Question

I have an equation I found on the internet but I would like a little help with how to find this answer on my own, err with OpenStudy. {The energy reaching Earth from the Sun at the top of the atmosphere is 1.36 x 103 W/m2, called the solar constant. Assuming that Earth radiates like a blackbody, what do you conclude is the equilibrium temperature of Earth?}

Answer 1

\[P \lambda = 2piHc^2/\lambda(e^(hc/lambdaKT)-1)\]

Answer 2

At equilibrium, the energy being radiated by the earth is equal to the energy coming in from the sun. From your knowledge of blackbody radiation, the power radiated by a blackbody depends on its temperature, right?

Answer 3

Some thing like
\[\frac{\text{Power}}{\text{Area}} = \sigma T^4\]
right?

Answer 4

yes, I understand that if the earth is at equilibrium and we treat it as a blackbody power radiated is the same as power absorbed.

I dont understand how to arrive at your equation, I'll look more into it. Thanks!

Answer 5

It's called the Stefan-Boltzmann Law. If you take the equation that you provided and integrate it over all possible frequencies from 0 to infinity then you'll arrive at the SB law.

Answer 6

Well, you gave it in terms of wavelengths, so if you can integrate it over all wavelengths, that'll do the trick.

Answer 7

Do you know how to do such an integral?

Answer 8

Hmmm, I wasn't going to integrate it over the entire spectrum, I was going to use the average wavelength 550nm. But your right, I should integrate it. I'll look at my notes for the SB law. Thank you so much. Its so tough to find people IRL to talk physics and problem solving with.

Answer 9

SB law looks like it is the way our instructor wanted us to do this problem. Thank you again

Answer 10

I'm not getting the correct answer :(
i'm typing ((1.36e3)/(5.6705e-8))^(1/4) =393K
the teacher says the answer is 278K

Answer 11

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Answer 12

Equilibrium condition: Energy in = Energy out

Energy in:
\[ (1.36\cdot 10^3)(\pi \cdot R^2) \]
Energy Out:
\[(\sigma \cdot T^4)(4 \pi R^2) \]
so
\[T = \left(\frac{1.36\cdot 10^3}{4\sigma} \right)^{1/4} \]

Answer 13

The 4 is because the energy absorbed from the sun is over a disc of area pi R^2, while the energy being emitted is over the surface area of the earth, 4 pi R^2.

Answer 14

That was the part I was missing!! dang! The shape of the earth when its absorbing energy is different from the shape when radiating. Thanks again Jemurray :D I'm really glad I found this community