By Steve Goddard
ESA’s Venus Express mission has been studying the planet and a basic atmospheric model is emerging.
Venus has long been the CO2 bogeyman of climate science. In my last piece about Venus I laid out arguments against the claim that it is a runaway greenhouse which makes Venus hot. This generated a lot of discussion. I’m not going to review that discussion, but instead will pose a few ideas which should make the concepts clear to almost everybody.
If there were no Sun (or other external energy source) atmospheric temperature would approach absolute zero. As a result there would be almost no atmospheric pressure on any planet -> PV = nRT.
Because we have a sun providing energy to the periphery of the atmospheric system, the atmosphere circulates vertically and horizontally to maintain equilibrium. Falling air moves to regions of higher pressure, compresses and warms. The greater the pressure, the greater the warming. Rising air moves to regions of lower pressure, expands, and cools. The amount of warming (or cooling) per unit distance is described as the “lapse rate.” On Earth the dry lapse rate is 9.760 K/km. On Venus, the dry lapse rate is similar at 10.468 K/km. This means that with each km of elevation you gain on either Earth or Venus, the temperature drops by about 10C.
It is very important to note that despite radically different compositions, both atmospheres have approximately the same dry lapse rate. This tells us that the primary factor affecting the temperature is the thickness of the atmosphere, not the composition. Because Venus has a much thicker atmosphere than Earth, the temperature is much higher.
dT = -10 * dh where T is temperature and h is height.
With a constant lapse rate, an atmosphere twice as thick would be twice as warm. Three times as thick would be three times as warm. etc. Now let’s do some experiments using this information.
Experiment # 1 – Atmospheric pressure on Venus’ surface is 92 times larger than earth, because the atmosphere is much thicker and thus weighs more. Now suppose that we could instantly change the molecular composition of Venus atmosphere to match that of Earth. Because the lapse rate of Earth’s atmosphere is very similar to that of Venus, we would see little change in Venus temperature.
Experiment #2 – Now, lets keep the atmospheric composition of Venus constant, but instead remove almost 91/92 of it – to make the mass and thickness of Venus atmosphere similar to earth. Because lapse rates are similar between the two planets, temperatures would become similar to those on earth.
Experiment #3 – Let’s take Earth’s atmosphere and replace the composition with that of Venus. Because the lapse rates are similar, the temperature on Earth would not change very much.
Experiment #4 – Let’s keep the composition of Earth’s atmosphere fixed, but increase the amount of gas in the atmosphere by 92X. Because the lapse rates are similar, the temperature on Earth would become very hot, like Venus.
Now let’s look at measured data :
http://www.astro.wisc.edu/~townsend/resource/teaching/diploma/venus-t.gif
http://www.astro.wisc.edu/~townsend/resource/teaching/diploma/venus-p.gif
Note that at one Earth atmospheric pressure on Venus (altitude 50km) temperatures are only about 50 degrees warmer than earth temperatures. This is another indication that atmospheric composition is less important than thickness.
Conclusions : It isn’t the large amount of CO2 which makes Venus hot, rather it is the thick atmosphere being continuously heated by external sources. It isn’t the lack of CO2 on Earth which keeps Earth relatively cool, rather it is the thin atmosphere. Mars is even colder than earth despite having a 95% CO2 atmosphere, because it’s atmosphere is very thin. If greenhouse gases were responsible for the high temperatures on Venus (rather than atmospheric thickness) we would mathematically have to see a much higher lapse rate than on Earth – but we don’t.
WUWT commentor Julian Braggins provided a very useful link which adds a lot of important information.
“The much ballyhooed greenhouse effect of Venus’s carbon dioxide atmosphere can account for only part of the heating and evidence for other heating mechanisms is now in a turmoil,” confirmed Richard Kerr in Science magazine in 1980.
The greenhouse theory does not explain the even surface temperatures from the equator to the poles: “atmospheric temperature and pressure in most of the atmosphere (99 percent of it) are almost identical everywhere on Venus – at the equator, at high latitudes, and in both the planet’s day and night hemispheres. This, in turn, means the Venus weather machine is very efficient in distributing heat evenly,” suggested NASA News in April 1979. Firsoff pointed out the fallacy of the last statement: “To say that the vigorous circulation (of the atmosphere) smooths out the temperature differences will not do, for, firstly, if these differences were smoothed out the flow would stop and, secondly, an effect cannot be its own cause. We are thus left with an unresolved contradiction.”
======================================================
An update for those interested in what Venus looks like at the surface.
On March 1, 1982, the Soviet Venera 13 lander survived for 127 minutes (the planned design life was 32 minutes) in an environment with a temperature of 457 °C (855 °F) and a pressure of 89 Earth atmospheres (9.0 MPa). The photo composite above shows the soil and rocks near the lander.
Here’s another Venera image that shows a hint of yellow atmosphere. – Anthony
Steve, thanks for the topic, it has been a great discussion.
dr.bill, I have more or less been trying to follow the discussion here. I will admit that my thermo-dynomics education was very weak, the prof had a heart attack and we got stuck with an organic chemistry teacher whose opening statement to the class was “I hate thermodynamics and you are going to pay…” nough said.
I remember PV=nRT as the ideal gas law.
Are you stating that since Venus is an “open” system the gas law does not apply? And if so what about the effects of gravity? What about the chart that shows that about 1 earth’s atmosphere the temp on Venus is similar to that of earth? Is the problem that the atmospheric gas on Venus is not being actively compressed or expanded?
Jeremy – yes I understand that g and Cp are bound to be different. My question was why is Cp for Venus be 85% of Earth’s. The point is that we cannot asser that the lapse rate is independent from composition.
Stevengoddard, Reur May 10, 2010 at 12:39 pm
I’ve not read the RC explanation but cannot see that the concept is a prime cause of uniform average surface temperature, (although it would help), for several reasons, but broadly:
Whilst the mass of the Venus atmosphere is ~92x more than the Earth’s, the nighttime cooling period is 117x longer. It does not seem possible that there is no heat loss over such a long period, even if we were to consider simple conduction alone. Because of the huge temperature and pressure range, it seems probable that there is very strong convection. Certainly, the observed upper winds that circulate the planet in ~4 earth days, and the polar vortices are violent, and not understood. (a lot of ‘work done’ energy BTW) There are inadequate data for what happens below. It seems to me that the (unknown) atmospheric dynamics are a likely candidate. Convection, (which validates your pressure hypothesis), may even be stronger at nighttime; see discussion below that I’ve extracted from my earlier comment:
[1] When facing the sun, she is said to receive at the surface about 10% of sunlight. Whether this is a midday or total facing area average, or the effects of scattering, I don’t know, but whatever, the surface receives SOME solar energy. This energy amount must be lost back to space because the planet is apparently in thermal equilibrium. The fundamental process for this should be convection and conduction. Additionally, since infrared photography etc of the surface has been accomplished there is at least one window for some infrared to escape directly to space.
[2] At nighttime she no longer receives any solar energy, but has capability to lose heat in the same way as on the daylight side, over a period 117 times longer than on Earth. What is more, because the upper atmosphere is no longer heated by solar infrared, (~40% of sunlight), the temperature gradient of the atmosphere should increase, inferring increased conductive/convective cooling. Yet, there is no change in surface temperature! Such a condition would require an impossible perfect insulation layer, [with] no geothermal energy, but clearly, this is not the case.
#
stevengoddard says:
May 10, 2010 at 8:39 am
GaryW
A much simpler empirical exercise would be to answer qualitatively this simple question.
What would happen to temperatures on Venus if you could strip away 99% of it’s atmosphere (making it like earth?)
Hint:
http://www.astro.wisc.edu/~townsend/resource/teaching/diploma/venus-t.gif
Steve, I’m not at all sure how a graph of atmospheric temperature versus altitude either simplifies or settles the discussion. The question at hand is not whether temperature varies with altitude but rather why it does. The strawman you present is that pressure is the determining factor in atmospheric temperature. Where I have a problem is that appears you are saying that the temperature of a gas is entirely determined by its pressure. That, in my mind, would mean that you would assign a specific temperature to a specific pressure in the senses that 100 PSIA CO2 would always exhibit some specific temperature X under all possible situations. That is what is coming across to me in your text.
Let me have a shot at this again. I think what you are trying to say that the adiabatic lapse rate of an atmosphere determines the basic slope of the environmental lapse rate. Other factors such as phase and chemical changes is atmospheric makeup modify that rate but a mixing atmosphere will always thermally settle to something close to the adiabatic lapse rate.
Further more, I believe you are saying that Venus’s surface level atmospheric temperature is higher than Earth’s because of the difference in depth of their atmospheres. The inherent adiabatic lapse rate of the mostly CO2 atmosphere on Venus provides its base temperature profile and since its atmosphere is deeper (and thus at a higher pressure) that lapse rate must force temperatures higher as the depth increases.
Also, I can see an argument that says the low pressure end of the linear part of the atmospheric lapse curve is determined by something like the temperature of the tropopause. The temperature at lower altitudes tends to follow the adiabatic lapse rate and if tropopause temperatures are similar then planets with deeper atmospheres will have higher surface temperatures.
Actually, I can pretty much allow that much. What appears to be missing from your descriptions is what factor causes atmospheric lapse rate to fairly well (but not exactly) follow adiabatic lapse rate. When you use PV=nRT as your basis, my problem is that in an unrestrained situation like a planetary atmosphere, it is entirely likely that a situation can be found in which nRT is constant but P varies. That is -40 degrees C on a mountain top and -40 degrees C in the arctic at sea level. Clearly the pressures will be different with constant temperature with the same mass of air but what will be different is the volume so that PV equals nRT. The total potential energy available from the arctic air mass will be a bit greater because of the higher pressure but the kinetic energy of each molecule will be the same.
All I can see that might cause an atmospheric temperature profile to follow the adiabatic lapse curve is vertical movement within the atmosphere. Somehow, that causes the PV=nRT thing to take over.
I may be getting closer to your point of view but still haven’t been able completely justify that last little bit.
>>> Dr Bill
>>>Nevertheless, the pressure at sea level NEVER changes in
>>>any SIGNIFICANT way from 1000mb, no matter where you
>>>are on Earth, or what time of the day or time of the year it
>>>happens to be.
Errrm. The International Standard Pressure is 1013.25 mb, not 1000. And since I have seen 936 mb, your claim that the pressure does not change significantly is rather wide of the mark.
.
Despite Steve’s best efforts this thread is going off track into unnecessary levels of detail and thereby causing confusion.
Here’s a renewed link to my 2008 article which supports Steve’s contentions in simple (ish) terms. I hope it helps some:
http://climaterealists.com/index.php?id=1562&linkbox=true&position=4
Dr.Bill
Since you are an expert in the field, could you please help me with my suggestion/enquiry here, concerning S-B emission levels of a body when it is immersed in a fluid? (and as to how it might affect claims here of 8,000 and 12,000 W/^2 from the surface).
>>>Spector.
>>If there are no rising or falling parcels of air then there can be no
>>>adiabatic cooling or warming.
Nonsense.
On a fine stable 1040mb day with zero convection and no downward flows there will still be an adiabatic lapse rate in the atmosphere (a dry one in this case). The lapse rate is always present, courtesy of gravity. The reference to ‘adiabatic temperature’ change is because we assume a theoretical parcel of air rising or falling in the calculations, even if the atmosphere is totally stable.
The temperature of the air at 35,000ft does not suddenly warm to 20oc, just because the atmosphere is not convective that day.
.
>>>Steve Goddard
>>>If the atmosphere (of Venus) were thinner (like earth) the
>>>temperature would be much lower. Does anyone actually doubt this?
>>>This isn’t theory. It is observational data.
But not if Venus had the orbit of Mercury – the thinner Venusian atmosphere would actually be much hotter than Earth.
You cannot leave the Sun out of this completely – a thick atmosphere does not equal a high temperature all by itself. Like the diver’s air tank, a thick atmosphere can cool down, if Venus had the orbit of Saturn, for instance.
What you need to concentrate on, Steve, is the fact that the presence of high levels of CO2 on Venus does not end up with a significantly higher lapse rate than Earth. If CO2 was the driver of temperature, the lapse rate on Venus should be much higher.
The thickness of the atmosphere on Venus is a complete distraction. This should be so whatever the atmosphere density on Venus. Even if Venus only had the same atmosphere density as the Earth, if CO2 was the driver of temperature and the Venusian atmosphere was still 99% CO2, then the lapse rate on Venus should be much higher than Earth – but it is not, and so CO2 is not the main driver of surface temperature.
Q.E.D.
.
GaryW
The reason that graph is important is because if you truncate off the bottom 50km, you have something similar to the earth – in spite of the fact that the composition of Earth’s atmosphere is very different.
http://www.astro.wisc.edu/~townsend/resource/teaching/diploma/venus-t.gif
Air that is dense transfers its energy more efficiently to a temperature sensor. The sensor will then read higher at a lower altitude than at a higher altitude as per the wet/dry lapse rate. The temperature is lower at higher altitude because of a lower efficiency of transfer of molecular heat energy NOT necessarily because of a lower temperature.
If you standardize pressure and then read temperature at different altitudes then you have eliminated one variable. I live at 800m and the summers are cooler than the plains because the thinner air is less efficient at transferring its heat energy too me. So the lapse rate is measuring the rate of energy transfer (collision rate) from air to sensor as well as air molecular energy.
Makes it confusing doesn’t it ?
My suggestion would eliminate one variable by eliminating the apparent temperature rise with increasing air density and revealing a truer reading of gas transfer energy.
Pardon me if this sounds like a lecture, but here goes anyway:
George Turner: May 10, 2010 at 3:13 pm
The 1st Law of Thermodynamics can be expressed as dU = dQ + dW, where U is the internal energy of the gas. Part of this is in the form of Kinetic Energy of Translation of the molecules (i.e. their ordinary speed as whole objects). That, and only that, is what determines the temperature of the gas. There are other “degrees of freedom” by which the molecules “store” energy, namely in their Rotational and Vibrational motion. These contribute to the heat capacity of the gas, but not to its temperature. If a molecule has a lot of internal degrees of freedom, it can “suck up” a lot of energy without getting hotter. That’s why some things are harder to heat up than others.
To change the internal energy of a gas, you have to either heat it (the dQ part) or do work on it (the dW part). When you spin up your jet engine, you are doing work on the gas by compressing it, and it will get hotter, because you are making the molecules mover faster by doing work on them, and the gas will stay hot as long as you keep spinning. I have no quarrel with that.
In the atmosphere, however, there is no external agent doing work on the gases. It is only the dQ part that matters, and that heating must occur at the bottom. If you heated some layer higher up in the atmosphere (which happens, for example, in the upper part of the stratosphere where the ozone layer occurs, and which is heated by the Solar Wind (high energy particles) coming in from above), it will NOT start heating the air layers BELOW it, because its density will decrease as it heats up, and it will thus tend to RISE because of the buoyancy effect of our gravitational field.
By the time you get to the top of the stratosphere, however, all but a tiny fraction of the mass of the atmosphere is below you, there is precious little buoyancy left anyway, and thus there is simply a cooling gradient (negative lapse rate, if you wish) downward from the tropopause. In any case, the stratosphere is hotter at the top because that is where the heating occurs, and the troposphere is hotter at the bottom because THAT is where the heating occurs.
Gail Combs: May 10, 2010 at 3:38 pm
The Gas Law DOES apply to our atmosphere, and to that of Venus, and I don’t see that there is anything wrong with your understanding of it. They are both, however, open systems, in the sense that they have no fixed volumes. They can expand and contract, depending on how much heating is done on them. That’s why the tropopause at the Poles is much lower than at the Equator. There is also, as I noted in my response to George Turner, no external agent applying forces to either atmosphere, as there is in the case of jet engines or compressors of any other sort, so it all depends on the heating.
To All
For what it’s worth, I am not trying to be a nuisance to Steve or anyone else, and I am sorry if my comments have created that effect. In my time being a reader of WUWT, and disregarding the trolls, it has been my impression that people want to know the reality of the Science involved in these issues. Many of these issues involve Thermodynamics, which I happen to know something about, partly because I teach it. I am not trying to pontificate. Everything that I have said here, and in earlier posts, can be found in standard textbooks. My purpose is mainly to point out that such knowledge exists, and that people can study it themselves once they are aware of its existence.
/dr.bill
Steve, I do understand what you are saying. You are saying that the troposphere is hotter at the bottom because of adiabatic heating, whereas I am saying that the troposphere is cooler at the top because of adiabatic cooling. I don’t see where we can go from here without a referee – or maybe a few beers. 🙂
/dr.bill
“The weak bands are linear, but with a small slope. That is why they call them “weak bands.””
Did you even look at the Figure in that paper? Did you even try to think about this? This is about the 4th time I’m going to try to get through to you on this one: at current concentrations, the strong bands are saturated at the middle, so increasing concentration causes the radiative forcing due to those bands to increase only logarithmically. The weak bands, as you point out, while increasing linearly are weak, so the logarithmic relationship of the strong bands dominates.
However, every time you double the concentration, you reduce the slope of the logarithmic relationship, but the slope of the linear relationship stays constant. If you had looked at Figure 1 in the linked paper (http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%281977%29034%3C0448%3AARCMSO%3E2.0.CO%3B2) you’d see that the weak bands contribute about 1/5th of the additional radiative forcing from a doubling of CO2, but about 1/3rd of the additional radiative forcing from increasing concentrations by a factor of 10, and, eyeballing the chart, it looks like once concentrations increase by a factor of 20 the weak bands could be contributing about as much to the increase as the strong bands did. A factor of 20 is puny compared to a factor of 2000 for increasing the CO2 in Earth’s atmosphere to 95%, and a factor of 200,000 for the CO2 in Venus’ atmosphere.
So once CO2 exceeds about 1% of the atmosphere, the linear weak band is dominating, and radiative forcing increases _much_ faster than if you assume the logarithmic relationship holds to infinity. (eventually the weak bands saturate too, so the linear regime is limited, but you only need to be linear for a few doublings before returning to log-land to double or triple the final temperature change.
Therefore, your estimate of temperature increase from a 100% CO2 earth is a very large underestimate – “Earth already has plenty of greenhouse gases, so my point is that even if earth went to 100% CO2, the maximum temperature increase would be less than 36C – nothing like Venus. (In reality, it would get cooler because of the loss of water vapour.)” And your estimate of the effects of a reduction of CO2 from 95% of the atmosphere of Venus to a couple % are a large underestimate.
(I’ll also note that the assumption that the climate sensitivity remains constant at 3 over orders of magnitude CO2 concentration changes is also crazy).
First of all Bob, I do not claim to be an expert in anything – except perhaps fly-fishing and getting into trouble with redheads, but I may be deluding myself there too (about the fly-fishing, that is – I definitely get into trouble with the redheads). 🙂 Like everyone else, however, I do know a few things about the topics I make my living at.
Your question is rather complicated, and I’m not sure how to reply to some parts of it. What I can say, however, is that the Stefan-Boltzmann and blackbody issues in general are all about surfaces. Sometimes, however, it is hard to define where the surface of an object is located. The Sun, for example, radiates as if it had an effective radiative temperature of a bit less than 6000K, but it is clearly much hotter than that on the inside, and the light it emits doesn’t come from just one thin layer. The same is true of the Earth. Some of our emissions come straight from the surface, but others come from the molecules in the atmosphere. In essence, very few objects can be classified as true single-temperature blackbodies, or even true greybodies. If we add up all the radiation coming from all levels of something like the Sun or the Earth, or better still – measure them, we can “fit” a curve to it all and thus obtain some kind of effective average temperature. That’s about as good as it gets.
In the case of your block of dry ice, you should note that solid CO2 generally sublimates directly to the gas phase when it melts, and thus you wouldn’t have much of a liquid coating to work with. If you had a block of regular ice, however, both the remaining ice and the liquid skin would emit radiation. If the liquid stayed the same temperature as the solid, they would radiate the same profile. If the liquid part warmed up a bit, it would have a different distribution, but the solid part would also radiate right through the liquid, and one would thus have a “composite” distribution.
It’s little wonder that this has given you headaches! 🙂
/dr.bill
Dr. Bill, excuse me but I am going to jump in here if I can.
I see your point but I also see Stevens point. That is why I’m now confused. Answer a question since you are TD versed and I’m a bit limited there. Take some molecules with a given, constant velocity. Every molecule has the same velocity, always. Take one cubic cm near Venus’s surface and take a one cubic centimeter from 50km up in the atmosphere where Earth like properties exist and the density is some 90 times less. Put a thermometer in each cubic centimeter. What are the two temperatures. Equal? Does the definition of temperature itself depend on density (number of molecules per volume) or only on the average velocity of the molecules.
I’ll reserve another question later depending on your answer.
I withdraw my comment
dr. bill,
If this is all standard science as you claim, why are we subjected to so much misinformation about Venus from the scientific community?
http://www.is.wayne.edu/mnissani/a&s/GREENHOU.htm
wayne: May 10, 2010 at 6:58 pm
Hi Wayne,
Temperature is defined in terms of the average Translational Kinetic Energy of a molecule, nothing more. An individual molecule doesn’t have a temperature – it only has mass and speed. Temperature is an “emergent property” that is valid only with large numbers of things. Pressure is a similar concept. Even in a cubic centimeter, however, there are generally way more than enough particles to satisfy this condition.
In any case, the expression is (1/2)mv² = (3/2)kT. Note that it isn’t just the mass or the speed that matters, but the combination. One consequence of this is that Oxygen molecules in the air are (on average) moving a bit slower than the Nitrogens. If there were some Hydrogens around, at the same temperature, they would by “flying” in comparison with the others. Some molecules, however, are always moving faster than average and others slower, even in a single-component gas. (Look up the Maxwell-Boltzmann velocity distribution.)
If we take your two identical cubic centimeters of molecules and put them anywhere, and keep them segregated, and don’t do anything to them, they will (initially at least) have exactly the same temperature. I’d actually go for a liter myself, unless you have a really tiny thermometer. Measuring changes the result. 🙂
/dr.bill
Steve, that’s from the website of a Liberal Arts College’s Interdisciplinary Studies department. It was written by someone with a bachelor’s in philosophy and a Ph.D. in genetics. It’s also from 1990.
If it was from an atmospheric sciences, or even earth and space sciences department, you might get away with a block quote like that as being a sign of scientists lying to people.
On a different aspect of this topic, it would be welcome if a geologist could comment on these surface images provided by Colin. Have a look at
Is this old lava flow?, why are the surface rocks flattened?
Hi Steve,
Personally, I’d call BS on all of that, and have been doing so on this particular issue for over a decade. As time goes by, however, I find that not as many of my colleagues are spouting such nonsense all the time. Those who do are generally on the receiving end of some kind of “benefits”. It’s possible, of course, that some people, scientists or otherwise, actually believe all that bull[snip] (save you the job CTM), but anyone who has actually looked at the matter realizes that it’s all a money and power-grabbing hoax.
I have lived through the “Coming Ice Age”, the “Population Bomb”, and about a dozen other major ones since becoming an adult. All they have in common, in my opinion, is the desire to funnel large amounts of money and resources into a small number of pockets, and shape the world into a certain mould, preferably without the presence of much of the 3rd world. Perhaps I’m just cynical, but that’s what my gut (and a lot of reading) tell me.
/dr.bill
Speaking of Venus, it was incredibly bright tonight – due to that very high albedo.
Keith Minto
Basaltic lava is pretty much the standard fare for planets besides earth. In order to form granites (continental crust) you need to have water in abundance.