Climate science - as easy as A, B, C.

We all accept that the actual average sea-level temperature of the Earth (288K) is about 33K higher than its 'effective temperature' i.e. what it would be if Earth had no atmosphere (255K). That's basic physics - average incoming solar radiation in W/m2 minus amount reflected as light (known figures) raises temperature of the 'surface' (as defined - see below) to whatever it needs to be to radiate the same amount of W/m2 back out to space.

The Consensus is that the entire 33K discrepancy is due to the presence of minor trace gases in the atmosphere, such as the +/- 2% water vapour or 0.04% CO2 ('greenhouse gases'). The surface converts short wave visible radiation from the Sun to longer wave infra red radiation, and this is trapped/absorbed or reflected by the 'greenhouse gases', which in turn warms up the surface even more, so it emits even more infra red in a vicious circle. Here is a typical article explaining this, which includes the energy budget diagram which doesn't add up as a bonus. Parts of that explanation are correct, parts are guesswork and assumptions, it is riddled with contradictions, leaves a lot of loose ends and unanswered questions, and overlooks some important basic physics.

Actually, if you sift out the basic physics from the guesswork and just apply the basic physics, it is quite easy to work out that the hard surface should be 45K - 50K warmer than the effective temperature. The formulas and calculations are the same whether or not there are any 'greenhouse gases', which suggest that they do not increase temperatures. As a matter of fact, there is only a 33K discrepancy, because water vapour and water in the troposphere have a moderating influence and reduce surface warming and temperature fluctuations (which accords with everyday experience and like-for-like comparisons of temperatures in humid and dry areas).

They could and should teach this as part of GCSE level physics, it wouldn't take more than two or three lessons.

A. Barry

You can easily work out that there are about 10,000 kg of air (101,325 Pascal ÷ 9.807 m/s/s) for every m2 surface. We know that 1 m2 of air at sea-level has a mass of 1.293 kg. So if atmosphere were same pressure and density all the way up, with a hard edge, it would be 7.7 km high.

You can guess intuitively that the atmosphere gets thinner as you go up and gradually tapers off into space and that there is no hard edge. So let's assume actual average density is half that at sea-level, a reasonable guess is that most of the mass of the atmosphere is up to an altitude of 15 km or so, and atmospheric pressure falls by about 7% for each km you go up, at least for the first five or ten km (which is not far off actual measurements).

If you want to calculate this properly, you use the Barometric Formula (or 'Barry', as I affectionately call it) which is based on actual ideal gas laws and gives you reasonably accurate predictions for pressures at different altitude, at least for the troposphere (which is all we really care about, i.e. is the bottom 11 km, others say bottom 13 km, it's thicker at Equator and thinner at the Poles). The formula is very clever. I can just about understand how they work it out, but I would struggle to reverse engineer it or explain how to derive it.

B. The lapse rate

In an intuitive way, you can also guess that the temperature at the top of the atmosphere is close to the temperature of the nearly empty vacuum of space, which is either close to 0K or has no measurable temperature at all, depending on your point of view. So temperature falls the higher you go, as anybody who has been up a mountain knows.

Remember that 'energy cannot be created or destroyed, it merely changes from one form to another'. Air at sea level as thermal energy (aka kinetic energy) and no potential energy (it can't fall any further down). Air higher up has the same amount of total energy - less kinetic energy and some potential energy. It is reasonable to expect the total amount of energy to be the same at different altitudes.

Once you accept this, you can work out the lapse rate. I'll show you how, just for fun and because it is important:

Potential energy in Joules = mass x gravity x height.
So J = m x g x h

Joules required to increase temperature of 1 kg of a substance by 1K = specific heat capacity ('cp' ) of that substance.
So T = J ÷ (m x cp)

We can simplify/merge those two equations as follows: T = (m x g x h)/(m x cp); cancel 'm' top and bottom; T = g x h/cp; divide both sides by h; T/h = g/cp.

(Thanks to Tallbloke for this short-cut - Wiki gives an explanation which is almost impenetrable to the layman, although it ends with exactly the same formula)

1 kg of air which is 1,000 metres higher up has got 9,807 more Joules of potential energy than 1 kg 1,000 metres lower down, so the air lower down must have 9,807 more Joules of kinetic energy (and vice versa).

How much warmer is the air 1,000 metres lower down?
Specific heat capacity of air = 1,006 J used/needed to increase 1 kg of material by 1K
T/1,000 = 9.807/1,006
T/km = 9.75K
Hence the predicted lapse rate = 9.75 K/km altitude.

[I think it makes more sense to a) use the specific heat capacity for constant volume rather than constant pressure, and b) to calculate J/m3 rather than J/kg. That means first using Barry to find pressure, at different altitudes and then finding the temperatures which balances Joules of kinetic energy and Joules of potential energy (basing calculations on density at different altitudes, not on pressure, so you need to know pressure and temperature to work out density, and the temperature is the thing you are looking for!), which is why the lapse rate I worked out was 8K/km. I'm not sure if my logic on a) and b) is 100% sound, but my method gives an answer which is closer to the real world observed typical rate of 6.5 K/km, so I'm happy with my method for now.]

C. What is the surface of the Earth?

The Consensus give a nod to Barry and the main reason for the lapse rate. They don't deny they exist, they give the formulas but then downplay them as irrelevances and draw no conclusions from them. It's like the road sign saying turn left and the sat nav saying turn left, but turning right anyway. It's all about radiation from the surface being reflected back down by those dastardly 'greenhouse gases'! Any other explanation is heresy!

The Consensus' most heinous and borderline criminal obfuscation is in their definition of the 'surface'. They define 'the surface' as the hard surface at sea level, or the surface of the oceans (two-thirds of Earth's surface is ocean).

Here are a few reasons why that is wrong:

1. If you calculate the 'effective temperature', you should also be looking at 'effective surface', which is whatever the sun light hits first, Earth has a lot of clouds, so the 'effective surface' is NOT at sea-level.
2. If you approach Earth at speed from space, you feel it when you hit the atmosphere. That is the real surface.
3. If you scale down the earth to the size of a football, the troposphere is only 0.2 mm thick. You think it's thick when you look up at an airliner, but it's only 11 km away, you could drive that far in a few minutes.
4. There are about 10,000 kg of air (mass) for every m2 surface (see above)
5. The sun only warms the top few inches of the hard surface (or water), that's a few kg of mass per m2. So if the atmosphere is 0.2 mm thick, the hard surface or sea-level is barely a couple of molecules thick.
6. When we talk about the surface of the ocean, we mean the top of the water, not the hard surface at the bottom.
7. When we talk about the surface of the Sun or a Gas Giant, we mean the surface of the atmosphere, not the hard surface lower down (to the extent there even is one). Why do we change the rules when looking at rocky planets with thinner atmospheres?
8. As far as heat distribution goes, we might as well treat the top few inches of the hard surface as the bottom part of the troposphere. It's usually the same temperature, and must be the same pressure.

Therefore, the real surface is the whole troposphere, not the hard surface or the surface of the oceans.

Summary

The troposphere is the surface, and as a whole and on average, is the temperature you would expect from incoming solar radiation = +/- 255K. This is what you would expect, and this is what you get. I don't see why anybody taking this view should be on the defensive in a discussion. It is those trying to say otherwise who are struggling.

The troposphere itself, it is not a constant 255K. There has to be - and there is - a lapse rate. There are different ways of calculating/predicting it, and it can be observed/measured, so it's about 33K warmer than 255K at the bottom (the hard surface); it's about 255K half-way up (as defined); and it's about 33K cooler than 255K at the top of the troposphere.

Bonus

This A-B-C easy GCSE-level approach also explains a lot of things which the Consensus explanation can't and doesn't (and by and large, just glosses over to save embarrassment), for example:

1. Why the top of the troposphere (or the peak of a very high mountain) is colder than the 'effective' temperature, i.e. colder than it would be if Earth had no atmosphere.

2. Why the day/night temperature range on Earth (+/- 15K) is so much smaller than the day/night temperature range on the Moon (+/- 300K).

3. Why, despite the 'greenhouse effect', Earth's day-time temperatures at the surface are lower than what they would be without an atmosphere.

4. Why the 'greenhouse effect' is much stronger at night (i.e. actual temperature minus expected temperature of the night side of Earth if it had no atmosphere) than in the day time (when the Sun is blazing down on us and there is plenty of radiation sloshing about).

5.  Why 'heat rises' is a truism only observed in enclosed spaces kept above the temperature of their surroundings (central heating in buildings; actual greenhouses/polytunnels) and why convection doesn't actually transfer heat upwards. For sure, there are thermals above hot surfaces (like square miles of dark tarmac at airports, which can make landing trickier than it need be on hot days), but for every molecule that goes up, one has to come down. One molecule converts kinetic energy to potential energy and the one coming down does exactly the opposite.

6. Why there is no need to get tied in knots over which methods energy (in its various forms - visible and infra red, kinetic energy, potential energy, latent heat of evaporation/condensation etc) is distributed in the troposphere (conduction, convection or radiation). All you need to know is that energy tries to distribute itself as evenly as possible (governed by the physical laws discussed here, or by winds/weather, if you want an everyday term for a complicated process).

7. Why the troposphere emits twice as much radiation towards the ground than it does out into space. This cannot be satisfactorily explained by 'greenhouse gases trapping and/or reflecting heat', it is because the troposphere at sea level is warmer and hence emits more radiation that the layer higher up, which is colder and so emits less radiation, see here.

8. Why water can't be a 'greenhouse gas', although this is based on observation rather than physics, which is really complicated with water vapour and water (see here).

9a. Why it is irrelevant that N2 or O2 are transparent to, and cannot absorb or emit infra red (even if this were true, which is questionable). They can certainly warm up, and they in turn keep the hard surface at the same temperature. So even if N2 and O2 aren't emitting infra red themselves, the hard surface converts that warmth back to radiation anyway. The total infra red leaving the hard surface is the same whether it is bouncing back and forth as infra red between hard surface and troposphere (the Consensus), or whether the hard surface has to convert kinetic energy from N2 and O2 back into infra red first (the actual explanation).

9b. Why, even if the troposphere were indeed completely transparent to and unaffected by infra red radiation, incapable of absorbing or emitting it, the temperature at the hard surface would be the same as it is now. The hard surface would quickly reach 255K (accepted by the Consensus) and it would warm up the troposphere by conduction and convection until the whole troposphere (the effective surface) were 255K on average (it being incapable of radiating heat to space, that's the Consensus). That 255K would be the average, there would still be a lapse rate and so the temperature of the hard surface would increase to 288K (and the top of the troposphere would be about +/- 222K). The hard surface would then be warm enough to emit the required amount of infra red straight through the troposphere and back into space (not being able to lose any more energy to the troposphere by conduction or convection).

10. Why there is a lapse rate on all planets with an atmosphere, even Gas Giants (Jupiter, Saturn), which have no hard surface (and if they do, radiation from the Sun never gets there) and which consist mainly of 'non-greenhouse gases' (mainly hydrogen and helium); why they are insanely hot at their centres; and why those Gas Giants are actually emitting more radiation to space than they get from the Sun.

11. Why the 'greenhouse effect' on Mars is barely measurable (max 5K), even though there is twenty-five times as much CO2/m2 surface area as there is on Earth.

12. Why you can predict Venus' hard-surface temperature fairly accurately using the same basic physics (here). All you need to know is distance from Sun and albedo, from which you work out 'effective temperature'; the height of 'effective surface'; the acceleration duty to gravity and specific heat capacity of the gases in the atmosphere, from which you work out the lapse rate up- and downwards. Whether or not the constituent gases are 'greenhouse gases' is entirely irrelevant. Knowing the mass of constituent gases in kg/m2 helps as well for cross checking.