Heat flux as a fraction of energy flux due to thermal re-radiation

[cwbullivant]

Homework Statement

Radioactive decay of elements in the Earth's interior results in a mean heat flux through the Earth's surface of 5x10^-2 W/m^2. What is this flux expressed as a fraction of the energy flux due to thermal re-radiation of absorbed solar energy? If radioactive decay were the only heat source for the Earth, what would the Earth's surface temperature be?

Homework Equations

$$ W_{p} = \frac{L_{\odot}}{4\pi r^{2}}(\pi R^{2})(1-A) $$

$$ L_{P} = 4\pi R^{2}\sigma_{SB}T_{P}^{4} $$

The Attempt at a Solution

To begin, I need to find the energy flux due to thermal re-radiation of absorbed solar energy. To do this, I figured, I could just plug the appropriate numbers into the first of the two given formulas. I'm not certain as to my answer doing that, using 3.839x10^26 W for L, 1.5x10^11 m for r, and .4 for A. I was initially not sure what value to use for the cross sectional radius R (where pi*R^2 is the cross sectional area of the planet); I used the radius of the Earth, 6.37x10^11 m, which gives an energy absorbed of about 10^17 W. I've heard figures for the Earth more around ~1300 W (though I don't recall the source for this), so I'm rather suspicious about this answer.

Where am I going off the right path here?

 

[willem2]

L/(4 pi r^2) with r the distance from the Earth to the sun gives the intensity of solar radiation if the sun were directly overhead. This is about 1300 W/m^2. You need to multiply this with the cross-sectional area of the Earth and divide by the surface area of the Earth to get the intensity average intensity (wich is lower because of nights and because the sun isn't always directly overhead)
After that you'll still need to multiply with (1-A). The answer should be in W/m^2

 

[cwbullivant]

You need to multiply this with the cross-sectional area of the Earth and divide by the surface area of the Earth to get the intensity average intensity (wich is lower because of nights and because the sun isn't always directly overhead)

What value of R should I be using to find the cross sectional area? The radius of the Earth doesn't seem like the appropriate thing to use, but it's all I've been able to think of.

 

[Chestermiller]

In my judgement, you should be using the second equation (per unit surface area), with the temperature equal to the average surface temperature of the earth. This would give you the heat re-radiated from the Earth's surface. Much of the solar flux that hit's the Earth's atmosphere is scattered back into space or absorbed by the atmosphere without reaching the surface.

Chet

 

[haruspex]

What value of R should I be using to find the cross sectional area? The radius of the Earth doesn't seem like the appropriate thing to use, but it's all I've been able to think of.

You correctly used the cross-sectional area of the Earth to get the incoming solar power. willem2 is saying that you then need to divide by surface area of the Earth to get the corresponding outgoing thermal flux.

Much of the solar flux that hits the Earth's atmosphere is scattered back into space or absorbed by the atmosphere without reaching the surface.

I believe that is supposed to be covered by the (1-A) term.

Not sure how to address the last part:

If radioactive decay were the only heat source for the Earth, what would the Earth's surface temperature be?

Can't treat the Earth as a black body, since that calculation gives -18C even with the solar input. On the other hand, if the Earth were a lot cooler then most of the moisture in the atmosphere would condense out and it would behave much more like a black body.

 

[Chestermiller]

I believe that is supposed to be covered by the (1-A) term.

Do you happen to know what the spectrally averaged albedo of the Earth is?

 

[haruspex]

Do you happen to know what the spectrally averaged albedo of the Earth is?

In the OP, cwbullivant has got 0.4 from somewhere.
Kiehl & Trenberth, 1997, (http://www.windows2universe.org/ear...earth_rad_budget_kiehl_trenberth_1997_big.gif) give
342 W/m² in from space
77 reflected by clouds and atmosphere
67 absorbed by atmosphere (and presumably reradiated back out)
30 reflected at surface
168 absorbed at surface

 

[cwbullivant]

In the OP, cwbullivant has got 0.4 from somewhere.
Kiehl & Trenberth, 1997, (http://www.windows2universe.org/ear...earth_rad_budget_kiehl_trenberth_1997_big.gif) give
342 W/m² in from space
77 reflected by clouds and atmosphere
67 absorbed by atmosphere (and presumably reradiated back out)
30 reflected at surface
168 absorbed at surface

0.4 was given in the text as the "average albedo".