In April 2011 NASA announced that the Mars Reconnaissance Orbiter had found a large, buried deposit of frozen carbon dioxide at the planet's south pole. It has a volume of about 12,000 cubic kilometers and holds up to 80% as much carbon dioxide as today's Martian atmosphere.

What would be the most efficient way to vaporize this deposit?

Laser beams? Hot air? Nuclear heat?

And when a part of the deposit is uncovered, will not sublime much of the CO$_2$ automatically in southern summertime?

A good reason for putting that much CO$_2$ into the atmosphere would be that liquid water could stay during summer days on the surface of the low regions on Mars.

Another good reason would be that oxygen producing cyanobacteria would have a better chance surviving the harsh conditions on Mars !

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    $\begingroup$ Explode or vaporize? Explosions tend to fragment things & throw them around, not vaporize them. Do you want to redistribute the frozen CO2 on the ground (hence explosion) or vaporize it & add it to the atmosphere as a gas? $\endgroup$ – Fred Aug 8 '18 at 9:35
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    $\begingroup$ By way of information. I get the impression you may be interested in terra forming Mars by putting more CO2 into the atmosphere. You may be interested in a recent article Sorry Elon Musk, but it’s now clear that colonising Mars is unlikely – and a bad idea. According to a new study published in Nature Astronomy there isn't enough CO2 on Mars $\endgroup$ – Fred Aug 8 '18 at 14:40
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    $\begingroup$ Once it's vaporized, what's going to keep the solar wind from blowing it all away in the absence of a megnetosphere? $\endgroup$ – Don Branson Aug 8 '18 at 19:16
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    $\begingroup$ @DonBranson Yes that's a shame, but that seems to happen over a long, long time, so i would say that's for the next generations to solve ! We have the excuse that we have not the means to prevent this right now. $\endgroup$ – Cornelis Aug 8 '18 at 19:29
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    $\begingroup$ @Fred The problem would be to get the frozen CO2 to the reactor. I think a tube system would be needed to both put heated CO2 into the deposit to vaporize and to get the vaporized CO2 out of the deposit to cool the reactor. $\endgroup$ – Cornelis Aug 11 '18 at 8:27

The research quoted in the comments indicates that this is indeed a bad idea, and wouldn't accomplish the intended result, but if it were decided that the CO2 should be vaporized, it's highly unlikely that active methods would be the most economical.

By "active methods" I mean that we have to generate the heat to do the vaporizing, by means such as combustion, nuclear reactions, etc. A "passive" method would be one that redirects or concentrates naturally available heat, such as solar energy, geothermal energy, etc.

In my answer to this question I summarize and provide links to sources regarding the concept of a "pole-sitter" orbit, a highly non-Keplerian orbit. Such a pole-sitter can use a solar sail to remain in view of a planet's polar region, as shown in Fig. 1 below. I've omitted the gravitational force due to the Sun and the centrifugal force from the heliocentric orbit because unless the pole-sitter is a long distance from the planet, those are relatively small forces. Also, the discussion is about the CO2 reservoir at the south pole; assume in my figures that north is down.

Electric propulsion can maintain a pole-sitter orbit too, but in this case we'd use the reflected sunlight from a solar sail, and there'd be no propellant required, a significant advantage if the project takes centuries.

In this case, the sail can be designed to concentrate its reflected solar energy onto Mars's south pole, increasing the average temperature there and eventually vaporizing the CO2. "Eventually" probably means decades or centuries; it depends on the sail size. Such a sail wouldn't need much of a spacecraft (in terms of mass) to control it, so the total areal density of the system would approach the limiting areal density of the sail and its support structure.

This low density is important for two reasons. One: this sail will be big. When you talk about vaporizing 12,000 cubic km of CO2, any method you use will involve something on a grand scale. The amount of sunlight you'd need to redirect to the pole will be very large, and that requires a large reflecting area. Lower density means less mass required to build the system. Two: the lower the areal density of the system, the higher in angular elevation above the pole's horizon a pole-sitter can get, so the energy distribution is better. In Fig. 2 below, this means the angle between the green "viewing angle" arrow and the planet's rotation axis is smaller. It might wind up that because of the increase in structural mass fraction as the sail goes beyond a certain size, a constellation of smaller sails might be better.

There is a problem the designers would face. The Sun-planet-pole-sitter geometry doesn't depend on the planet's obliquity, the tilt of its equatorial plane with respect to its orbit plane, so the pole-sitter is always a bit farther from the Sun than the planet. This is required to have all the various force vectors (gravity, centrifugal force, sail force, etc.) add up properly. That geometry rotates with Mars's revolution around the Sun, to keep the pole-sitter on the anti-sunward side of Mars. Mars's obliquity is ~25 degrees, just a bit larger than Earth's, and that obliquity does not rotate with Mars's revolution around the Sun. For part of a Martian year the pole would be oriented more or less toward the pole-sitter, yielding a good illumination geometry as shown in Fig. 2. But half a Martian year later the pole would be more oriented toward the Sun, and the pole-sitter would see the polar region close to (or even beyond) the limb of the planet, as shown in Fig. 3. (Oops. I just noticed I forgot to switch directions of the "Sunlight" arrow in Fig. 3. It should point to the left, not the right.) This makes for a poor, or even impossible, illumination geometry. The energy input for vaporization might be very seasonal.

enter image description here

There would be several trades that would need to be analyzed when designing this system, such as the "one big sail or many smaller sails" trade. I haven't done any of those! (As Gomer Pyle would say: Soo-PRISE, soo-prise!) I'll wait until one of my clients pays me to do that job.

The geothermal option is intriguing. There'd be some large-scale engineering in that as well. Would there be geologic hazard implications of cooling a fair chunk of Mars's mantle?

You might note I tend to avoid methods that throw large masses of rocks, dirt, dust, and natives into the air, i.e., huge explosions. But there are methods that would involve such, including nuclear explosions within the CO2 reservoir, or impacting a small asteroid or comet onto the reservoir, etc., and those methods have the advantage of getting the project finished in relatively short order—and they also have some distinct disadvantages. (side benefit to the comet: lots of volatiles, including water!) Altering the orbit of a few-km-sized comet or asteroid to an impact might wind up being less expensive than the pole-sitter solution...but you'd better not have anyone living on Mars when you do that!

  • $\begingroup$ I really love the graphics! Did you make those yourself? $\endgroup$ – uhoh Aug 10 '18 at 5:56
  • $\begingroup$ Thank you for the extensive answer. Unfortunately the good illumination geometry happens when the south pole is in constant darkness To evaporize the CO2 would demand much radiation concentration. $\endgroup$ – Cornelis Aug 10 '18 at 9:45
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    $\begingroup$ @uhoh Yep, I did generate those myself. It was something of a kludge. I generated them in PowerPoint, but found I couldn't import a PowerPoint file into SESE. So I printed it as a PDF, and found I couldn't import a PDF either! So I actually hard-copy printed it, scanned it into a JPEG file, and imported that. Sheesh! $\endgroup$ – Tom Spilker Aug 10 '18 at 19:16
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    $\begingroup$ I just noticed...you could download my graphics, copy the "Lots of distance" symbol I generated, scale it—and use it as an integral sign! ;-) $\endgroup$ – Tom Spilker Aug 10 '18 at 19:23
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    $\begingroup$ @Conelisinspace Actually, adding energy to the area, even relatively small amounts (i.e., fractions of the normal annually-averaged insolation), at any time of the Martian year will shift the equilibrium toward more evaporation. During the south polar winter, if you put half or a third of the summer solar insolation onto just the reservoir area, you will greatly decrease the amount of winter-time CO2 condensation from the normal quantity, resulting in a net evaporation when averaged over a year. $\endgroup$ – Tom Spilker Aug 10 '18 at 19:35

First of all, vaporizing 12k cubic kilometers of CO2 on another planet is not going to be "economical" (in the sense of "cheap") at all! Even if the deposit had a "switch" that you just need to touch to make it happen, a robotic mission to Mars to push that switch would costs millions :) This is going to require a lot of work.. but how much exactly?

How much energy we need

I would start by calculating the energy needed to vaporize that CO2 deposit. We need enough energy to heat the CO2 up to the temperature it will vaporize, plus the energy to actually make it vaporize (enthalpy). To calculate this, we need to know the current CO2 temperature, pressure, have a look at the CO2 phase diagram and also know the total mass we want to vaporize.

According to this article, the mass of that deposit is 2.4 x 10$^1$$^6$ kg (and the volume is actually 14800 cubic kilometers according to a revised estimate).

As for the temperature, in that article, they say that the mean annual surface temperature is 155K. We'll make a simplifying assumption here and assume all deposit is at that temperature (it could actually be higher, so easier to vaporize: see Geothermal gradient and this answer here).

In the article, they mention that Mars average surface pressure is 610 Pa. I will make a very much simplifying assumption and say that all the CO2 ice we are going to vaporize will be at that pressure (imagine as if we vaporize it layer by layer and somehow maintaining atmospheric pressure constant...).

It is also mentioned in other articles [source?] that CO2 actually sublimates on this conditions (155 K, 610 Pa). If you look around for the CO2 Phase diagram, it will often will look like this: enter image description here

Unfortunately, our region of interest, with very low pressure and temperature is not shown. Also, finding the enthalpy of sublimation in those conditions has proven to be difficult. The best near value I found is in this article, which mentions another article with a value of 613 kJ/Kg latent heat of sublimation @ 121 K. This is 34 K off of what we need, and unfortunately i have no estimate on much the value could be far off. (Warning! Our final estimate could be way off!)

Anyway, assuming this is correct enough, we need 2.4 x 10$^1$$^6$ * 613 = 1.47x 10$^1$$^9$ kJ to vaporize the CO2 deposit.

Go Nuclear!

As a comparison, the Tsar Bomba, the most powerful nuclear weapon ever created had a yield of 210 Pj (PJ = 10$^1$$^5$ J). So we would need roughly 70 of those to vaporize all the CO2 ice (assuming we could redirect ALL the nuclear weapon output as heat into the ice - a very much unrealistic efficiency estimate).

Unfortunately, a single Tsar Bomba weights 27000 kg, and none of the current orbital launch systems seems able to deliver that weight to Mars. Judging by the list, the best bet seems to be the Falcon Heavy which can carry 16800 kg to Mars. Let's assume that we can built smaller Tsar Bomba with linear reduction in their output, i.e. a small Tsar Bomba weighting 16800 kg and delivering 130 PJ. We would need about 113 of those "small Tsar Bomba", so 113 Falcon Heavy launches.

According https://en.wikipedia.org/wiki/Falcon_Heavy#Launch_prices, you can get a fully expendable Falcon Heavy launch for 150 million. This means that you can launch all those bombs for just 16950 million dollars, less than 17 billion.

This could be cheaper than developing massive solar sails or giant 100km radius mirrors.

Also, consider:


  • Elon Musk is in favor of terraforming Mars by thermonuclear weapons, so you might get discount prices

  • Maybe you don't need to vaporize it all, you just need to vaporize a part of it and then you will have destabilized Mars equilibrium so that the rest will vaporize by itself (i.e. as a simple example, the first article mentions three layers of water ice that stabilize the deposit.. what if you vaporize enough so that all those layers are gone?)


  • We assumed that all the energy of the bombs goes into vaporizing the CO2 ice

  • We just assumed that ice just vaporizes and goes away (i.e. we did not really model how the already vaporized ice would change the conditions for vaporizing the rest,nor we did model the surrounding atmosphere, water, rocks, etc.)

  • We did not factor the development/manufacturing costs for those bombs

  • You might need additional machinery/considerations to coordinate the bombs so that they hit the deposit in a coordinated manner (you don't want to vaporize just a bit of it.. and see it deposit back to ice before the next bombs hits)

  • We did not consider what's on top of that deposit (a layer of water ice I guess), and we need to vaporize that as well.

But after all, this is just a rough baseline....have fun and don't try it at home :)

  • $\begingroup$ By the way, the energy estimate can be used also for other methods: - i.e how big should be the asteroid? - how big should be a power plant on Mars to provide energy to vaporize the deposit? $\endgroup$ – BlueCoder Aug 16 '18 at 10:21
  • $\begingroup$ Thank you for the answer ! How did you calculate the total amount of energy needed to vaporize the CO2 deposit, and what are the e's ? $\endgroup$ – Cornelis Aug 16 '18 at 17:26
  • $\begingroup$ The e's are just engineering notation. They just mean 10 elevated to something. i.e. 7,4e8 means 7,4 x 10^8. $\endgroup$ – BlueCoder Aug 16 '18 at 18:05
  • $\begingroup$ To vaporize CO2 (meaning: to transform it from solid to a gas), one usually needs to spend energy to turn it into a liquid and then into a gas. In our case, the starting temperature and pressure is so low that it will pass directly from solid to gas (it sublimates) - also it seems that conditions on the surface are just those needed to make it sublimate, you don't need to rise its temperature, you just give it the energy to change state (sublimation enthalpy). Sublimation enthalpy for a very similar condition is 613J/kg, hence the calculus and the result. $\endgroup$ – BlueCoder Aug 16 '18 at 18:09
  • $\begingroup$ @Conelisinspace, thanks for the edit, I got rid of the e's as well :) $\endgroup$ – BlueCoder Aug 16 '18 at 18:14

Why not just hit it with an asteroid? In other words, use asteroid redirect maneuvers to send an asteroid into the deposit thereby sending debris into the atmosphere.

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    $\begingroup$ Are there asteroids right now that could be redirected toward Mars ? $\endgroup$ – Cornelis Aug 12 '18 at 7:59
  • $\begingroup$ @Conelisinspace I edited my post after Peter made his comment. Sorry about not clarifying that. I'm sure there are, but I haven't really done much of the research to know for sure. $\endgroup$ – Chris Aug 12 '18 at 9:37
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    $\begingroup$ @Conelisinspace There might not be asteroids, but there are comets. theskylive.com $\endgroup$ – Chris Aug 12 '18 at 9:53
  • $\begingroup$ @Chris Could you calculate how big the asteroid will have to be to evaporize the whole CO2 deposit ? Then we could evaluate if it would be economical to redirect an asteroid with that size because that demands huge amounts of energy $\endgroup$ – Cornelis Aug 12 '18 at 10:22
  • $\begingroup$ Thank you for the interesting link, but i think it will be sheer coincidence to get the chance to redirect an asteroid or comet that precise that it will hit the CO2 deposit, $\endgroup$ – Cornelis Aug 12 '18 at 12:46

Looking at the amount of CO2 we want to vaporize, we can expect to need a lot of energy. If we kindly steal the energy requirement BlueCoder calculated, we can see that this is the same order of magnitude as the amount of energy stored in the global uranium-238 reserves.[1] This would favour a passive method over an active one.

Of the passive methods, my personal favourite would be the artificial magnetosphere.

Computer models by NASA show that the placement of an 1-2 Tesla magnet at L1 would be able to shield Mars from some of the incoming solar wind.[2] Which, is the reason Mars lost most of its atmosphere in the first place. With this shielding, the atmosphere will slowly thicken over time, slowly increasing the temperature. Which in turn will cause more evaporation, starting a chain reaction until the CO2 deposits are depleted.

The proposed method for creating an artificial magnetic dipole at Mars’ L1 Lagrange Point. Credit: NASA/J.Green

Although it wouldn't be the fastest approach, it would be very cost effective. And as an added bonus, create some protection for the new colonists.

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    $\begingroup$ Got any sources for this? Sounds significantly interesting, and you have a picture, so I'm assuming that came from somewhere. $\endgroup$ – Magic Octopus Urn Aug 16 '18 at 18:52
  • $\begingroup$ Why should the atmosphere thicken with this shielding ? And when you read the very interestig article in the answer of @Bluecoder, you would know there is enough CO2 stored at the south polar cap to increase the pressure enough to create liquid water ! $\endgroup$ – Cornelis Aug 17 '18 at 9:18
  • $\begingroup$ @MagicOctopusUrn, it was presented at a workshop, from which an abstract can be found here: hou.usra.edu/meetings/V2050/pdf/8250.pdf. I cited the news article, since they elaborate a bit more. $\endgroup$ – Martini Aug 17 '18 at 13:04
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    $\begingroup$ @Conelisinspace, The reason why shielding from solar wind would help is that the atmosphere of mars is currently in equilibrium, with small gas sources equalling the atmospheric erosion. Of the atmosphere erosion, solar wind is one of the largest contributors. By preventing this, the equilibrium will shift to a thicker atmosphere with a higher temperature. This in turn provides more sublimation, shifting the equilibrium even further. $\endgroup$ – Martini Aug 17 '18 at 13:26
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    $\begingroup$ @Martini In this answer space.stackexchange.com/questions/25562/… it is mentioned that Mars loses about 100 grams of its atmosphere every second. That is 3156 tonnes a year. If that is the same as the yield of the outgassing it would take millions of years to contribute to the pressure of the atmosphere in a significant way ! $\endgroup$ – Cornelis Aug 17 '18 at 15:09

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