89. The Greenhouse Effect on Mars

In my previous three posts I have explained how the Greenhouse Effect (GHE) works on Earth, and how it is affected by changes to the carbon dioxide (CO₂) concentration in the atmosphere. The problem with studying the GHE on Earth, though, is that its operation is complicated by the presence of large amounts of water vapour in the atmosphere. As water vapour also has a broader absorption band and higher atmospheric concentration than CO₂, this means that changes in the CO₂ concentration are less important than they would be otherwise. If we want to understand and measure the GHE just due to carbon dioxide, then we need an environment with high levels of CO₂ but low levels of other greenhouse gases. In this respect one of the best places to study is Mars. 

The Atmosphere of Mars

The Martian atmosphere has some similarities with Earth but many differences. It contains many of the same gases (nitrogen, oxygen, water vapour, argon, CO₂), but the proportions are vastly different. The atmosphere of Mars is 96% CO₂ with about 2% nitrogen and 2% Argon (although different pages on Wikipedia give slightly different values such as 95% CO₂ and 3% nitrogen). There are also trace levels (< 0.1%) of other gases such as water vapour (210 ppm) and oxygen (0.15%). 

The other main difference in terms of the atmosphere is the pressure at the surface. At 610 Pa this is only 0.60% of the surface pressure on Earth, and as about 96% of this is CO₂, this means that there is a surface density of 3610 mol/m² of CO₂ on Mars compared to only 150 mol/m² on Earth. So any outgoing radiation from the surface of Mars has 24.06 times as much carbon dioxide gas to penetrate, in order to escape into outer space, compared to on Earth. What this means in practice is that the Greenhouse Effect in the Martian atmosphere should be easier to analyse because it can only have one source - CO₂.

 

The Energy Balance for Mars

Mars is also approximately 52% further from the Sun than is Earth, so one might expect that to mean that its surface is colder. This is true, but not as much as it should be based on distance alone. 

The solar radiation flux entering the Martian atmosphere is only 586 W/m² compared to the 1360 W/m² that irradiates the Earth. As this radiation is spread over a surface area (4πr²) that is four times greater than the cross-sectional area of the planet (πr²) in each case, this means that the average radiation flux at the Martian surface is 143.5 W/m². Yet this is only 22% less than the 184 W/m² that reaches the Earth's surface. This is because nearly 50% of incident radiation on Earth is either reflected by clouds or is absorbed by ozone and water vapour in the upper atmosphere. But there are no clouds, ozone or significant water vapour on Mars.

Then there is the issue of surface albedo or Bond albedo. This is the proportion of incident radiation that is reflected by the planet back out into space without being absorbed, either from clouds or the surface. For Earth this is about 31%; for Mars it is only 25%. This means that the average surface absorption on Mars is 108 W/m² compared to 161 W/m² on Earth. So while Mars only receives 43% of the solar radiation that Earth does, after absorption and reflection the Martian surface receives 67% of the radiation that the Earth's surface does. That is a relative increase of more than 50% for Mars which partially compensates for its greater distance from the Sun.

Calculating the Surface Temperature of Mars

If we now invoke the Stefan-Boltzmann law (see Eq. 13.1 in Post 13),

I = σT⁴

(89.1)

where I is the surface radiation flux, σ is the Stefan-Boltzmann constant, and T is the surface temperature in kelvins, we see that a surface radiation flux of 108 W/m² equates to a mean surface temperature on Mars of 209 K (or -64°C). Yet the true mean temperature is thought to be about 215 K (or -58°C). The difference is due to the Greenhouse Effect. 

For comparison, on Earth a solar flux at the surface of 161 W/m² would equate to a mean surface temperature of 231 K (or -42°C), yet the true mean temperature is about 289 K (or +16°C). So the GHE on Earth adds 58°C of warming while on Mars it only adds 6°C. Yet there is almost twenty-five times more carbon dioxide on Mars (3690 mol/m²) than on Earth (150 mol/m²), so you would expect the greenhouse effect due to CO₂ to be stronger. But how much stronger? The answer is: not very. In fact it is significantly less than the actual measured value on Earth.

Calculating the Strength of the GHE on Mars

In Post 87 I showed how the width of the CO₂ absorption band at 15 µm can be determined using the known concentration of CO₂ (N_o), its scattering or absorption cross-section (σ_s), the quantized frequency of rotation of the CO₂ molecules (B), and the temperature (T). From this it is possible to estimate the critical CO₂ concentration needed to absorb most of the infra-red radiation (N_th). The term N_th can be estimated to be 0.5 mol/m² based on the value of σ_s, while N_o will be about half the total CO₂ concentration, so N_o = 1845 mol/m². From this we can estimate the maximum number of excited rotational states in the absorption band, J_th, using (see Post 87)

(89.2)

where k is Boltzmann's constant, h is Planck's constant and Z is a normalization constant (see Eq. 87.2 in Post 87) equal to 185.5 in this case. The ratio term kT/hB has the value 185.3 (as hB = 0.1 meV and T = 215 K) and is always approximately equal to the Z value. 

The result we obtain is that J_th = 36.3, which, given that the spacing of the bands is 0.2 meV, means that the absorption band has a width of 14.52 meV and extends from a wavelength of 13.78 µm to 16.44 µm. This compares to a calculated range of 14.00 µm to 16.14 µm for the same band on Earth, although the measured width on Earth is actually found to be from about 13.35 µm to 17.35 µm. So even though the calculated width of the 15 µm absorption band for Mars is slightly larger than the equivalent for Earth, it is not significantly greater. But it is significantly less than the measured band width on Earth.

 

Fig. 89.1: The electromagnetic emission spectrum for the surface of Mars at a mean temperature of 215K (blue curve) together with the absorption profile due to CO₂ between 13.78 µm and 16.44 µm (red curve).

The impact of the high Martian atmospheric CO₂ concentration on the radiation feedback is demonstrated in Fig. 89.1 above. The red curve indicates the proportion of the outgoing infra-red radiation (blue curve) that is reflected by the CO₂ molecules and it amounts to f = 12.5%. This is slightly more than the 10% seen for the GHE due to backscattering from CO₂ on Earth (see Post 87) despite the absorption band being further from the peak in the emission spectrum due to the lower surface temperature on Mars.

Calculating the Temperature Rise

The radiation feedback of  f = 12.5% shown in Fig. 89.1 equates to a 14.3% increase in the surface radiation and also of T⁴ (because of Eq. 89.1), which then equates to a 3.4% increase in T. So if the initial surface temperature on Mars without the GHE was 209 K, the temperature with the GHE included will be 3.4% greater, or 216.1 K. This means that the temperature rise at the surface due to the Greenhouse Effect is expected to be 7.1 K, which is pretty close to the observed value of around 6 K (or 6°C). The reason for the small difference could be the limited availability of accurate Mars temperature data, or the uncertainty in the value of the CO₂ absorption cross-section, σ_s.

We know how much radiation from the Sun is arriving at Mars to high accuracy, but knowing how much is being absorbed by the planet surface is more difficult as this depends on an accurate measurement of the Bond albedo. However, conventional astronomical telescopes should be able to measure the reflected radiation to pretty good accuracy as well. That allows us to estimate the expected mean surface temperature without the GHE to fairly high precision. The problem is knowing what the actual surface temperature is with the GHE in operation. This is a difficult enough calculation to do on Earth where we have over 16,000 weather stations measuring the surface temperature on a daily basis, and numerous satellites in orbit. Sadly, none, or very little, of this exists for Mars.

So far in this blog post I have assumed a value of 215 K for the mean surface temperature of Mars, but some reports have put it as high as 225 K (or as low as 210 K). In which case Z = 194.15 and J_th = 37.0. This leads to a 15 µm band stretching from 13.76 µm to 16.47 µm, and a feedback factor of f = 13.0%. Under these circumstances the warming from the Greenhouse Effect increases slightly, but only to 7.7°C. 

 

A Comparison with Earth

For comparison, it is instructive to hypothecate the extent of warming on Earth if its atmosphere also contained 3690 mol/m² of CO₂. In that case Z = 249.3 and J_th = 41.6, which leads to a 15 µm band stretching from 13.62 µm to 16.67 µm, and a feedback factor of f = 14.1%. The resulting predicted temperature rise due to CO₂ would be 10.74 K, which is 3.26°C less than the 7.48°C rise currently predicted for Earth as was shown in Post 87). So a twenty-five fold increase in the CO₂ concentration would only result in a 3.26°C temperature increase, although as I showed in Post 87, masking by water vapour would probably reduce this by 75% to only 0.8°C. 

It is a point of note that the density of CO₂ molecules on Mars (3690 mol/m²) is more than double the combined density of all the greenhouse gases on Earth (1560 mol/m²), yet it results in a temperature rise of 5°C - 7°C that is almost ten times less than the 58°C observed for Earth. This is mainly because most of the GHE on Earth is due to water vapour as the width of the CO₂ absorption band is so much less than that for water vapour. Even increasing the concentration of CO₂ on Mars by a factor of twenty-five cannot appreciably change this.

Summary and Conclusions

What I hope I have shown in this post is that Mars is a good test bed for studying the Greenhouse Effect (GHE). Knowing only its albedo, the atmospheric concentration of CO₂, and the intensity of radiation arriving from the Sun, it is possible to accurately predict the temperature rise due to the Greenhouse Effect. This I have predicted to be about 7°C, in close agreement with the current estimate based on observational data (6°C). And this is despite the significant uncertainty over the true measured value of the mean surface temperature on Mars.

The reduced GHE on Mars relative to the Earth occurs despite its much higher (i.e. 24 times greater) atmospheric CO₂ concentration. This in turn suggests that the increasing levels of atmospheric CO₂ we are currently seeing on Earth will produce only slight temperature increases in the future. 

Yes, Mars has its own complicating factors. Heat retention on Mars is limited by the thin atmospheric blanket compared to Earth. This means that the planet does not retain heat very well, but conversely it means that the atmosphere will warm quickly when heated by the Sun. For this reason it may be better to consider Mars under direct solar illumination in daytime. Under these conditions the peak solar flux at the surface of Mars will be four times greater than stated above, or 432 W/m². This will equate to a peak surface temperature of 295 K (or +22°C). Yet the actual maximum temperature is reported to be around 303 K to 308 K (or +30°C to +35°C). So on this measure the warming from the Greenhouse Effect in daytime near the equator appears to be in the range 8°C to 13°C. Yet the predicted value based on a calculation of the width of the 15 µm absorption band is found to be 10.9°C, in other words in the mid-range of the observed values. Once again this is still much less than the total warming seen on Earth and comparable to the contribution to Earth's GHE seen just from CO₂.

15. The truth about sea level rise

One of the most emotive and alarmist claims made by climate scientists is that global warming will lead to a catastrophic sea level rise (SLR) that will submerge major cities and lead to an unprecedented humanitarian crisis and global extinction event.

Fig. 15.1: Britain's favourite polar bear - Peppy.

On the face of it this seems quite plausible, even likely. We see images of collapsing ice shelves and retreating glaciers on an almost daily basis. We see icebergs the size of cities being calved off from Antarctica and then polar bears looking forlorn on icebergs the size of a lifeboat, like something from a Fox’s glacier mint advert. So what is the reality?

Fig. 15.2: Life imitating art.

There are two principal ways that sea levels might change: either the amount of water in the sea changes, or the existing sea water changes its density. The former can happen if ice caps on Greenland or Antarctica melt. It cannot happen through the melting of sea ice because of Archimedes' Principle as I explained in Post 2, nor for the same reason can an increase in sea ice change the sea level. One alternatively mechanism is through increased evaporation of condensation, but that requires the humidity of the atmosphere to change. As for density changes, these are governed mainly by changes in the temperature of the sea water.


Scenario 1: Thermal Expansion

In the case of rising sea temperatures it is not the current global temperature that is important per se, but the temperature history and the thermal budget of the Earth. Global temperature rises can only raise sea levels directly (excluding from ice melting) by thermal expansion of the sea water. As I pointed out in Post 2, the coefficient of thermal expansion by volume for water is 0.000207 per degree Celsius or 207 ppm/°C. So a column of water 1000 metres or 1 km high will increase in height by only 20.7 cm if its temperature increases by 1 °C (or 1 K where K is the unit of absolute thermodynamic temperature - the kelvin).

But there is another factor we need to consider - the total heat or thermal energy required to do this. This is because this energy needs to come from somewhere, and once used it remains trapped in the water. It is, therefore, energy that has been sequestrated, the effect of which is to create an imbalance between the amount of energy the Earth receives from the Sun, and the amount it emits back out into space. This difference can only come from the net energy imbalance of the Earth’s energy budget.

In the last post we saw that this imbalance is currently estimated to be as much as (and no-one is saying it is more than) 0.9 W/m². If this figure is true (and as I pointed out, there is enormous uncertainty over its accuracy), and if it had been constant over the last 100 years (which is very unlikely), it would imply that the Earth has absorbed a total of 4.591x10¹⁴ joules of energy every second in that time period, or 1.45 x 10²⁴ J in total. Of course that is the upper limit of what is likely. No-one seriously thinks the Earth’s energy imbalance has always been 0.9 W/m². So the average over the last 100 years must be considerably less, and probably less than half.

Whatever the value, though, this heat will increase the temperature of the oceans. The question is, by how much, or to what depth?

As the specific heat capacity of water is 4200 Jkg^-1K^-1, the amount of energy required to heat 1 kg of water by 1 K (or 1 °C as these temperature changes are the same) will be 4200 J. If our 1 km high water column has a cross-section of 1 m², then it will contain 1000 tonnes of waters. Therefore, the total energy required to raise the temperature of the entire column by 1 °C will be 1,000,000 times greater than 4200 J, in other words 4.2 x 10⁹ J. As 70.8% of the Earth’s surface is covered by the oceans, the total volume of water down to a depth of 1 km will be 3.61 x 10¹⁷ m³. The total mass will be 3.61 x 10²⁰ kg, and the total heat capacity will be 4200 times higher still at 1.52 x 10²⁴ J/°C. So the mean temperature rise of the oceans down to a depth of 1 km over then last 100 years will be (at the absolute maximum) 1.45 x 10²⁴ ÷ (1.52 x 10²⁴) = 0.95 °C.

So, if we assume that the Earth’s energy imbalance over the last 100 years has been 0.9 W/m² everywhere and at all times, and if we assume that this heat has all ended up in the ocean, and it has heated the top 1000 m only, then the temperature rise of that top layer of water will be 0.95 °C. Some may find this number suspiciously close to the value claimed for global warming of about 1 °C per century. In other words, that climate scientists have worked backwards. They have assumed that the oceans must heat up by the same amount as the land over the same period, and to a depth of up to 1000 m, and worked out the amount of heat required to do this. From this they have inferred an imbalance in the Earth’s thermal budget rather than measured it.

Whatever the sequence of events, the resulting sea level rise (SLR) will be 197 mm (i.e. 0.95 x 207). If the warming layer of the ocean is thinner (say 500 m) but the surface energy imbalance is still 0.9 W/m², its mean temperature rise will be greater (an unlikely 1.90 °C for a 500 m thick layer), but the SLR will be the same, in other words a massive 1.97 mm per year. So what is clear is that the maximum sea level rise that can occur depends on the energy imbalance and not the water depth. In reality, the surface energy imbalance may be less than 0.9 W/m² (G. L. Stephens et al., Nature Geoscience 5, 691–696 (2012) suggest 0.6 W/m² as I pointed out in Post 13) and has in all likelihood got worse over time. So while it may be 0.9 W/m² now, it has probably averaged less than half of 0.9 W/m² over the last 100 years as global temperatures have risen. In which case the SLR has probably been less than 100 mm over the last century (or less than 66 mm if Stephens et al. are correct). But what about the future?

Assuming that the surface energy imbalance remains at 0.9 W/m² for the foreseeable future, and if we now assume that only that portion of the surface energy imbalance over the oceans is actually absorbed by the oceans, then the heat absorbed by the oceans each year will be 28.4 MJ/m². If this were absorbed by a column of water 1000 m deep it would result in a temperature rise of 6.76 mK. If it were instead absorbed by a column of water only 100 m deep it would result in a temperature rise of 67.6 mK. Either way, the resulting thermal expansion would be 1.40 mm per year. Even over 100 years this is a long way short of the 10 m rise some doom-mongers are projecting, and on its own is unlikely to pose a major threat to human civilization or the planet.

But thermal expansion is just one component of the overall problem, because not all the 0.9 W/m² (or 0.6 W/m²) need end up in the oceans. Some may instead end up melting the ice caps.


Scenario 2: Melting Ice Caps

In this scenario there are a number of factors that we need to identify and address. Firstly, where is the ice? If it is floating on the ocean surface then it cannot add to the sea level increase when it melts, despite 9% of the ice being above the water line. This is because of Archimedes’ principle as I explained in Post 2. It can, though, sequestrate energy and actually reduce warming elsewhere. The only ice that can increase sea levels when it melts is ice that is on land. This is found mainly on Greenland (3 million cubic kilometres) and Antarctica (30 million cubic kilometres).

Then there is the question of the ice temperature. Before it can melt it needs to be heated to 0 °C, yet the mean temperature of the ice on Antarctica is about -50 °C. In order to raise the ice temperature to melting point would require 3.20 x 10²⁴ J of energy (the specific heat capacity of ice is 2108 Jkg^-1K^-1 and its relative density is 0.92).

Next, you need to melt the ice. This will require an energy input of 1.01 x 10²⁵ J (the specific latent heat of ice is 332 kJ/kg); and then you need to heat it to the ambient temperature of the Earth, about 15 °C, otherwise it will cool the oceans and your global temperatures will go down. This requires another 1.91 x 10²⁴ J of energy. So the total energy required is 1.52 x 10²⁵ J. Given that the amount of power available to achieve this is at most only 0.9 W/m², that means it would take at least 1100 years to occur. But even that assumes that all the power from the Earth’s energy imbalance could be channelled somehow into melting the ice caps and nothing else.

In reality most will initially go into the oceans. As Antarctica and Greenland comprise only 3% of the Earth’s surface area, a more realistic estimate is that it would take up to 30,000 years, by which time we would be in the next ice age. Recent studies of Greenland appear to confirm this as they show Greenland has lost less that 0.05% of its ice in the last 10 years, but this may just be cyclical. 

One final point of note: while the total volume of ice on Antarctica is 30 million cubic kilometres, almost a third of this is below sea level. The net result is that if all the ice on Antarctica and Greenland were to melt, sea levels would rise by 65 metres not 91 metres. Yet over 30,000 years this will amount to a rise of only about 2 mm per year.


Scenario 3: Evaporation

The final possibility is that the sea water might just evaporate into the atmosphere leading to a loss of volume in the sea rather than a gain. Approximately 0.4% of the atmosphere by volume is water vapour. But if all this water were to suddenly condense out of the atmosphere it would only add 3.6 cm to the depth of the oceans. Given that the ambient temperature at the surface of the Earth is 15 °C, while the humidity at 0 °C would be expected to drop to near zero, this suggests that an increase of 1 °C in the surface temperature of the air would lead to a decrease in sea levels of around 2.4 mm. That is a pretty crude estimate, though, and assumes that the vapour pressure of water increases proportionately with temperature from its freezing point. A more accurate estimate can be made using the Clausius-Clapeyron equation.

Fig. 15.3: Schematic of phase boundaries on a P-T diagram.

This equation (see Eq. 15.1) relates the the slope of a phase boundary in a pressure-temperature diagram to the thermodynamic temperature, T, the molar latent heat for the phase change, L, and the change in molar volume across the boundary ∆V. An example of a phase diagram is shown in Fig. 15.3 above.

(15.1)

For a change from liquid to gas (as in evaporation) the term ∆V should be the difference in molar volumes between the water in the liquid phase and the vapour in the gas phase. However, as the volume of the vapour at the relevant pressures we are likely to encounter (i.e. around atmospheric pressure) is so much greater than it is for water (in fact by more than a factor of 1000) we can use the ideal gas law in Eq. 15.2 to substitute the molar volume of water vapour V for ∆V on the basis that the molar volume of the liquid is negligible.

(15.2)

In Eq. 15.2 the term V is the volume of one mole of water vapour at a pressure P and a temperature T. The term R is the molar gas constant where R = 8.314 Jmol^-1K^-1 and n is the molar density in mol/m³. The approximation of V for ∆V allows us to make a substitution from Eq. 15.2 into Eq. 15.1 to generate Eq. 15.3 which is now a function of only two variables, P and T.

(15.3)

This allows us to relate fractional changes in the pressure of the gaseous phase to fractional changes in temperature along the phase boundary. The differential in Eq. 15.3 implies that for small changes in P and T the following relation holds

(15.4)

while Eq. 15.2 yields the following relation between P, T and n.

(15.5)

Equating Eq. 15.4 with Eq. 15.5 gives the result for the fractional change in molar concentration of the vapour that will occur due to  evaporation across the phase boundary for a temperature change ∆T.

(15.6)

The relation in Eq. 15.6 allows us to estimate the change in water vapour concentration n as the temperature T changes. So for example, if the temperature T = 288 K and the latent heat of evaporation of water is 40.8 kJmol^-1K^-1, then a temperature rise of ∆T = 1 K will yield a fractional change of water vapour concentration of 0.0557. That in turn implies a total fall in sea level over the period of the temperature rise (which is about 100 years) of 2.0 mm (= 0.0557x36). Reassuringly, this is not that dissimilar to our original estimate of 2.4 mm, thus demonstrating two important points. Firstly, that the result is robust and consistent. Secondly, that our original back-of-the-envelope approximation, much loved by physicists everywhere, did not let us down.

The advantage of calculating the fractional change in n using Eq. 15.6 rather than calculating ∆n directly is that it means that we can avoid the complication of working out the relative humidity. The Clausius-Clapeyron equation, strictly speaking, only applies to closed systems in equilibrium, i.e. at 100% relative humidity. In open systems, such as the Earth's atmosphere above large oceans, the humidity is always less than the maximum. But Eq. 15.6 effectively removes the issue of relative humidity as it just introduces an additional scaling term that applies more or less equally to n and ∆n. Therefore it cancels out in Eq. 15.6. All of this may be somewhat pedantic, however, as the sea level fall due to evaporation is a full two orders of magnitude less than the previous two effects considered.


So the conclusion is this: prophesies of apocalyptic rising sea levels and submerging cities are still just alarmist nonsense. The physics proves that there is currently not enough energy available to achieve this on the timescale that some climate scientists predict, at least not yet. Thermal expansion and melting ice caps will each add not much more than 2 mm per year to sea levels. A recent paper by Anny Cazenave et al. (Advances in Space Research 62(7) 1639-1653 (2018) ) puts the sea level rise (SLR) from all sources at about 3.5 mm per year for the period 2005-2015  (see Fig. 15.4 below). These numbers do appear to be more consistent with the surface energy imbalance of 0.9 W/m² reported by Trenberth and co-workers rather than the 0.6 W/m² of Stephens et al., particularly the thermal expansion component.

Fig. 15.4: Possible breakdown of different contributions to sea level rise (1993-2015) from Cazenave et al.

One of the striking features of Fig. 15.4 in my view is the low contribution to sea level rise from ice melt in Antarctica compared to that from glaciers. Three explanations spring to mind. Firstly, there is probably more warming in the Northern Hemisphere because that is where the heat is being generated. Secondly, the glaciers in Europe are very close to the source of that heating. And thirdly, the ice in Antarctica is much colder than that in alpine glaciers, and so requires more heat to melt it. So, as I pointed out in the last post, this could mean that alpine glaciers will continue to recede, not because of CO₂ emissions, but because of local human industrial activity that leads to surface heating of the local environment, and thus a temperature rise of more than 0.3 °C above pre-industrial levels.