Thank you for creating this video! I do have a question, though. For c, Why do we say that I(in) = I(out) when the temperature of the Earth's surface is increasing?
It is not that the T is increasing but rather that a different equilibrium temperature is reached. When temperature is constant (in equilibrium), we can say Iin=Iout.
Also, what is the distinction between power radiated by the atmosphere, vs. solar power absorbed at the surface of the Earth? I would have assumed that the values would have been the same: all power that isn't reflected off the atmosphere, S/4 * (1 - albedo) = power radiated by the atmosphere, and = the power absorbed by the Earth since it is effectively a black body, however the Stefan Boltzman Law seems to contradict my line of thought. Thank you!
Ordinarily we say that the solar radiation is either absorbed or reflected by the atmosphere, but for this model, we are saying that the solar radiation is either reflected or transmitted through the atmosphere, and then it is absorbed by the Earth and re-emitted in the infra-red which can then be absorbed by greenhouse gases in the atmosphere.
Ahh that makes a lot of sense. So the first part of the question (a) refers strictly to SOLAR power per unit area absorbed by the Earth, whereas the total power per unit area absorbed by the Earth = solar + atmospheric radiation, because we treat the atmosphere as a grey body radiator as well, which absorbs the infrared rays re-emitted by the Earth and re-radiates them in all directions. Thank u sir!
At around 23:50 I am just a little confused: for b (ii) the power / unit area absorbed by the Earth's surface is found to be 402 W / m^2, but the solar constant coming through the atmosphere is just 344 W / m^2 - how can the power per unit area absorbed be greater than the the power per unit area entering the atmosphere? It seems to me as if there is a discrepancy in the method utilized to solve for power per unit area absorbed by the Earth's surface: in part a, the radiation emitted by the atmosphere is not added, and in part b it is. Which is the correct way to answer these questions? Any help would be greatly appreciated. Thanks! - Max
Not all the 344 makes it through some is reflected. 248 from the sun goes straight through the atmosphere. The remaining radiation comes from the atmosphere itself, treating it as a grey body radiator. The model is a simplification and assumes that all the energy in the atmosphere is obtained from the power radiated by the Earth, and none of it comes directly from the solar radiation.
At 19:38, we're looking at the radiation (absorbed by) and then emitted from Earth and then the radiation (absorbed by) and emitted from the cloud BACK to Earth. Why aren't we considering the radiation reflected from Earth to the clouds and the radiation reflected back from the clouds?
There are different ways to look at this. If it reflects from the Earth, it will remain at that higher frequency and will, therefore, travel back through the clouds without being absorbed. The fraction that is reflected from Earth is very small as the Earth is nearly a perfect black body radiator and you are asked to assume this in the question. Good question.
No that would be at the surface...an average temp. of about 15 degrees (seems reasonable right?). We want the surface temperature but our original model gave -15 degrees, this model makes a correction.
Just to clarify: the area (A) in P=eσAT^4 is the area of the radiating body. While the area (A) in I = Power/A is the area of where the light/radiation is incident. Is that correct?
Yes, surface area of the radiating body in the first equation, and area through which the radiation passes or is incident. If we consider radiation from say the earth near the surface, where it does not have a chance to spread out, we can effectively divide out the surface area in the first equation to give us an expression for intensity.
The incoming power is from the sun, and the outgoing power is from the Earth. Both are essentially blackbodies, so the hotter one emits wavelength of a shorter frequency.
It is just in the wording of the question. In part a), we determined the intensity radiated by the atmosphere, and the intensity of the solar radiation striking the earth. In part b) we recalculate the intensity radiated by the atmosphere, the solar intensity stays the same, and we work out the Earth's equilibrium temperature by saying that all the incoming radiation (solar + atmosphere) must balance all outgoing radiation from the Earth.
This might be a stupid question but hopefully Sir you can help. I understand modelling the incident radiation over the disc of the earth (pi r^2) but why isn't it half the surface area of the Earth (4 pi r^2)/2? Thanks.
At 19:02, you mention how we treat the atmosphere like a "black body radiator", but aren't we modelling with the assumption that the atmosphere is a "grey body radiator"? If it were a black body radiator, wouldn't it simply emit all the radiation back down to the Earth?
Some radiation passes through the cloud and the rest in absorbed and then re-emitted in agreement with the cloud being a grey body radiator but the earth is assumed to be black.
The nice thing about using intensities is that you are assuming radiation is all directions, but we are finding the temperature at a location where the intensity in equals intensity out.
Thank you so much for this video! I would like to ask in part b., is there the assumption that the area of the atmosphere is the same as the area of the Earth's surface? Otherwise this could not be simplified from a power equation to an intensity equation, is that correct?
Areas need to be the same, but if you think about it that will almost always be the case because the area of transfer must be the common for incoming and outgoing radiation.
Perhaps you are skipping around the video because earlier I predict using the climate model that does not include the GHE that the temp. is -15, and this turns out to equal the temp. at the "midway" point of the atmosphere.
you're a blessing to society
Thank you.
Thank you for creating this video! I do have a question, though.
For c, Why do we say that I(in) = I(out) when the temperature of the Earth's surface is increasing?
It is not that the T is increasing but rather that a different equilibrium temperature is reached. When temperature is constant (in equilibrium), we can say Iin=Iout.
Also, what is the distinction between power radiated by the atmosphere, vs. solar power absorbed at the surface of the Earth? I would have assumed that the values would have been the same: all power that isn't reflected off the atmosphere, S/4 * (1 - albedo) = power radiated by the atmosphere, and = the power absorbed by the Earth since it is effectively a black body, however the Stefan Boltzman Law seems to contradict my line of thought. Thank you!
Ordinarily we say that the solar radiation is either absorbed or reflected by the atmosphere, but for this model, we are saying that the solar radiation is either reflected or transmitted through the atmosphere, and then it is absorbed by the Earth and re-emitted in the infra-red which can then be absorbed by greenhouse gases in the atmosphere.
Ahh that makes a lot of sense. So the first part of the question (a) refers strictly to SOLAR power per unit area absorbed by the Earth, whereas the total power per unit area absorbed by the Earth = solar + atmospheric radiation, because we treat the atmosphere as a grey body radiator as well, which absorbs the infrared rays re-emitted by the Earth and re-radiates them in all directions. Thank u sir!
At around 23:50 I am just a little confused: for b (ii) the power / unit area absorbed by the Earth's surface is found to be 402 W / m^2, but the solar constant coming through the atmosphere is just 344 W / m^2 - how can the power per unit area absorbed be greater than the the power per unit area entering the atmosphere? It seems to me as if there is a discrepancy in the method utilized to solve for power per unit area absorbed by the Earth's surface: in part a, the radiation emitted by the atmosphere is not added, and in part b it is. Which is the correct way to answer these questions? Any help would be greatly appreciated. Thanks! - Max
Not all the 344 makes it through some is reflected. 248 from the sun goes straight through the atmosphere. The remaining radiation comes from the atmosphere itself, treating it as a grey body radiator. The model is a simplification and assumes that all the energy in the atmosphere is obtained from the power radiated by the Earth, and none of it comes directly from the solar radiation.
Thanks for helping me understand! #GOAT
At 19:38, we're looking at the radiation (absorbed by) and then emitted from Earth and then the radiation (absorbed by) and emitted from the cloud BACK to Earth. Why aren't we considering the radiation reflected from Earth to the clouds and the radiation reflected back from the clouds?
There are different ways to look at this. If it reflects from the Earth, it will remain at that higher frequency and will, therefore, travel back through the clouds without being absorbed. The fraction that is reflected from Earth is very small as the Earth is nearly a perfect black body radiator and you are asked to assume this in the question. Good question.
@@donerphysics I see! Thank you so very much for the clarification!
At 17:03, this new surface temperature, is that also technically at a height of 5km above Earth's ground (or into the atmosphere)?
No that would be at the surface...an average temp. of about 15 degrees (seems reasonable right?). We want the surface temperature but our original model gave -15 degrees, this model makes a correction.
Just to clarify: the area (A) in P=eσAT^4 is the area of the radiating body. While the area (A) in I = Power/A is the area of where the light/radiation is incident. Is that correct?
Yes, surface area of the radiating body in the first equation, and area through which the radiation passes or is incident. If we consider radiation from say the earth near the surface, where it does not have a chance to spread out, we can effectively divide out the surface area in the first equation to give us an expression for intensity.
Why is the output power (P emitted) at a different wavelength to the input power (P absorbed)?
When in the video are you referring to?
@@donerphysics 01:10 thanks!
The incoming power is from the sun, and the outgoing power is from the Earth. Both are essentially blackbodies, so the hotter one emits wavelength of a shorter frequency.
How comes in part b we accounted for radiation emitted by atmosphere and it part a we didn't?
It is just in the wording of the question. In part a), we determined the intensity radiated by the atmosphere, and the intensity of the solar radiation striking the earth. In part b) we recalculate the intensity radiated by the atmosphere, the solar intensity stays the same, and we work out the Earth's equilibrium temperature by saying that all the incoming radiation (solar + atmosphere) must balance all outgoing radiation from the Earth.
Thank you!
This might be a stupid question but hopefully Sir you can help. I understand modelling the incident radiation over the disc of the earth (pi r^2) but why isn't it half the surface area of the Earth (4 pi r^2)/2? Thanks.
No it is a good question. The Earth is rotating so the average intensity takes into account night time.
At 19:02, you mention how we treat the atmosphere like a "black body radiator", but aren't we modelling with the assumption that the atmosphere is a "grey body radiator"? If it were a black body radiator, wouldn't it simply emit all the radiation back down to the Earth?
Some radiation passes through the cloud and the rest in absorbed and then re-emitted in agreement with the cloud being a grey body radiator but the earth is assumed to be black.
In the c ii part of the question you assumed that all energy radiated by the atmosphere goes to the earth. Is this a fair assumption to make?
The nice thing about using intensities is that you are assuming radiation is all directions, but we are finding the temperature at a location where the intensity in equals intensity out.
In the last problem (c), why do we treat the Earth's surface as a blackbody? Is it because no emissivity value is given for it?
The emissivity of the earth's surface is somewhere around 0.96, which is pretty close a perfect blackbody's emissivity of one.
thank you so much for this amazing content!!
You're so welcome!
Thank you so much for this video! I would like to ask in part b., is there the assumption that the area of the atmosphere is the same as the area of the Earth's surface? Otherwise this could not be simplified from a power equation to an intensity equation, is that correct?
Areas need to be the same, but if you think about it that will almost always be the case because the area of transfer must be the common for incoming and outgoing radiation.
Some segments in the video are stamped not adjacent to each other
Could you please tell me the times for these occurrences?
You seemed to sneak in that the -15 degrees was at 5km, no explanation.
Perhaps you are skipping around the video because earlier I predict using the climate model that does not include the GHE that the temp. is -15, and this turns out to equal the temp. at the "midway" point of the atmosphere.
How did you get 288k?
where?
look at 21:46 Earth's temp is 288K
Given in the problem. It is an estimate the Earth's average surface temperature (i.e. 15 Celcius)
Very helpful!!!! Thankyou!!!! Have my exam in a few days!
how did it go?
@@ashwinvishwakarma2531 It went well, thanks for asking.