- #1

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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

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- Thread starter jamesfirst
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- #1

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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

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Not sure I understand your version of the equation. Please explain the "Te4" bit.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

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

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T^4

Sigma = stefan-boltzmann constant ??

Sigma = stefan-boltzmann constant ??

- #4

DEvens

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- 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

pbuk

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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

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http://c21.phas.ubc.ca/article/simple-earth-climate-model-additional-concept-explanations

is this the right equation ???

- #7

pbuk

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Have you tried the equation in your first post?

http://c21.phas.ubc.ca/article/simple-earth-climate-model-additional-concept-explanations

is this the right equation ???

- #8

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yeah I get like 394 kelvin

- #9

pbuk

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- #10

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i dont o.0....

im confused... can you explain it please.

im confused... can you explain it please.

- #11

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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.yeah I get like 394 kelvin

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

pbuk

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- #13

pbuk

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You are overthinking this.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.

- #14

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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.

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

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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

pbuk

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

- #17

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Exactly. From https://en.wikipedia.org/wiki/Solar_constant: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

"The

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.)You are overthinking this.

- #18

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

- #19

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

- #20

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area of the earth ?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?

- #21

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Look at it from the perspective of rays coming from the Sun. What does Earth look like?area of the earth ?

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a 2d circleLook at it from the perspective of rays coming from the Sun. What does Earth look like?

- #23

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Right. So given the flux density, what total flux hits the Earth?a 2d circle

- #24

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1367 * Pi (radius of earth) ^2Right. So given the flux density, what total flux hits the Earth?

- #25

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Right.1367 * Pi (radius of earth) ^2

The question requires you to assume the whole Earth's surface is at the same temperature. If that temperature is T and the radius is R, what's the total flux being emitted?

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