LURCH said:
You have an Astrobiology course?! I'm a bit jealous.
It's a very cool class, taught by Debra Fisher, one of the key astronomers in the detection of exosolar planets. And we get lots of great guest lecturers too. Nick Platts, Frank Drake, Geoff Marcy, and others.
Here's my attempt to throw some numbers at the temp problem.
Earth’s average temperature: 287K.
Therefore, Earth radiates:
<br />
\begin{array}{l}<br />
E = \sigma T^4 \\ <br />
E = \left( {5.7 \times 10^{ - 8} W/\left( {m^2 } \right)} \right) \times \left( {287} \right)^{} = 387\,W/m^2 \\ <br />
\end{array}<br />
On average, the Earth intercepts its 2-D cross section of the Sun’s flux:
Earth's cross section:
<br />
\pi r^2 = \pi \left( {6378000m} \right)^2 = 1.28 \times 10^{14} m^2 <br />
Solar flux at Earth's distance:
<br />
\frac{{3.8 \times 10^{26} W}}{{4\pi \left( {149580000000m} \right)^2 }} = 1352\,W/m^2 <br />
Earth receives:
<br />
1.28 \times 10^{14} m^2 \, \times 1352\,W/m^2 = 1.7 \times 10^{17} W<br />
Earth's surface area:
<br />
4\pi r^2 = 4\pi \left( {6378000m} \right)^2 = 5.1 \times 10^{14} m^2 <br />
So each square meter of Earth's surface receives an average of
<br />
\frac{{1.7 \times 10^{17} W}}{{5.1 \times 10^{14} m^2 }} = 338\,W/m^2 <br />
Earth’s albedo is 0.367, so therefore each square meter absorbs:
<br />
\left( {338\,W/m^2 } \right) \times \left( {1 - 0.367} \right) = 214\,W/m^2 <br />
So Earth is absorbing 214 watts per square meter, but radiating 387 watts per square meter?
My logic is probably flawed. The greehouse effect probably accounts for the difference, not internal heat escaping. Any thoughts?
