What is the most economical way to vaporize the CO₂ deposit near the south pole of Mars?

Question

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 !

Answer 1

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.

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!

Answer 2

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.

Answer 3

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:

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:

PROS:

• 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?)

CONS:

• 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 :)

Answer 4

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.

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.