# Solar Energy and Green House Effect Question

1. May 3, 2008

### TFM

1. The problem statement, all variables and given/known data

Negelcting internal heat sources, the surface Temperature of the Earth is calculated to be 255 K. Its actual mean surface Temperature is 288 K. The additional heating results from the greenhouse effect. Regarding this as an internalk heat source, calculate that heat source as a fraction of the energy absorbed from the Sun.

2. Relevant equations

3. The attempt at a solution

I am not sure what I am supposed to do exactly, do I just work out the differnect in Temperature and put it over the final temperature (ie $$\frac{33}{288}$$) or do I need to work out the energy in Joules using equations?

TFM

2. May 3, 2008

That sounds right to me. The energy in joules would be very difficult to calculate and would be very large because the mass of the surface of the earth would be huge.

3. May 3, 2008

### kamerling

I think you are supposed to work out how much extra the earth radiates because it's at 288 K and not at 255K, and divide this by the energy that the earth gets from the sun.

You can do everything in units of W/m^2.
The heat radiation from the earth for each square meter is Ct^4. (C an unknown constant)
The intensity of sunlight is 1366 W/m^2 but you can't use that because the sun isn't always directly overhead and it isn't always daytime and there are clouds. You can use an unknown constant for the amount of sunlight received per m^2. (lets call this E)

you know that C(255^4) = E

4. May 4, 2008

### TFM

Would this formula be of any use:

$$(1-A)*L_{sun}\frac{\pi r^2}{4\pi D^2}$$

Where A is the Albedo, r the radius of the Planet, D the distance from the planet to the Sun, $$L_{Sun}$$ is the Sun's Luminoscity.

?

TFM

5. May 5, 2008

### TFM

Is the Equation in the post above required for this question?

TFM

6. May 5, 2008

### kamerling

I don't think so. You can use the incoming energy of the sun as an unknown constant. there are some simplifications made here, because the albedo does depend on having an atmosphere, as does convection, evaporation etc. Then there's the day-night and seasonal cycles. There is years of work here.

But I think the only thing you need here is the formula for black body radiation.

7. May 6, 2008

### TFM

Does this look right then?

$$I = \sigma T^4$$

where sigma is the Stefan Boltzmann Constant, 5.67*10^-8

$$I_{from sun} = 5.67*10^{-8) * 255^4$$ gives 240

$$I_{from greenhouse} = 5.67*10^{-8) * 288^4$$ gives 390

Would you then:

390-240 = 150

and thus make thge answer: 150/240?

TFM

8. May 6, 2008

### kamerling

I think that would be it