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The surface temperatures of Mars are about -87C to 5C, which is much colder than that of Earth's. If Mars has 95% carbon dioxide, which is a Greenhouse gas, why is the surface of Mars so cold? Shouldn't it trap heat and render it hot?

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    $\begingroup$ You should be careful drawing conclusions from percentages. Mars's atmosphere is a lot sparser than Earth's. $\endgroup$ – called2voyage Oct 9 at 11:28
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    $\begingroup$ Venus only has 1.1% more CO2, why's it so hot? (spoiler: it ain't because its closer) "Average surface pressure‎: ‎93 bar" ... that's 93 times that of Earth. - Density, not percentage. $\endgroup$ – Mazura Oct 9 at 23:31
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    $\begingroup$ @Mazura: Well, it's partly because it's closer; if Venus and Earth had exactly the same atmosphere, Venus would still be warmer (although not by nearly as much) due to its closer proximity to the sun (and it's been postulated that this might actually be the reason Venus got tipped into a runaway greenhouse effect, while Earth hasn't [yet]). $\endgroup$ – Sean Oct 10 at 0:59
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    $\begingroup$ Sounds like a setup for a Chuck Norris joke. "Chuck Norris has already been to Mars, that’s why there are no signs of life." "Chuck Norris roundhouse kicked Mars and knocked the atmospere off." $\endgroup$ – CrossRoads Oct 10 at 14:36
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    $\begingroup$ @CrossRoads: Pretty much. The Sun is playing Chuck Norris, doing repeated roundhouse kicks, with Mars undefended due to low gravity and no magnetic field. $\endgroup$ – Nathan Hughes Oct 10 at 19:48
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Firstly, Mars has a mean distance from the Sun of 1.524 AU, so by the inverse square law the energy it gets from the Sun is about 40% of what the Earth gets.

But the main reason that Mars is so cold is that its atmosphere is very thin compared to Earth's (as well as very dry, see below). From Wikipedia Atmosphere of Mars:

The atmosphere of Mars is much thinner than Earth's. The surface pressure is only about 610 pascals (0.088 psi) which is less than 1% of the Earth's value.

In comparison, the mean surface pressure on Earth is 101,300 pascals. So the atmosphere of Mars is barely more than a vacuum compared to Earth's.

So even though the Martian atmosphere is over 95% carbon dioxide, there simply isn't enough of it to trap much heat.

Although carbon dioxide is a greenhouse gas, the predominant greenhouse gas on Earth is actually water vapour. However, water is usually cycled in and out of the atmosphere very quickly in response to temperature and pressure changes. Carbon dioxide is a problem because it stays in the atmosphere for a long time, and its presence shifts the equilibrium temperature upwards from that of the plain water cycle.

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    $\begingroup$ But Earth's atmosphere only has 0.04% CO$_2$, so that's an order of magnitude less than Mars, in absolute values. $\endgroup$ – pela Oct 9 at 12:12
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    $\begingroup$ @pela True, and that's why I added that info about water. But I suppose I shouldn't make that a "BTW" section. $\endgroup$ – PM 2Ring Oct 9 at 12:16
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    $\begingroup$ @pela <cue joke about blind astronomer> :) $\endgroup$ – PM 2Ring Oct 9 at 12:54
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    $\begingroup$ @pela: So what would the temperature of Mars be if it had no atmosphere, or if the atmosphere was composed entirely of non-greenhouse gasses like nitrogen or argon? $\endgroup$ – jamesqf Oct 10 at 16:40
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    $\begingroup$ @jamesqf It would be roughly 5 K colder, which is the difference between Mars' effective temperature and its average surface temperature. For Earth, that difference is 33 K (see my answer below for references). $\endgroup$ – pela Oct 11 at 13:33
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I'm just going to expand and deepen on what the other answers already said.

In the following I contrast the atmospheric transmission ($T$) and absorption ($A$, which is $A=1-T$) of Mars and Earth. The Mars plot (top) is from Prof. J. Irwin via this review by P. Read et al. 2015 and the terrestrial data (bottom) is from wikipedia.

The plots of $A$ and $1-T$ area easily matched by eye when comparing $T=0$ and $A=1$ for Mars and Earth. To make this superclear, I've added arrows to mark the $CO_2$ absorption bands.
Additionally, the plot for Mars has the energy spectra of the incoming solar (T=5800 K) and outgoing infrared (T=216 K) radiation inserted. Earth's outgoing infrared radiation from the surface would be at T=300K, and in this graph at nearly the same position. For a strong greenhouse effect, an atmosphere must absorb the majority of this outgoing peak.

Mars and Earth atmospheric windows

From this we learn a few things:

1.) Looking at Earth, we see that there are several important greenhouse gases in the atmosphere, but the most significant one is water.
2.) A significant portion of Earth's greenhouse peak is absorbed by $H_2O$+$CO_2$, while only a small fraction is absorbed by the martian atmosphere, visually explaining what already @pelas answer noted, that Mars has a weak greenhouse effect.
3.) If comparing the width of the $CO_2$ absorption bands very carefully, then one will note, that the $CO_2$ bands for Earth are broader than on Mars, although Earth's atmosphere has much less total $CO_2$ in it! This is an effect of the so-called 'pressure broadening' of atmospheric lines, enhancing their greenhouse capability significantly. Pressure broadening becomes important in atmospheres at pressures of about $0.1-1 \; bar$ (rough rule-of-thumb). This is another reason why the low mass/pressure of the martian atmosphere cannot create a strong greenhouse effect, as the pressure is around $0.01 \; bar$ on the surface.

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    $\begingroup$ While it has little to do with the question, I'm curious why Methane is such a strong greenhouse gas with a relatively small absorption spectrum in the chart above. Nice details by the way. $\endgroup$ – userLTK Oct 9 at 14:10
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    $\begingroup$ @userLTK There is not enough methane in the atmosphere to saturate those absorption bands (yet). The optical depth $\tau_{\lambda}$ at a given wavelength $\lambda$ and height $z$ is $\tau_{\lambda}=\int_{\infty}^{z} dz' n(z') \kappa(z')$, with $n(z')$ being the local number density and $\kappa(z')$ the opacity per mole, which is again a function of the temperature. Only at $\tau_{\lambda}>1$ is the photon mean-free path smaller than the atmospheric scale (which means the photon can't escape), and we see that both the absorption strength (given by the opacity) and the amount of gas contributes. $\endgroup$ – AtmosphericPrisonEscape Oct 9 at 14:42
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    $\begingroup$ @MichaelS: Right, this band at $15\mu m$ is opaque, so the outgoing radiation gets trapped there. But this is only a small fraction of the total black body energy, so it doesn't contribute to a strong greenhouse effect. $\endgroup$ – AtmosphericPrisonEscape Oct 10 at 10:18
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    $\begingroup$ It might improve this answer is you explicitly mention that the bottom graphs are Earth, the top is Mars, and the Mars absorption graph is "upside down" compared to Earth graphs (Earth graphs "absorption and scattering", while Mars measures "transmission"). Great answer! $\endgroup$ – Yakk - Adam Nevraumont Oct 10 at 17:22
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    $\begingroup$ @Yakk-AdamNevraumont: Thanks, but that's literally my first two sentences. $\endgroup$ – AtmosphericPrisonEscape Oct 10 at 18:06
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Mars does have a greenhouse effect, only somewhat weaker than Earth's.


Mars' atmosphere is very dilute, with a with a surface pressure only 0.6% of Earth's. So even if 95% of it is CO2, that's not a lot. However, it is actually a higher absolute abundance of CO2 molecules than on Earth, which only has a CO2 abundance of 0.04% (by volume; e.g. NOAA, corresponding to roughly 0.06% by mass).

The exact calculation depends on how the atmosphere decreases in density with height (the "scale height"), but for an order-of-magnitude calculation, we can use the total masses of the atmospheres $M_\mathrm{atm,Earth} = 5.15\times10^{18}\,\mathrm{kg}$ (Trenberth & Smith 2004), and $M_\mathrm{atm,Mars} = 2.5\times10^{16}\,\mathrm{kg}$ (NASA). The total amount of CO2 in Mars' atmosphere, compared to Earth's, is then $$ \frac{M_\mathrm{CO_2,Mars}}{M_\mathrm{CO_2,Earth}} = \frac{95\%\times M_\mathrm{atm,Mars}}{0.06\%\times M_\mathrm{atm,Earth}} \simeq 7.9, $$ i.e. almost an order of magnitude higher. The fact that Mars is smaller than Earth means that this number is probably somewhat larger.

Roughly 10–20% of the radiation emitted by Mars' surface is absorbed in the atmosphere (Haberle 2015). One way to quantify the greenhouse effect is the difference between the effective temperature and the planet's average surface emission temperature. For Earth, the difference is 33 K, while for Mars it's a much lower 5 K (but must have been much higher in the past in order for liquid water to exist).

So why is the effect so much larger on Earth? Well, CO2 is not the only agent of the greenhouse effect. Other gases — e.g. methane, nitrous oxide, and ozone — add as well. But actually the largest contribution comes from water vapor. Exactly how much is a matter of heated debate, I think, but it's probably not controversial to say that at least 1/3 of the effect is due to water (RealClimate says 36–66%).

Water molecules are less reflecting than CO2, but constitutes 0.3% of Earth atmosphere. In contrast, only 0.03% of Mars' atmosphere is water vapor (e.g. Trokhimovskiy et al. 2015).

In summary, the smaller greenhouse effect, and the fact that Mars is roughly 1.5 times as far from the Sun as Earth, and hence receives less than half of the incident radiation, is the reason that Mars is so cold.

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    $\begingroup$ "However, it is actually a higher absolute abundance of CO2 molecules than on Earth, which only has a CO2 abundance of 0.06% (by mass)." citation needed? AFAIK there is still less CO2 per unit area and in the air column on mars than on earth, though it may still be in the same order of magnitude, and actual measurements for the mass for both earth and mars for CO2 is scarce from google searches. $\endgroup$ – opa Oct 9 at 20:50
  • $\begingroup$ @opa References for the abundance of CO$_2$ in Earth's atmosphere are plentiful, see e.g. NOAA (it's 0.04% by volume, so roughly 0.06% by mass). As for the comparison, I just did the calculation $\rho_{\mathrm{CO}_2,\mathrm{Earth}} / \rho_{\mathrm{CO}_2,\mathrm{Mars}} \simeq 0.06 / (0.95\times0.006) \sim 10$. I used the Mars-to-Earth pressure ratio, which is slightly wrong — I should instead use the density ratio, which actually increases the ratio a bit. $\endgroup$ – pela Oct 10 at 9:51
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    $\begingroup$ @opa I added references and calculations :) $\endgroup$ – pela Oct 10 at 10:52

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