Register to reply 
Earth black body temperature wrong? 
Share this thread: 
#1
May3007, 07:58 AM

PF Gold
P: 5,458

The blackbody temperature of Earth is supposed to be 18C or 255K. This can be derived by reworking Stefan Boltzmann law to:
T_{K} = (S*(1a)/4*rho) ^{^0,25} See equation under 5.9 in which S is solar influx, we use 1367 w/m2 a is albedo: 0.3 rho is the Stefan Bolzmann constant: 5.667E8 The factor 1/4 is caused by the difference in influx surface. The cross section of Earth (pi*r^{2}) is intercepting the solar radiation but it has been equally divided over the surface of the Earth (4*pi*r^{2}). 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? 


Earth sciences news on Phys.org 
#2
May3007, 08:14 AM

P: 747

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



#3
May3007, 09:13 AM

P: 355

[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
May3007, 09:35 AM

PF Gold
P: 5,458

Earth black body temperature wrong?
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
May3007, 09:38 AM

P: 355

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



#6
May3007, 10:27 AM

PF Gold
P: 3,081




#7
May3007, 10:40 AM

PF Gold
P: 5,458

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
May3007, 11:08 AM

P: 747

Andre, whatever happens energy must be conserved.



#9
May3007, 11:12 AM

P: 355

Flux intercepted by circle area = Flux reemitted 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 reradiated 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
May3007, 11:17 AM

PF Gold
P: 3,081

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



#11
May3007, 11:41 AM

PF Gold
P: 5,458

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. 


#12
May3007, 11:45 AM

P: 355

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
May3007, 11:50 AM

PF Gold
P: 5,458

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



#14
May3007, 12:07 PM

PF Gold
P: 5,458




#15
May3007, 12:35 PM

HW Helper
PF Gold
P: 1,198




#16
May3007, 12:48 PM

PF Gold
P: 5,458

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. 


#17
May3007, 01:17 PM

HW Helper
PF Gold
P: 1,198

In that case, if [tex] S[/tex] is the solar constant (area perpendicular to sun rays), the energy abosrbed by the earth is [tex]S \pi r^2 [/tex], where r is the radius of the earth. Now, since the earth is assumed as a black body, the energy emitted by the earth, [tex] \sigma T^4 4 \pi r^2 [/tex] must be the same. From this, solve for T. I don't know why you are splitting the earth into various parts? 


#18
May3007, 02:44 PM

PF Gold
P: 5,458

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 01  2930 5960  8990 degrees lattitude is attached. 


Register to reply 
Related Discussions  
What is absolute body temperature?  General Physics  4  
Low body temperature  Biology  17  
Body Temperature and Exercise  Medical Sciences  10  
Temperature of body in space  Introductory Physics Homework  7  
Black hole black body  Astronomy & Astrophysics  2 