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## The Earth’s Energy Budget – Part One

In the first post about CO2 I included a separate maths section which showed the energy budget for the earth and also derived how much energy we receive from the sun. A comment today reminded me that I should do a separate article about this topic. I’ve seen lots of comments on other blogs where people trip up over the basic numbers. It’s easy to get confused.

Don’t worry, there won’t be a lot of maths. This is to get you comfortable with some basics.

### Energy from the Sun

It’s quite easy to derive how much energy we expect from the sun, but the good news is that since 1978 there have been satellites measuring it.

The solar “constant” is often written as S, so we’ll keep that convention. I put “constant” in quotes because it’s not really a constant, but that’s how it’s referred to. (And anyway, the changes year to year and decade to decade are very small – a subject for another post, another day).

The first important number, S = 1367 W/m2

Note the units – the amount of energy per second (the Watts) per unit area (the meters squared). By the way, sorry America, the science world moved on. We won’t convert it to ft2..

Just for illustration here’s the satellite measurements over 20 years:

For anyone a little confused, note that different satellites get different absolute measurements, it is the relative measurements that are more accurate.

### Comparing Apples and Oranges? Surface Area vs Area of a Disc

The sun is really long way away from the earth – about 150M km (93M miles). We measure the incoming solar radiation at the top of the atmosphere in W/m2.

So how much total energy can be absorbed into the earth’s climate system from this solar radiation?

Solar radiation received against a "2d disc". From Elementary Climate Physics, Taylor (2005)

Hopefully the answer will become more obvious by looking at the image above. The solar radiation from a long way away strikes the effective 2d area that the earth cuts out.

A 2d area – or a flat disc – has area, A = πr2

Therefore, the total energy received by the earth = Sπr2

[Radius of the earth = 6.37 x 106 m (6,370 km) so Energy per second from the sun = 174,226,942,644,300,000 W  also written as 1.74 x 1017 W]

It’s a really big number, so to make everything easier to visualize, climate scientists generally stay with W/m2, rather than numbers like 1.74 x 1017 W.

Now the real surface area of the earth is actually, Ae= 4πr2 (not πr2)

(Area of earth, Ae= 510M km2, or 5.1×1014m2)

Why isn’t the energy received by the sun = S x 4πr2?

Look back at the graphic – is the sun shining equally on every part of the earth every second, for all 24 hours of the day? It’s not. It’s shining onto one side of the earth. It’s night time for half the world at any given moment.

So think of it like this – the absolute maximum area receiving the sun’s energy on average can only be half of the surface area of the earth – 2πr2 (=4πr2/2)

But that’s not the end of the story. Picture someone where the sun is right down near the horizon. It’s still daytime but obviously that part of the earth is not receiving 1367W/m2 – they are receiving a lot less. In fact, the only spot on earth where someone receives 1367W/m2 is where the sun is directly overhead. So the effective area receiving the solar constant of 1367 W/m2 can’t even be as high as 2πr2.

So if the idea that solar radiation only strikes an effective area of πr2 is still causing you problems, this is the concept that might help you.

The earth radiates out energy in a way that is linked to the surface temperature. In fact it is proportional to the fourth power of absolute temperature.

As we think about the earth radiating out energy, it might be clearer why we labored the point earlier about the area that the sun’s energy was received over.

Take a look at that graphic again. The energy from the sun hits an effective 2d disc with area = πr2.

The earth radiates out energy from its whole surface area = 4πr2.

So to be able to compare “apples and oranges”, when climate scientists talk about energy balance and the climate system they usually convert radiation from the sun into the effective radiation averaged across the complete surface of the earth.

This is simply 1367/4 = 342.

The second important number, incoming solar radiation at the top of atmosphere = 342 W/m2 (averaged across the whole surface of the earth).

Some energy is reflected but before we consider that note that this doesn’t mean that each square meter of the earth receives 342 W/m2 – it’s just the average. The equator receives more, the poles receive less.

### Albedo

Not all of this 342 W/m2 is absorbed. The clouds, aerosols, snow and ice reflect a lot of radiation. Even water reflects a few percent. On average, about 30% of the solar radiation is reflected back out. A lot of slightly different numbers are used because it’s difficult to measure average albedo.

The third important number, solar radiation absorbed into the climate system = 239 W/m2

This is simply 342 * (100% – 30%). You see slightly different numbers like 236, 240 – all related to the challenges of accurate measurement of albedo.

Some of the radiation is absorbed in the atmosphere, and the rest into the land and oceans.

### The Equation

Energy radiated out from the climate system must balance the energy received from the sun. This is energy balance. If it’s not true then the earth will be heating up or cooling down. Even with current concerns over global warming the imbalance is quite small. And so, as a starting point, we say that energy radiated out = energy absorbed from the sun.

Energy radiated from the earth, Ee = S (1- A) / 4  in W/m2

where A = albedo (as a number between 0 and 1, currently 0.3)

### Conclusion

The solar constant, S = 1367 W/m2

The solar radiation at the top of atmosphere averaged over the whole surface of the earth = 342 W/m2

The solar radiation absorbed by the earth’s climate system = 239 W/m2 (about 28% into the atmosphere and 72% into the earth’s surface of land, oceans, ice, etc)

Therefore, the approximate radiation from the earth’s climate system at the top of atmosphere also equals 239 W/m2.

These numbers are useful to remember.

Update – new post The Earth’s Energy Budget – Part Two

Update – new post The Earth’s Energy Budget – Part Three

### 49 Responses

1. Dave asked on a different article – before this was written:

Still not sure how x number of watts received in an average square meter can be emitted by some greater number of square meters. It really seems like the conditions of the initial equation have been altered.

It seems as though the calculations assume the earth is static when it receives the energy, but dynamic when it radiates that energy.

I look forward to an additional post on this vexing riddle.

Hopefully this post has answered the question.

If not, consider a different thought experiment. If you applied heat to one end of a metal bar, how would you calculate the temperature rise of the whole metal bar?

You might measure how much energy per unit area was being input. And you would multiply that energy per square meter (or square mm or whatever was appropriate) by the area that the heat was applied to.

That would tell you total incoming energy.

Then you would work out how much radiation emitted from the total surface area as a function of average temperature.

The equilibrium temperature would occur when outgoing energy = incoming energy.

It might be complicated as the bar might not be all at the same temperature.

But the simple point is that the energy in is applied to a small area. While the energy radiated out must be considered over the whole area of the bar.

2. […] OLR (outgoing longwave radiation) at the top of the atmosphere. We know this is about 239 W/m2 (see The Earth’s Energy Budget – Part One). Using the boundary conditions, we solve the radiative transfer equations for each slice, and the […]

3. I noticed the solar constant at 1367. It would appear it was lower in the late 1800’s.

4. […] is reflected by clouds and the earth’s surface. Check out the numbers in The Earth’s Energy Budget. This is all measured by […]

5. Hi!
Thanks for compiling all this in a very neat way. It’s all been up here and there in various places before, but I think your compilation is pretty good. I have a (silly) question though about the Albedo number (0.3 in your example). If some of this is due to scattering from clouds, ice and snow, then inherently that means that part of the sun rays must have passed through a fraction of the atmosphere – in the case of ground snow and ice; the whole atmosphere actually – and caused some heating of it on the way in and back out. Comments? Do you have any idea where this value (0.3) comes from? Has there been any exact satellite measurements done and how is the effect I just mentioned taken into consideration?

Another (silly) comment: I remember reading once, years and years ago, before global warming was on everyone’s lips (I’m 59…), that part of the additional warming of the atmosphere is due to the heat radiating from the core of the earth. Total heat loss from the earth is 4.2 × 10^13 Watts, which is of course a lot smaller than the figure for solar heating, but I guess the effect of this heat loss is really to heat up the very lowest layer of the atmosphere. I may remember this wrong so I will appreciate any comments from anyone. So, actually, taking this heat loss into account the top of the atmosphere ought to radiate just slightly more energy out than comes in…

Cheers
Kenneth

6. Kenneth:

Albedo
The albedo is a complex subject, the questions are very sensible and the effects you mention are taken into consideration.

If you take a look at
Positive Feedback, Albedo and the End of All Things you will see some more detail about how albedo varies with latitude, with location and throughout the year.

Albedo is primarily measured using satellites. Satellite data tells you the total reflection from clouds and aerosols in the atmosphere, as well as the reflection from the ground. The satellites give you a composite number, but by cross referencing that against other data like clear skies, low aerosol measurements and so on, the satellite data can generate much more information.

Then of course, lots of detailed research has been done into the albedo of different substances at different angles by doing more direct measurements.

Geophysical heat
The heat from within the earth is very low and insignificant. However, I can’t lay my hands on a reference right now.

In reality, as a further post will probably cover, the energy budget for the earth is not really “closed”.

That means, we can’t take the two complete measurements – incoming and outgoing – and find that they match to within 0.1W/m^2 for example.

The margins of error in incoming absorbed solar radiation versus outgoing longwave (OLR) is in the order of at least 5W/m^2.

The energy budget is “closed” by either assuming the two numbers match and using that to find the values we don’t know (like absorption of solar energy within the atmosphere) or by attempting an approximate out of balance number via the increase in ocean heat content (OHC).

This OHC increase leads to a number – recently calculated – of around 0.9W/m^2 more energy in than out. Well, even that number is probably prone to restatement.

The fact is: “We can’t account for the lack of warming at the moment, and it’s a travesty that we can’t” as Kevin Trenberth said in his widely misunderstood statement.

This comment was about the problems of measuring energy in the earth’s climate system. There are a lot of gaps.

7. […] 18, 2010 by scienceofdoom In Part One we looked at a few basic numbers and how to compare “apples with oranges” – or […]

8. […] radiated out will match the incoming energy. See The Earth’s Energy Budget – Part Two and also Part One might be of […]

9. […] The Earth’s Energy Budget we looked at “comparing apples with oranges” – why we need to convert the TSI or […]

10. […] of 1367W/m2 can be balanced by the earth’s OLR of 239W/m2 then take a look at Part One and Part […]

11. […] value of incoming – reflected solar radiation, divided by 4 for spatial geometry reasons, see The Earth’s Energy Budget – Part One. The total of 1.2 x 1017 W also equals the TSI of 1367 W/m2 x “disc area” of the earth […]

12. […] For an explanation of why the value is divided by 4, see The Earth’s Energy Budget – Part One […]

13. […] Part One – which explained a few basics in energy received and absorbed, and gave a few useful “numbers” to remember […]

14. […] 17, 2010 by scienceofdoom This could have been included in the Earth’s Energy Budget series, but it deserved a post of its own. First of all, what is “back-radiation” ? […]

15. […] If we average the incoming solar radiation that is absorbed by the earth’s climate over the surface of the earth we get around 239 W/m2. (See The Earth’s Energy Budget – Part One). […]

16. […] Note as well that we are looking at averages in this diagram. The solar radiation absorbed in any one places is very rarely 240 W/m² – at night it is zero, and at midday in the tropics it is closer to 1000 W/m². If you want to understand why the average value of solar radiation absorbed is 240 W/m² take a look at Earth’s Energy Budget – Part One. […]

17. Heat flow from the interior of the earth. A median value per unit area would be about 60 milliwatts per square metre. See Stacey, 1992, Physics of the earth, p 292, where the author refers to Sclater et al, 1980, Revs. Geophys. Space Phys. 18, 269-311.
However there are restricted and not well sampled regions with very much higher heat flow (mid-ocean ridges are the most extensive of these, but all volcanoes come into this category), which makes it harder to estimate the total than the median. However even a mean flux per unit area an order of magnitude greater than the median would still be small compared with other earth’s surface and atmospheric energy fluxes. Additionally, there is no reason to suppose that the total terrestrial heat flow varies much on the century time-scale, so it can probably safely be neglected when considering contemporary atmospheric temperature change.

18. […] so the radiation is effectively captured by the surface area of the sphere. This area = 4πr². See The Earth’s Energy Budget – Part One for more explanation of […]

Without the sun the earth temperature would be -454 degrees F. Currently as a result of the suns heat the average temperature is 513 degrees F warmer than that. Global warming in the last 50 years is about 1.33 degrees F yielding a ratio of .0026.
Heated air from fossil fuel combustion (.025 W/M) enters into the radiative budget in the section labeled convection where 7 per cent of the 51 per cent of 340 W/m (12.138W/M) is returned to the outgoing flux. This yields a ratio of .0021.
The similarity of these two ratios indicates to me that the 15 terawatts of heat energy added to the environment by burning fossil fuels is not so small as to be insignificant in the global warming picture (as is suggested in the literature). Where am I going wrong? Thanks in advance.

20. Erv Unger:

The annual global energy absorbed from the sun is about 3.8 x 10^24 J (see Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part One).

If 15TW added through energy creation is the right number (it seems low, but I have no idea) = 15 x 10^12 J and is therefore insignificant in comparison to fluctuations in solar radiation that are too small for us to measure.

21. […] Absorbed solar radiation = (1-0.3) x S/4, where S = solar constant of 1367 W/m². The “0.3″ is the reflected radiation due to the albedo of the earth and climate system. So only 70% of solar radiation is actually absorbed on average. The term 1/4 appears because solar radiation is not directly overhead all points on the globe at all times. For the detailed explanation see The Earth’s Energy Budget – Part One. […]

22. SoD, I’ve just come across this explanation of the earth’s energy budget and thank you, it’s a nice simple explanation. That said, I don’t understand it. Can you answer a couple of questions to clarify for me? Note I am just an average Joe with no training in physics or maths, so my question simply represents a practical take on the model you use to illustrate the matter. And I hope I explain myself sufficiently that you understand my question!

In this article, Energy Budget Part 1, you seem to be saying that the budget is derived by comparing energy measured as outgoing at the top of the atmosphere versus energy measured as incoming but averaged over the surface of the earth. This does not make sense to me, as I would have thought we want to measure the outgoing energy (ie after all absorptive/reflective etc effects) and compare it to the incoming energy (ie before all absorptive/reflective etc effects). I mean by this that it is the ToA measurements that count – how much is available incoming at the ToA versus how much is outgoing. Not how much is available at the ToA outgoing versus how much is available incoming at the surface.

To my way of thinking, any point in space must have available to it a discrete amount of energy. How an object at that point might react to that energy will be dependant on the surface area, reflectivity/absorptivity, angle of incidence etc, but those are all effects to be applied to the available energy. Put another way, the available energy is an independant property to the properties of the object.

Now, the rays from the sun produce, for our purposes here, a constant amount of radiation with a constant amount of energy. Thus, it’s not relevant whether half of the earth’s surface is in darkness. The amount of energy available is the same regardless of the shape of the earth or its movements, providing we know the surface area that is exposed to the radiation. Another way to think of this is that if we absent the earth from our system, the total energy available from the sun’s rays is the same.

So, all we need know is the surface area of the object presented to the sun’s rays. And that area is the area of the hemisphere represented by the outer shell of the atmosphere.

I don’t see why we should consider the amount of incoming energy needs to be divided by anything. It is what it is. A thought experiment illustrates what I mean. Presuming we have a sphere of the same size as the earth with no atmosphere. If I erect panels that absorb 100% of the available incoming energy, each 1 square meter in area, and each arranged such that they cover each square meter of that sphere’s surface and each exactly perpendicular to the sun’s rays, what is the energy collected by those panels? That is the available energy, and in the case of the earth that would be 1367 W/m2.

In calculating how that sphere might deal with that energy in order to calculate its heat I need to know all sorts of things, including the angles at which the rays strike the surface, but the available balance is what is available BEFORE I start those calculations, before I include those various parameters.

Perhaps I can say all this a little more simply: if you consider the earth and its atmosphere and all the effects thereof as a black box with tunable properties, and you wish to observe the effect on temperature of that black box as you adjust those parameters, then isn’t the budget based upon the available energy arriving at the surface of the black box (ie ToA for our earth system) compared to the energy leaving that same surface?

Where am I wrong with this thinking?

23. Graeme M:

I think I understand your issue, but I’m not totally sure. Let me try and discover it with some questions/points – and if I’ve got it (your question) wrong, it’s not intentional..

Let’s look at the total energy absorbed by the earth from the sun over 1 (average) day.

Ein = Incoming Solar Flux x (1-albedo) x Area x Number of seconds in one day

⇒ Ein = 1367 x 0.7 x π x r2 x 3600 x 24

The bit that you are having trouble with is why use Area = π x r2 and why not use Area = 2 x π x r2? Is this correct?

[Note: 2 x π x r2 is the surface area of a hemisphere, and 4 x π x r2 is the surface area of a sphere]

Assuming I’m correct in understanding your issue I will jump ahead – suppose that we had a flat disc, of albedo = 0.3, orbiting the sun at a distance of 150M km away. (And the disc surface always faces the sun).

What would be the energy, Ein’, absorbed by the flat disc in one day?

The answer, Ein’ = 1367 x 0.7 x π x r2 x 3600 x 24.

So how is it that we can curve this disc into a convex shape (the earth) and somehow it collects double the energy from the sun? (If I have understood your issue correctly).

The answer is – we can’t. Curving (stretching) the disc into a convex shape doesn’t change the total energy absorbed – it just gets spread over a larger surface area.

I’ll stop here so you can comment.

24. SoD, I am impressed by the speed of your response! I thought my question would simply languish in cyberspace… I appreciate the effort.

Now, you’ve sort of grasped where I am coming from, but not quite.

First a caveat. I may be completely misunderstanding what is meant by the ‘earth’s energy budget’. I had thought that the ‘greenhouse effect’ simply describes the fact that at thermal equilibrium the earth must radiate the same energy it receives, and that the presence of an atmosphere influences how this occurs. That is, the greenhouse gasses slow the radiation of heat from the earth in proportion to their extent in the atmosphere.

I also may be confusing terminology…

Anyway, the budget presumably is measuring the energy in compared to the energy out.

Where I struggle is here:

The outgoing radiation is measured by satellites. I assume that for the sake of models/theory, this is considered to be at the ToA. In other words, when it is measured, outgoing energy is net of all effects of the earth’s surface and atmosphere. Why then for incoming energy are you using a figure that represents the amount absorbed by the earth system? Surely the budget should be comparing outgoing radiation net of all effects with gross incoming radiation NOT the amount absorbed.

That gross amount would be what we can measure at the ToA, NOT at the surface. You are introducing initial discounts based on atmospheric effects, surface absorptivity and so on.In a sense you are double-counting.

My thinking is that if the correct measurement is gross available incoming solar flux, and you wish to calculate that across the area at ToA, then we would use a hemisphere representing the ToA, not the earth’s surface. And every point at which we receive that incoming flux is free of any effects of physical matter – that is, at any point on that hemisphere, if we measure the solar flux, it must be the same in terms of its energy content. Thus available gross incoming flux is that arriving at every point on our theoretical hemisphere.

That disk business appears to me a red herring. I grasp the concept, but the shape of the object surely cannot affect the gross incoming flux because that is a property of the radiation itself. The balance available is the energy times the surface area exposed.

Let me pose a question. Imagine a theoretical hemisphere of the same size as the earth (or more exactly the limit of the earth’s outer atmosphere) at the same distance from the sun. If I place a measuring device at any point on that theoretical hemisphere, what is the energy measured by that device? Note there is no actual physical object, the theoretical hemisphere simply represents each point at which we measure the solar flux. If we placed a measuring device at a point representing every square metre of the theoretical hemisphere, how much total energy would we collect in a unit period? To my mind, that value is our gross available incoming balance.

25. Graeme M:

Anyway, the budget presumably is measuring the energy in compared to the energy out.

This is correct. But “energy in” = “energy absorbed”.

“Energy in” does not equal “energy irradiated”.

Where I struggle is here:

The outgoing radiation is measured by satellites. I assume that for the sake of models/theory, this is considered to be at the ToA. In other words, when it is measured, outgoing energy is net of all effects of the earth’s surface and atmosphere. Why then for incoming energy are you using a figure that represents the amount absorbed by the earth system? Surely the budget should be comparing outgoing radiation net of all effects with gross incoming radiation NOT the amount absorbed.

Some of the solar radiation is reflected. Therefore, it is not part of the energy budget of the earth.

If 99% of solar radiation was reflected it would be a totally different energy budget from the case where only 1% was reflected.

That gross amount would be what we can measure at the ToA, NOT at the surface. You are introducing initial discounts based on atmospheric effects, surface absorptivity and so on.In a sense you are double-counting.

The energy budget is nothing to do with the surface or the inappropriately-named “greenhouse” effect. It is a TOA measurement.

The only effect of atmosphere and surface are in the total energy absorbed. We can measure this (because we measure the reflected solar radiation).

That disk business appears to me a red herring. I grasp the concept, but the shape of the object surely cannot affect the gross incoming flux because that is a property of the radiation itself. The balance available is the energy times the surface area exposed.

No this is not correct and easily disproved.

The challenge is explaining it so your mental model can accept it.

Here is a picture of energy received:

The sun’s energy comes from a long way away so the rays are parallel. That is, the energy in each 1/4 is equal.

The surface area of the planet (surface or TOA) is not equal in each 1/4. Therefore, the energy absorbed per unit area is not equal.

Does this picture demonstrate that energy absorbed is proportional to cross-sectional area not surface area of a sphere?

..If I place a measuring device at any point on that theoretical hemisphere, what is the energy measured by that device?..

If you place a device pointing towards (= “normal to”) the sun you will measure 1,367 W/m2.

If you place the device “normal to” the surface of the planet you will measure 1,367 W/m2 x cos θ – where θ = the angle between the surface and the direction of the sun – so, when the device is pointing directly to the sun you will measure 1,367. And when the device is at an angle of 45′ you will measure 967, when the device is at an angle of 60′ you will measure 683, and when the device is at an angle of 90′ you will measure 0.

[Note, corrected the angles from the original comment a few minutes ago]

26. Thanks again SoD. The key point is clearly that energy in = energy absorbed. I had assumed that energy in = energy irradiated. I’ll now have to go think about that for a bit.

The issue of the disc versus the sphere is due to me thinking that each point receiving radiation receives the same as would the total area of measaurement (ie a square metre) – clearly rather mixed up there.

The simple explanation for me is that if I held a square meter board facing the sun and slowly moved it until it was perpendicular it would progressively present less effective surface area to intercept the radiation.

Again thank you for the time, very much appreciated.

27. Oops, I clearly meant parallel to the sun’s rays, not perpendicular… You know what I mean!

28. science of doom says:

“The solar radiation absorbed by the earth’s climate system = 239 W/m2 (about 28% into the atmosphere and 72% into the earth’s surface of land, oceans, ice, etc)”

Do you agree that the full 239 W/m^2 gets to the surface one way or another?

29. Just to make things a bit more complicated – it would be quite ok, in my opinion at least, to present the balance as such that the incoming light is the total power of the sun reaching earth and outgoing is the 239 W/m^2 plus the reflected amount (a total of 342 W/m^2). It’s just that the reflected part doesn’t interact with the earth much and due to that it’s more practical to approach the budget in the way SoD presented it (roughly absorbed shortwave vs emitted longwave).

If you put a detector in orbit to measure all the energy coming from the earth – you’d average out at 342 W/m^2 for the surface of the earth per example provided by SoD.

30. RW:

That’s a strange question.

Do you mean does all of the solar radiation always get absorbed by the surface at some time?

If that is your question, the answer is no. Some of the radiation absorbed by the atmosphere in any given second will be radiated back out to space without ever making it to the surface.

31. RW,

That’s not a very healthy way to look at energy. You can’t really break it down to parcels so that this piece of energy was here and went there, made a few rounds around the fountain and came back again. It works with radiated energy to some extent, but once you get down to an environment where energy is exchanged in other ways as well, it gets quite messy.

If you mean whether the downwards longwave radiation at the surface would be higher or lower if atmosphere didn’t absorb any shortwave then it’s not quite as simple as taking 45 W/m^2 from one place and putting it to another. My guess would be that DLR would increase if the atmosphere stopped absorbing shortwave, but it’s probably a tad to complicated problem for me.

32. science of doom says:

“That’s a strange question.

Do you mean does all of the solar radiation always get absorbed by the surface at some time?

If that is your question, the answer is no. Some of the radiation absorbed by the atmosphere in any given second will be radiated back out to space without ever making it to the surface.”

My question is does the full post albedo ultimately get the surface one way or another (directly or indirectly)? I know that technically some of the post albedo is absorbed by the atmosphere and some of that gets radiated out to space without ever getting the surface, but this amount is already included in the albedo of 0.3 as incremental reflection. This is why measured outgoing LW via satellites is typically about 250 W/m^2 or more. The +10 or so W/m^2 is the amount absorbed by the atmosphere that is emitted back out to space without ever reaching the surface or entering the system.

By the full post albedo, I means the full 239 W/m^2 referenced in Trenberth’s 2009 paper. Does this 239 W/m^2 get to surface?

33. RW:

I know that technically some of the post albedo is absorbed by the atmosphere and some of that gets radiated out to space without ever getting the surface, but this amount is already included in the albedo of 0.3 as incremental reflection.

No it is not “already included in the albedo of 0.3..”.

The albedo is solar radiation reflected by the atmosphere or surface without being absorbed. Reflection is totally different from absorption and re-emission.

For example, a wavelength of 0.5 μm reflected by the atmosphere will be 0.5 μm. A wavelength of 0.5 μm absorbed and re-emitted will depend on the temperature of the atmosphere. A typical wavelength of emission will be 10 μm.

This is why measured outgoing LW via satellites is typically about 250 W/m^2 or more. The +10 or so W/m^2 is the amount absorbed by the atmosphere that is emitted back out to space without ever reaching the surface or entering the system.

This is not correct. Well, provide a source and I will take a look.

From year to year the emission of longwave radiation by the whole planet is, within the errors of absolute accuracy of satellite measurement, equal to the absorption of solar radiation by the whole planet. (See note at end).

By the full post albedo, I means the full 239 W/m^2 referenced in Trenberth’s 2009 paper. Does this 239 W/m^2 get to surface?

The answer is the same as before.

I encourage you again to take a Heat Transfer Textbook and go through some worked examples. This might clear up most of your confusion about the first law of thermodynamics. I have previously provided an online reference.

Note: This doesn’t mean the planet is necessarily in radiative balance on decadal timescales, but any imbalance is smaller than the errors in measurement accuracy of satellites.

• Ah ha. I think this is crux of many of our disagreements.

The 239 W/m^2 referenced in Trenberth’s 2009 paper is the amount of solar energy that ultimately gets down the surface one way or another.

Look at Table 2a for the Global values for “this paper”. Listed there is an ASR (Absorbed solar radiation) of 239.4, an OLW (outgoing long wave) of 238.5 and – most importantly a “Net Down” of 0.9 W/m^2. NET Down means 239.4 or the full post albedo gets down the surface and 238.5 W/m^2 is leaving at the TOA for a net imbalance of 0.9 W/m^2 at the surface.

If you don’t think I’m interpreting this table correctly, look at the surface components in table 2b for “this paper”. Listed there is solar absorbed 78.2 W/m^2 and 161.2 W/m^2. 78.2 + 161.2 = 239.4, which the exact amount listed for absorbed solar radiation. The amount listed for “NET Down” is also 0.9 W/m^2.

According to the definitions for the terms in the tables, “NET Down” specifically means “net energy absorbed at the surface”.

Furthermore, table 2b also lists 333 W/m^2 of ‘back radiation’, but ‘back radiation’ is specifically defined as only “LW downward radiation to the surface”. It is not defined as downward emitted LW that last originated from surface emitted even though this is what the diagram makes it look like.

If you add up the numbers, there is 1 W/m^2 extra in the ‘back radiation’ of 333 W/m^2 (78 from the Sun + 80 from latent heat + 17 from sensible heat + 157 from ‘back radiation’ that last originated from surface emitted = 332) to show the “NET Down” of 0.9 W/m^2. 157 + 239 from the Sun = 396 W/m^2 – the net flux and amount emitted at the surface.

34. The source of our disagreements is already outlined in the latter part of the discussion in The Mystery of Tau – Miskolczi – Part Four – Emissivity.

I don’t intend going through that all over again.

• Me either.

BTW, if you don’t think my interpretation of the data in the tables is correct, contact Trenberth and ask him. I think you’ll find he’ll confirm what I’m saying is right.

35. This is probably why most people don’t understand the constraints Conservation of Energy puts on the boundary between the surface and the TOA – they don’t realize the full post albedo of about 240 W/m^2 is getting to and ultimately ‘forcing’ the surface.

This is also why, as I’ve tried to point out, Trenberth’s diagram has confused so many people.

• RW

It seems to me SoD has already given great answers to your concerns. But I can’t help offering a couple of thoughts (imprecise as they may be). First, you might say that ALL of the energy in the diagram comes from the sun. Without the sun’s radiance, the earth would be a cold frozen rock. Trying to trace what’s from the sun and what’s not kind of loses a little meaning. But more importantly, and more to the point, I think the Trenberth diagram makes the most sense if you consider the arrows as showing the initiation and termination points of the actual energy carrying photons. When a photon reflects, it just changes path – there’s no change in energy. However, when a photon is absorbed it sacrifices its energy (its existence) into the rotational, vibrational, kinetic or electronic energy level of an atom or molecule. At that point it’s gone, done, finito. It no longer exists. Later, if that same molecule undergoes a relaxation and emits a photon, its not the same photon that was absorbed previously and isn’t related to it.

So to try to trace how much of the back radiation is actually from the sun is an impossibility. The back radiation consists of photons that begin their existence in the atmosphere. It’s not possible to find which ones came from the sun, because none of them did. The only downward traveling photons which actually originated at the sun and which reach the Earth’s surface are the 168 W/m2 absorbed and 30 W/m2 shown in Trenberth’s 1997 diagram.

If you consider the arrows this way, and it’s actually what they represent physically, it’s not disingenuous or misleading to lump together all the downward atmospheric radiation into one arrow. They all have the same source and destination, so should be in the same arrow. I believe you object to calling it all “back radiation” thinking that implies all the energy first originated at the earth’s surface. But that’s not what it means to those who understand what’s going on. So suggest it be called something else, like “Downward Atmospheric Radiation”. But don’t suggest that some of it needs to be broken off and shown as being from the sun because in terms of the genesis of all those photons, they have been birthed in the atmosphere as a result of energy previously absorbed from above and below.

• “So to try to trace how much of the back radiation is actually from the sun is an impossibility.”

I know it is. You’re missing the point. I suggest you go and read my and SoDs exchange he linked. I realize now this thread was probably not the best place to ask SoD that question.

“So suggest it be called something else, like “Downward Atmospheric Radiation””

Funny you mention this, because I have specifically suggested this on another thread.

36. See the computation for a gray ball in our orbit at http://climatewiki.org/wiki/Category:Essential_Physics . That’s simply the energy density of a point in our orbit . Rather than reduce the problem to 2D , it simply sums the energy coming from all directions over the sphere . I don’t think much in terms of energy because that just gets calculated and summed in the middle of the expression from temperature to temperature . I get a temperature of 278.7 given the 5.411e-6 of the total surrounding sphere the sun covers and assuming an effective temperature ( total radiant energy equivalent ) of 5778 . That matches your 342 W%M^2 .

The absolute numbers ARE important . They translate into effective temperatures for the sun and consequently for the earth . The uncertainty seems to be as much as 2.5 W%M^2 , from about 278.52 to 278.98 , about 0.46 degrees , more than half the 0.8 to be explained . It matters which satellite is right .

Are there public interfaces to the data bases those fuzzy graphs are plotted from ?

In my newsletter , http://cosy.com/y12/NewsLetter2012.html , I put out the challenge to calculate the maximum and minimum temperatures for any uniformly ( isotropically ) colored ( spectrum ) ball in our orbit .

37. Hi.

What are the errors in ‘Buget equation’ we can consider

38. As I understand it we do have good direct measurements of CO2 levels,

So if we had equally good direct measurements of how much energy was actually coming in and going out, we could then see whether / to what extent the energy budget moves in step with CO2 as hypothesised.

Which would presumably settle virtually all arguments on the topic ? – one way or the other.

But that doesn’t seem to have happened. Would I therefore be correct in concluding that this is because we do NOT have the required good energy -in and -out measurements ?

• Yes, you are correct. The precision and accuracy of satellite measurement of incoming and outgoing radiant energy are not sufficient for the purpose of determining global annual radiative imbalances of less than 1W/m². The closest thing we have is measuring the rate of change of ocean heat content, which should be directly proportional to the radiative imbalance. But historic OHC data are problematic and the ARGO float system may still have teething problems.

• I guess the next question then is : given that better quality radiation balance and OHC data would cleanly resolve the CO2 question, why don’t we prioritise them? Divert funds from modelling maybe?

• Punksta,

On satellite measurements:

There is a high priority on measurements. The CERES and AIRS projects are massive.

Getting the absolute measurement resolution of net radiation to better than 1W/m2 is unachievable even with these massive projects.

The earth is a big place and you need to be scanning everywhere for longwave radiation. Then the reflected solar is even more difficult because the reflection direction varies.

The net radiation is solar insolation incoming – reflected solar – outgoing longwave radiation. So you are subtracting two different absolute measurements from one other absolute measurement. Much more of a challenge than if there was one net parameter to directly measure.

Solar insolation is pretty constant, but even there we have much better knowledge of changes than the absolute value. This is generally true for this kind of instrumentation project – the absolute values are harder to know to a 1W/m2 accuracy than the changes with respect to time and location.

There is a general idea in the blog world that climate scientists mostly sit in front of computer models and believe what the models say, but having read well over a thousand papers I see that there is a huge effort (and \$ expenditure) in experiments and measurements.

I couldn’t tell you if the balance between model effort and measurement effort was right, but neither can people who learn climate science from reading blogs.

On ocean measurements:
I’ve read a lot less about the oceans than satellite measurements, but still plenty of papers. Once again, the oceans are massive and getting the level of accuracy in enough places is even more difficult than for satellites.

• Punksta.
I have the same question. If 93% of surplus energy goes into the oceans, and OHC is a reliable measure of this, then it will also be a good measure of the difference between incoming and outgoing radiation. An there is a very close correspondence between OHC and sea level change (SLR). So I think sea level change is the best way to estimate change in net radiation historically. And that this will tell us something of climate sensitivity. And I wonder why this discussion is so neglected.

• Nobodyknows,

And I wonder why this discussion is so neglected.

How many papers do you think there have been on this subject in the last 5 years?
And how many papers do you think there should be?