Its weird that someone would say that earth is 33° “warmer than it should be”,  especially when they already reduced the amount of heat by 30% before the calculation started. My approach is that everything is exactly what it should be, and if someone says that there is heat coming from an ice-cold atmosphere, my first thought is that someone miscalculated. .

If we start by taking a look at the differences between the 2 dimensional spherical surface of the absorbing and emitting blackbody, and the 3 dimensional volume of Earth absorbing solar radiation.

To find the emitted intensity of the surface of the inner spherical shell, we need to account for absorption of the received intensity from solar irradiation, TSI, over the surface area of the disc that is the shadow of earth .

Then we need to account for absorption in depth of the volume within the system. Actually, two shells of concentric volumes needs to be accounted for. Atmosphere and solid earth, which gives:

simple

Which is equal to a surface temperature of 286.6 Kelvin.

When the surface emission has been found, we can use the Stefan-Boltzmann equation for radiative heat transfer from the solar constant to the surface:

simple

But the the transferred heat from the solar constant must follow the same laws as the emitted radiation, so with the inverse square law we find:

simple2

Which gives us the effective temperature of 256.3 Kelvin.

If we use radiative heat transfer for the emitted intensity of the surface and the effective temperature:

simple3

We can see that it fits the temperature at the tropopause with satisfying accuracy.

The first transformation of solar radiation through the volume of the outer shell, using the full value of TSI, gives a result close to direct irradiation at zenith on the surface:

simple14

And using the solar constant and the effective temperature for radiative heat transfer to the system gives a good match to the total solar irradiation, direct and diffuse:

simple4

Both are logical results, as the transformation through the outer shell gives a value that should represent what arrives at the surface from the bottom of the volume of the atmosphere, and the heat transfer to the troposphere should include both what is diffused/scattered in that volume, and what arrives at the surface.

We now have a main structure of the system determined by only small modifications of the blackbody model, heat transfer and the inverse square law.

Finally, I want to address the difference between the emitted effective temperature, and the true blackbody temperature of a perfect absorber and emitter of the solar constant.

If we use the solar constant to find the effective temperature that should be emitted from a perfect blackbody, we get:

simple15

The difference is:

simple16

If the difference is assumed to be the amount of work performed by the system to keep the atmosphere in place, according to the first law of thermodynamics where deltaU=Q-W, we can use this for gravity. Consider gravity as the force that acts in a point at the center of mass in a parcel of air, that has a surface area of 1m², and the mass of 1kg, laying statically above the surface of Earth. The force acting in the point at the center of mass in that parcel of air is:

simple5

If we use units of Nm^2, that are used for thermal resistance, stress and pressure, the force acting on the surface of the parcel is:

simple6

The source strength of the body emitting the power needed to raise the force radiated into the surroundings of the source according to the inverse square law is:

simple8

The source strength matches the surface emission of heat in the model. The difference between the true blackbody intensity of earth and the effectively radiated intensity, and the missing heat in the emitted effective temperature is found to be equal to the force of gravity. It also balances the emitted intensity of heat from the surface.

The earth core is said to have a temperature of about 6000 kelvin, I assume that it has the same temperature as the surface of the sun.

If we use the same approach as for the surface and atmosphere, the internal layers decrease by 3/4 for each surface and another 3/4 for the volume above that surface, as heat travel from the core. If the core is 5780 Kelvin and there are 4 pairs of surfaces and volumes.

19

1200px-Earth-crust-cutaway-numbered.svg

 

Greenhouse effect? Yeah, and pots of gold at the end of the rainbow. With unicorns and flying pigs, maybe he-man badgers and aura-photographing as well.

I think that we should stop making shit up, and use what we know is true.

 

 

simple13.png

 

There is one more way of doing it.

E=I-pi=T

If we use the relationship:

14

There is another solution:15

 

/LIT

iss040e090540

 

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