# Homework Help: How to calculate surface temperature of earth?

1. Sep 10, 2015

### jamesfirst

Assumption: Earth is a perfect blackbody. This means that the emissivity (ε) is 1.

Earth’s radius = 6371 km

Incident solar radiation = 1367 W/m2

Temperature in space = 0 K

I'm given the above data. I tried using the E= σεTe4 but that doesnt include the radius of the Earth :S

Im really confused... help

2. Sep 10, 2015

### haruspex

Not sure I understand your version of the equation. Please explain the "Te4" bit.
Your equation requires a value for sigma. Please define that sigma and how you might calculate it from the given information.
There are some simplifications you will need to make, namely, that the incoming radiation somehow gets spread uniformly over the Earth, so that the whole Earth is at the same temperature.

3. Sep 10, 2015

### jamesfirst

T^4

Sigma = stefan-boltzmann constant ??

4. Sep 10, 2015

### DEvens

The value you will calculate using Stefan-Boltzmann will be pretty wildly wrong even if you get the right $\sigma$ and $\epsilon$. Several reasons such a model is pretty naïve.
- As you mentioned, it does not include curvature of the Earth. The W/m^2 has to be adjusted.
- The Earth rotates. So the W/m^2 has to be adjusted to account for that as well.
- Any possible $\sigma$ and $\epsilon$ will be some kind of average of effective value. Different ground cover, ocean, cloud, snow, trees, desert, etc. Often time-varying rather drastically.
- It does not take into account such things as the "green house effect." Thermal radiation gets, to some extent, reflected off components of the atmosphere. This includes CO2, but also water vapor, clouds, etc. etc. IIRC, the net of this is something approximately 35C. (From memory. I might be wildly wrong on that.)
- The atmosphere does a lot of things. It holds heat, moves it around, convects it.

But naively: You take the power that comes in. You divide by $\sigma$ and $\epsilon$. You take some factors to account for only half the Earth being in sunlight, and the curvature (roughly a factor of 4, but you should work that out). Then you take the fourth root to get a temperature. If I recall, you get something round about -20C or there about.

5. Sep 10, 2015

### MrAnchovy

It is common to include superfluous information in problems to make people think: it seems to have worked.

If you have asked to find the surface temperature of the Earth it is safe to assume it is constant: what does that imply?

6. Sep 10, 2015

### jamesfirst

7. Sep 10, 2015

### MrAnchovy

8. Sep 10, 2015

### jamesfirst

yeah I get like 394 kelvin

9. Sep 10, 2015

### MrAnchovy

I understand how you got that answer, but if you explain the steps you might realise what you have not taken into account.

10. Sep 10, 2015

### jamesfirst

i dont o.0....
im confused... can you explain it please.

11. Sep 10, 2015

### haruspex

The meaning of the given irradiation needs to be clarified. Is it an average over the whole Earth's surface, over 24 hours, or, the local value when the sun is overhead? Or something else? From what I see on the net, it is the value of the 'solar constant'. That makes it the overhead sun value, at top of atmosphere.
But as I wrote, you then need to suppose the incoming power gets spread uniformly over the Earth's surface, so the average power per unit area will be a lot less.

12. Sep 10, 2015

### MrAnchovy

Your answer is too high, so you seem to be assuming that the area over which the 1367 W/m2 is relevant is larger than it is. Why could this be? The radius of the Earth is a hint (you can use it if you want, but you don't actually need it).

13. Sep 10, 2015

### MrAnchovy

You are overthinking this.

14. Sep 10, 2015

### jamesfirst

This is the question

A perfect blackbody is an object that perfectly absorbs and emits radiation across all wavelengths. A very simple 'climate' model is to assume that the Earth is a perfect blackbody and that the temperature is uniform over the surface. Using this model and 6371km for the Earth's radius, calculate the surface temperature of the Earth in Celcius assuming that the incident solar radiation is 1367 W/m2 (a possible value for the varying solar constant) and the temperature of space is 0K

15. Sep 10, 2015

### jamesfirst

what do you mean ... ==;;
So without using the Earth's radius. I just use σεT^4 and get 394 kelvin ??
can you stop asking rhetorical questions and just tell me... really stressed out here

16. Sep 10, 2015

### MrAnchovy

They are not rhetorical, I want you to answer them: that's how it works here.

I'll try once more: from what area does the Earth radiate heat?

17. Sep 10, 2015

### haruspex

Exactly. From https://en.wikipedia.org/wiki/Solar_constant:
"The solar constant, a measure of flux density, is the conventional name for the mean solar electromagnetic radiation (the solar irradiance) per unit area that would be incident on a plane perpendicular to the rays, [at our distance from the Sun]"
So, first thing to calculate is the total flux that lands on the Earth. Don't worry about the exact value (it won't matter), just take the radius of the Earth to be R. In terms of R, what is the total flux striking the Earth?
Next, what is the flux per unit area of the Earth's surface?
Not at all. This is precisely how this problem needs to be approached. (Except, I didn't need to mention "at top of atmosphere", since we are assuming none.)

18. Sep 10, 2015

### jamesfirst

how do you calculate this ? do you use I=Isun (R/d)^2 ?

19. Sep 10, 2015

### haruspex

I assume your 'd' is distance from the Sun. That's already taken into account in the number you are given - it's the flux per unit area at our distance from the Sun. What area do you need to multiply that by to get the total flux striking the Earth?

20. Sep 10, 2015

### jamesfirst

area of the earth ?