slices instead 180. If you do it numerically you should check this. ] Also, for your slice calculations did you use albedo of .3 or .32 ? it will make huge difference.

4. May 30, 2007

### Andre

Sneez, What the original formula is doing is dividing the influx energy over the surface of the sphere before converting it to heat. What I'm doing is converting the incoming radiation to heat first for every degree lattitude and then calculate the average temperature. That's done numerically avoiding cowardly the intergration math.

5. May 30, 2007

### sneez

yes, i undestand and what I said applies. Are you SURE that its 180 and not 179 slices by some mistake?

6. May 30, 2007

### mheslep

Interesting. Andre - may I suggest the albedo constant is the place for further inquiry as I believe Billiards is correct w/ respect to Gauss's Law and a black body radiator- the shape of the surface is not important, one simply sums all the energy flux through the surface to derive the BB temp. The http://en.wikipedia.org/wiki/Albedo" then its shape w/ respect to the incoming energy distribution does matter. To assume the albedo is constant everywhere (.3) implies the surface & atmosphere must be lambertian everywhere as seen by the incoming solar flux. I'm not familiar w/ the source of the .3 figure, but the actual 'terrestrial albedo' must be a complicated function of surface type and angle of incidence.

Last edited by a moderator: Apr 22, 2017 at 5:53 PM
7. May 30, 2007

### Andre

Okay once more to the basics.

Spectral energy hits earth. That energy is turned into heat on the spot of impact according to the basic Stefan Bolzmann law implying the operator 4th square root first. Then we get to the average temperature by calculating the weigted average temperature second.

Now suppose that the hitting energy is 16 and we operate the square root first then we get 4 then we do the area average temp by dividing by 4. The result is one.

Now suppose we do the area averaging first on the hitting energy we get 16/4 = 4 before we do the square rooting then we end up with 2. That's the difference.

The sequence of the operators appears to be wrong in the scholar equation. That's the problem. Temperature first, then averaging, not averaging the energy first before converting it to temperature.

8. May 30, 2007

### billiards

Andre, whatever happens energy must be conserved.

9. May 30, 2007

### sneez

Flux intercepted by circle area = Flux re-emitted from spherical area (conservation of E)
solving for Te we get effective temperature of sphere or what temperature sphere would have if all intercepted energy was re-radiated back to space.

If you play a scientist andre, do it right and post clearly all steps of what you talking about because this is confusing. Either I do not understand or you cannot do simple algebra, i suppose its the first one. If you do honestly want somebody to check your work be decent enough not to waste our time with half baked examples which are not clear.
You are not schooling us Andre even though you like to play that role, that simple Teff derivation is so simple and the laws that its derived from are clear, unlike your counterexample.

10. May 30, 2007

### mheslep

Sneez - pls drop the ad hominem and let him and us work through the problem.

11. May 30, 2007

### Andre

okay I have two handicaps here, I'm neither familiar with the English math terms nor latex. But we'll give it a try.

We take slice of the earth pie between the lattitude of 'a' degrees and 'a+1' degrees. The surface of the cross section of that slice intercepting the solar flux is average length of the top and bottom length: 2*R(cos(a) + cos(a+1))/2 times effective height (cos(a)*(60 nautical miles)). We multiply this surface with the influx and divide it over the effective surface of the slice:
pi*R*(cos(a)+cos(a+1)*absolute height( 60 nautical miles (without the cos(a)!). Now we use the basic stefan boltzman law to calculate the temperature of that particular spheric slice.

We summarize these steps for a=0 to 89, and also sum the temp times surface area divided by sum of surface area (weighted average) to get 251.9 kelvin.

I'm really interested to see what is wrong with that.

Last edited: May 30, 2007
12. May 30, 2007

### sneez

No ad hominem comitted. From a first post im interested with the new idea. However, after 3 posts of clear avoidance of posting the problem, it smells like not science. Since andre is one of few ppl interested in truth Im reminding him of his own medicine. Hope its taken that way.

13. May 30, 2007

### Andre

I was not avoiding btw but just having limited time available. And I have a habit of triggering others to duplicate and scrutinize.

14. May 30, 2007

### Andre

I'm sure hat there is no violation of this principle, a uniformly heated earth to 255K is probably emitting the same flux as an Earth with temperatures differentiated with lattitude from ~270K to <100K. But the average of the latter is not the same.

15. May 30, 2007

### siddharth

Do you mean that the Temperature isn't uniform? When we model the Earth as a black body, then we assume that the temperature is uniform.

16. May 30, 2007

### Andre

okay but what temperature? The actual average temperature is apparently not the same as the temperature following from an evenly radiating shpere.

Atrtached is a snapshot of the spreadsheet (bottom part) I had to redo it because I worked it out first elsewhere and can't find the original file back. Somehow the rework gives me a new value: 252,5 degrees K. I''ll compare them later.

#### Attached Files:

• ###### BB-revolution.GIF
File size:
22.2 KB
Views:
183
17. May 30, 2007

### siddharth

But, the point is that if you're going to use Stefan's law, you must assume that the temperature is uniform throughout the sphere. So, what do you mean by an average temperature?

In that case, if $$S$$ is the http://www.britannica.com/eb/article-9068556/solar-constant" (area perpendicular to sun rays), the energy abosrbed by the earth is $$S \pi r^2$$, where r is the radius of the earth. Now, since the earth is assumed as a black body, the energy emitted by the earth, $$\sigma T^4 4 \pi r^2$$ must be the same. From this, solve for T.

I don't know why you are splitting the earth into various parts?

Last edited by a moderator: Apr 22, 2017 at 5:53 PM
18. May 30, 2007

### Andre

The question is, what would the average temperature of Earth be as a grey body with an energy absorption of 30%.

For that the current model apears to simple since it assumes interchangeability of the factors emission and temperature. It divides the radiation evenly on the Earth surface before it is converted to temperature.

I'm trying to see what happens if you convert the radiations to temperature first for each degree of lattitude and then see what the average temperaure is. An example of four slices for 0-1 - 29-30 -59-60 - 89-90 degrees lattitude is attached.

#### Attached Files:

• ###### earth-slices.GIF
File size:
12.9 KB
Views:
227
19. May 30, 2007

### Andre

I see I can upload zips

So here is the excel sheet to scrutinize.

#### Attached Files:

• ###### blackbody.zip
File size:
13.6 KB
Views:
79
20. May 30, 2007

### sneez

Dont you have to modify albedo as well ? The constant albedo is consistent only with the circular area of intercepted radiation.

Im still working and will be for a longtime , but I will have a crack at this when im off.