Astronomy - Ch. 9.1: Earth's Atmosphere (5 of 61) What is the Atmospheric Energy Balance?

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In this video I will explain how the energy balance works in Earth's atmosphere. It involves some of the Sun's energy not reaching Earth because the Sun's does not hit Earth perpendicularly. Some of Sun's energy are reflected from the atmosphere and surface, and some are radiated from the atmosphere and surface. Some are absorbed directed by the atmosphere, and others are absorbed by the atmosphere after radiating off the surface of Earth.
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[@madayanmuttal886]

Fabulous explanation. No one explanation with this kind of clarity!

[@MichelvanBiezen]

Thank you. Yes we tried to put a series of videos together that will help us to understand our atmosphere better. There is a lot of misinformation out there.

[@TheShepherdw]

BASED on your fantastic presentation I can NOW work out the following from the IPCC diagram Figure 7.2 | Schematic representation of the global mean energy budget of the Earth (upper panel) IPCC Sixth Assessment Report * Energy Reflection and Absorption: Approximately 29-30% of the incoming solar energy is reflected back into space, with 23% (100/340) reflected by clouds and 7% (25/340) by surfaces like ice and snow. About 23% (80/340) is absorbed directly by the atmosphere. * Surface Warming and Energy Redistribution: Around 48% (160/340) of the incoming energy reaches the Earth's surface, warming it. This warming leads to various processes: * About 5% (21/340) of the energy is transferred back to the atmosphere through conduction. * Around 25% (82/340) is transferred via evaporation, significantly contributing to atmospheric energy. * Atmospheric Absorption: In total, approximately 28-29% of the energy in the atmosphere is from absorption (both from the sun to the surface (80) and from the surface back to space (25), and about 30% (82+21) is from convection currents and evaporation. This adds up to about 59% (208/340) of the total energy input being absorbed by the atmosphere. HOWEVER WHEN COMES TO THIS * Radiation and Greenhouse Effect: Of the energy that warms the Earth's surface, 12% (assume 40) is radiated back into space without being hindered by atmospheric molecules. I CANT WORK OUT THE % OF INCOMING ENERGY AND RADIATED ENERGY THAT ARE ABSORBED BY GREENHOUSE GASES IN THE ATMOSPHERE FROM THE IPCC DIAGRAM Of the energy that warms the Earth's surface after Conduction and Evaporation = 57Wm2 (160- 21-82) What happens to it? It looks like a recycling of heat by greenhouse gases GH Downwood back radiation is 342 W/m2 Up Radiation is 398 W/m2 Out going to space 239 W/m2 That doesn’t make sense !!! Hope you can help

[@MichelvanBiezen]

I have been studying this for the last 15 years and I found that trying to model the atmosphere like that will not give us the answers we are looking for. There are too many variable and too many assumptions we have to make which cannot be accurately determined. My expertise is in data analysis, so I have been studying the temperature data from the 7000 weather stations around the world and that provides a very different picture compared to the narrative that is being pushed. As a scientist I tried to make sense out of it all and I am in the process of writing a book

[@donaldklemen7532]

Enjoying your lectures! 1) Your area ratio, does that account for reflection and lens effects vs various frequencies? 2) The solar wind shock wave wraps around the earth. This also should affect sun energy transmitted. The ionized gas of the shock has a higher temperature due to KE of the particles. it may also have some heat on the other side of the earth. Some sunlight passes, but Ions probably provide some protection. I would assume the ionized shock wave provides some heat radiation. 3) In later lectures, what about earth conduction, and, at night, the cooling effect. Once these are included, what would be the Co2 sensitivity for a rotation of the earth at apahilion and perihelion? Thanks Don! Indy!

[@theultimatereductionist7592]

I hate to criticize your fantastic lectures. However, whenever showing % of anything, you really should define a variable, even just an X(1), X(2), etc. where X(k) is an energy flux, watts per square meter, and then say what you are taking (48%, 5%, etc) of WHICH variable: i.e. X(1) = energy flux from son, X(2)=average energy flux over earth accounting for angle that earth surface makes with the sun so X(2)= some % of X(1). I mean, are 59% (radiates to space), 5% (goes to atmo), 25% (evaporation), fractions of the 340 W/m^2?

[@richardeastman9846]

Less light at the poles, yes, but what about less energy at dusk, between night and day? And what about halving the answer to take the night side into account?

[@MichelvanBiezen]

When you write "what about", what is the question you are trying to ask? ("What about" doesn't indicate the question)

[@richardeastman9846]

@@MichelvanBiezen When you were explaining that the higher latitudes receive less solar energy than the equatorial regions you drew a hemisphere and computed the total sunlight received by that hemisphere, my question was does this operation adjust total solar energy for obliqueness as the poles or approached? or is all the energy lessening due the obliqueness in all directions taken into account in you (less energy reaching the equator at dusk as well as less as the poles are reached? In other words, at any moment is the sunlight is as attenuated by angle at the dust regions at lower latitudes as attenuated as moving closer to the poles and did your computation make the complete adjustment for all directions? I asked because you mentioned the diminished enery towards the poles, but not towards dusk regions.

[@fractalnomics]

I think this model is completely wrong in the same way 'the geocentric model' was wrong. I think all the gases absorb and emit and there are no special ones. If this is so, what would the model look like? How would we draw it to show it balances.

[@MichelvanBiezen]

Your assertion does not match the the basic science that is well known. (I know the "science' has a bloodied nose lately, because of the "political" hijacking of "science", ) but the absorption capability of greenhouse gases is well established.

[@tsehayenegash8394]

Professor, what is the the kinetic and potential energy per unit mass i(what is the the minimum and maximum value of potential and kinetic energy per unit mass ) in the troposphere and stratosphere? thank you professor

[@MichelvanBiezen]

since the atmosphere is made up of gases we tend to consider the thermal energy per mol. U = nCv T where n = # of mols Cv = molar heat capacity at constant volume T = temperature in Kelvin

[@tsehayenegash8394]

@@MichelvanBiezen I appreciate your help professor, I get an idea to do my research.

[@DavidSiegelVision]

I'm still uncomfortable with 5-6 percent "absorbed" but apparently you're going to cover that later. However, you have zero percent going directly into and heating the oceans, and that misses 50 percent of absorbed energy. What's happening? Does sunlight not go into the oceans to heat the water?

[@MichelvanBiezen]

Actually, a big part of the energy reaching the surface goes into evaporation which is the cause of the hydrological cycle. The rest goes into heating the surface (which includes the ocean water) and off course at night that heat is released back into the atmosphere.

[@MichelvanBiezen]

It turns out that a large portion of the heat reaching the surface causes the oceans to evaporate (as is shown in this video) which is the cause of the hydrological cycle which influence the climate tremendously. The remainder is either reflected, of absorbed, both by the land and by the oceans. At night this energy is released at a greater rate than it is absorbed and the ocean temperature drops.

[@DavidSiegelVision]

@@MichelvanBiezen I think you're missing direct ocean heating by SW radiation. I believe half of all radiation absorbed by the surface goes into the oceans and gets transported, some laterally and mostly poleward.

[@DavidSiegelVision]

@@MichelvanBiezen You seem to believe that the tropics are always in radiative balance, but that's clearly not true. The tropics have a surplus of net energy while north of 45 degrees there is a net deficit. Heat goes into the oceans. Not much comes out overnight. Most of it goes into currents to be transported northward. this is a huge amount of energy, on the order of HALF of all energy absorbed by the earth in the tropics goes into the oceans for poleward heat transport. This, I believe, is what you're missing. Something has to power the AMOC and the El Nino, and it's poleward heat transport of temperature gradients in the water.

[@hansvetter8653]

Great presentation! Thank you Prof. van Biezen. May I ask you two questions? ... 1) if the globe is seen as a 'black body radiator' with its average surface temperature of 288°K (59°F) ... what about its total radiated infrared energy of 390.1 W/m2 ? ... I do not find this in your presented values ... 2) the climate apocalyptics claim the existence of a so called "atmospherical IR back-radiation" (it's in the IPCC-report with a fantastic value of 324 W/m2 ... ?!? ... holy moly ... give me a break ... ;-) But I would assume that H2O & CO2 is absorbing and immediately re-emitting infrared photons, which will not reach the surface of the earth, because these ir-photons will be re-absorped & re-emitted from H2O & CO2 molecules multiple times on its paths ... and therefore all these ir-photons get scattered in all directions ... can you elaborate a bit on this so called "back-radiation" ? ...

[@MichelvanBiezen]

Studying the data record of the thousands of weather stations around the world, reveals that the empirical data does not support the model that you reference in your statement # 2. Your description of the absorption and re-emittance of the photons offers a good picture of what is going on in the atmosphere.

[@hansvetter8653]

@@MichelvanBiezen ... thank you for your answer Dr. van Biezen. You also use the term "black body" in your talks, but I always wondered if that describes reality in the sense of the Stefan-Boltzmann-Law well enough. So your view is that the globe emitts in total 48% of 340 W/m2, which is 163 W/m2 by conduction & radiation through its surface into the atmoshere. Is in that scenario the Stefan-Boltzmann-Law still applicable? ... would'nt it say that with a value of 163 W/m2 energy emittance the temperature of the earths surface, according to the Stefan-Boltzmann-Law, "must" be 231.5°K (-43°F) ? ...

[@enderwiggin1113]

@@MichelvanBiezen "Studying the data record of the thousands of weather stations around the world, reveals that the empirical data does not support the model that you reference in your statement # 2." Hey, Prof. Biezen! I'm not entirely sure what you meant by this - does this refer to the 342 W/m^2 (which after all is well supported by measurements for decades scienceofdoom.com/2010/07/17/the-amazing-case-of-back-radiation/ ) or does it refer to the silly claim of Hans Vetter that the photons won't reach te surface?

[@muhammadhamzajaved9297]

5% is absorbed by the atmosphere via convection or conduction?

[@MichelvanBiezen]

5% via convection, molecules picking up energy from the surface on contact and then rising up in the atmosphere.

[@andyl147]

But surely the input of the sun is spread out over a hemisphere, not the full sphere and so the ratio of 1/4 is incorrect. I'm confused.

[@charlesbrightman4237]

It would seem that as the atmosphere, ionosphere, magnetosphere, etc, vary over time, that would also have an effect on the energy interactions concerning the Earth.

[@geraldotieno6010]

Wow!!! Great

[@MichelvanBiezen]

Thanks!!

[@patricksilva2345]

Dear Biezen, when you presented the solar constant in the top of atmosphere, which is 1361 W/m2 and then present the average solar intensity at the surface of the earth about 340 W/m2 saying that, this value is less just because the curved surface of the earth which influence the angle of incidence seams a little confused. If we are talking about average it would be the same because even the atmosphere border (if exists) is curved. I think the decrease in its intensity is due to its path in the way in atmosphere, which causes reflection (diffuse) and absorption. Please, tell me if I am right! Thank you!

[@MichelvanBiezen]

Since the Earth is curved the energy deposited on the surface per square meter is less at the poles compared to the equatorial regions. That is why it is colder at the poles.

[@patricksilva2345]

@@MichelvanBiezen Thank you! But, I am just asking if the atmosphere is not the main reason of the decreasing of sun light intensity, instead of only the shape of the earth?

[@tsehayenegash8394]

Hello professor. I have a question. How stratospheric Quasi biennial oscillation (QBO) would affect temperature in the troposphere? please help me

[@MichelvanBiezen]

The effect of the QBD on the climate is very complicated and difficult to tie to specifi temperature changes in the troposphere. What can be said is that the QBD is not as significant as the jet streams over the polar and temperate areas and the prevailing (or lack of) high pressure systems in the polar regions that push cold are down (or not), and the el nino and la nina phenomena over the Pacific Ocean as well as other decadal osciallation. I have not seen any impact significant correlation to the temperature records in the weather station data from the QBD.

[@tsehayenegash8394]

@@MichelvanBiezen Thank you professor

[@derrickbecker9856]

How do these percentages change by latitude?

[@banushasathi3526]

Thank you so much

[@charlesbrightman4237]

The amount of energy from the Sun reaching and leaving this Earth seems would also have to take into account the ionosphere and magnetosphere, which would also affect electromagnetic radiation energy one way or another. And just as light 'bends' as it changes mediums and angular deflections (rainbow effect), the angle of incidence of the energy from the Sun on the Earth and it's surrounding atmosphere and energy barriers would also seem have to be taken into account.

[@abeywarc]

I have a question, maybe its a very dumb one... We are discussing radiation that approaches the earth's surface through various layers of the atmosphere, which we consider as being one of the major sources of energy received by the earth as a system. If we get the planet earth as our system and its boundary would be earth and its atmosphere, then for the energy balance the incoming radiation energy and the outgoing radiation energy must be balanced out. Once equilibrium is reached, the earth reaches its steady-state temperature. What about all the absorption of radiation by the earth's terrestrial surfaces, plant leaves for photosynthesis, oceanic uptake of energy as radiation is absorbed by oceans, etc...? How is that absorption factor in to the energy balance? If we consider the atmosphere as the system; the boundaries are the earth's surface and the mesosphere, then: In - Out - Absorption = 0, right?

[@MichelvanBiezen]

Some energy is locked in plants, trees, etc., and then returned when they die and decompose. It is a small portion of the total energy equation.

[@abeywarc]

Michel van Biezen absorption of radiation heat by the oceans also return back and therefore in equilibrium? Thanks for the quick reply 👍🏼👍🏼👍🏼

[@MichelvanBiezen]

Oceans are good heat sinks as water has a very large heat capacity. But over time, things tend to balance out. The Earth has gone through HUGE temperature swings in the past, FAR GREATER than what we are experiencing today many, many times.

[@charlesbrightman4237]

Since the Earth is NOT a perfect sphere, you induce errors into the equations when you assume such.

[@hansvetter8653]

Does'nt earth radiates more energy towards the universe compared to the energy coming from the sun down to earth? Solar constant is 1361 W/m2 ... divided by 4 ... makes around 340 W/m2 energy coming from the sun in average to the whole surface of the earth ... But according to the Stefan-Boltzmann-Law the earth radiates at an average temperature of 59°F (15°C, 288K) around 390 W/m2 into space ... which is 50 W/m2 more than the incoming energy from the sun ... If you take into account that only 48% of the energy coming from the sun down to earth and causes warming ... which is just around 162 W/m2 ... the difference to the energy radiated by the surface of the earth towards space growths up to 228 W/m2 ... What have I missed in this simple calculation? ...

[@MichelvanBiezen]

If the Earth simply heated and cooled via the Stefan Boltzmann law, the Earth would continue to cool until incoming radiation and outgoing radiation was equal and that means the Earth would be about - 18 C (- 2 F) on average. It is much more complicated than that.

[@hansvetter8653]

@@MichelvanBiezen ... ok. Actually I've found the value of 390 W/m2 in the IPCC-report. So I assumed they see the earth as a block body radiator. That's why I've asked. After watching your videos I had doubt if the earth can be seen as an black body radiant ... and according to your answer it is not. So thanks. How can I learn more about it? ...

[@Kintabl]

Why are you applying incoming solar radiation that hits only half of Earth's surface at the time, to a whole surface of the Earth? By doing that, you reduce solar power to 25% of what should be.

[@MichelvanBiezen]

Only the side exposed to the sunlight receives sunlight (half the Earth's surface at any point in time). While the entire surface of the Earth radiates heat at all times.

[@Kintabl]

@@MichelvanBiezen Yes, but i'm talking about incoming sunlight. Half of the Earth's surface get much more energy from sunlight than other half that receive none at any poit of time. You guys should correctly distributed energy from sunlight to entire Earth's surface. 50% of surface gets entire energy from sunlight, 50% doesn't get nothing. In that 50% of the Earth's surface that get sunlight, equator get more energy than areas at high latitude. But no. You guys just put sunlight to entire Earth's surface at all time, and by doing this, you get that the sun can only warm earth's surface to -18°C. Sorry, but this is not science.

[@MichelvanBiezen]

That is accounted for by using the cross sectional area rather than the surface area. Check the equations used and you'll see that they correctly represent the amount of energy the Earth receives from the Sun.

[@Kintabl]

@@MichelvanBiezen And what is that at 1:28 ? Why are you dividing 'solar constant' by 4. It's clearly that you apply Sun's energy to entire Earth's surface. That's why you get 340W/m^2. It would be more correct to apply 1361W/m^2 to half Earth's surface, and you would get 680W/m^2 which is 30°C. But areas at equator receive full solar energy at 1361W/m^2, which is 86°C. Sun can clearly heat some areas of Earth's surface enough to create climate. But no, you guys rather spend all this time to figure it out how CO2 warms the planet from -18°C to 15°C.

[@MichelvanBiezen]

That is not what the videos in this playlist are indicating. Watch the remaining videos for a full understanding.

[@evelynwaperi7369]

Thank you so much

[@MichelvanBiezen]

You're most welcome