11
$\begingroup$

Introduction

For more backstory, see here and here

The 438th Harmonious Congress of the People of Mars was perhaps the most anticipated since the early years of the settlement. In a rare action, the Supreme Council of Harmony agreed to broadcast some of the proceedings of the Congress on tele-projector for all the tens of millions of Martians to witness. The occasion was, of course, significant. The Chief Designer was presenting to the Council, and indeed to all the people, a new 100 Year Plan for the Revivification of Mars.

There was a great buzz and anticipation in the air. Rumors flew that some great milestone had been met in the construction of an atmosphere. Already, if you received permission to surface walk outside the airlocks, you could see the fruits of the Bureau of Revivification. In the lowest latitudes, slender pine trees thrust skyward from the red dirt. In more seasonal climes, grass bloomed in great billows of green during spring. Running water could be seen for part of the year anywhere within 30 degrees of the Equator.

People had not been outside for generations stretching back to Old Earth. The first colonists had dug into dormant volcanoes and the cliff faces of vast chasms. Over the centuries, millions of miles of passages and corridors were extended under the surface. No one walked on the surface for centuries, save a few scientists perhaps. But in the past few decades, the air pressure had gotten so high-pressure suits were barely thicker than regular clothes. A rupture was no longer catastrophic.

All Mars waited with anticipation the speech from the Chief Designer. What would she propose? What was the next step for Mars? Was it possible that a Green and Blue Mars, a truly habitable Mars, would soon be a reality?

Question

The Martian air pressure is above 10 kPa. The massive amounts of carbon dioxide available on the surface of the planet have all been vaporized. Nuclear driven oxygen synthesis has greatly sped the conversion of this carbon dioxide to oxygen, and widespread plant life is contributing its part. It will not be long until oxygen levels are 50% that of Earth; equivalent to 5 km altitude on Earth. High enough to be considered habitable and breathable to humans.

The Chief Designer and her team have decided that it is time to raise the atmospheric pressure on Mars. In order to do this, they will need to generate about 40 kPa of air pressure from some inert gas.

Given the energy cost of transporting an inert gas from somewhere else in the solar system, and the energy cost of any chemical reactions needed to put it in the atmosphere, what is the least energy expensive way to add 40 kPa of air pressure to Mars?*

For example, if the best gas is diatomic nitrogen, then the cost of transportation from a source in the outer solar system as well as the cost of turning whatever nitrogen compounds can be found into the diatomic gas must be considered.

Considerations

  • The Earth was hit by a large bolide 500 years ago. It is still glowing. Earth's former atmosphere and oceans are not available to be moved to Mars.
  • Any other resources in the solar system are available.
  • The O$_2$ and CO$_2$ information in the question are presented as facts; they are not relevant to the discussion.
  • Technology level is near-future but mostly irrelevant. The correct answer will give an energy cost in Joules (or Calories, I suppose, if you like to be contrarian).
  • Energy cost only has to consider the cost of moving the materials; a function of mass and whatever combinations of delta-v's will get it from its current location to Mars. The cost of rockets and fuels and such can be transparent.
$\endgroup$
  • 1
    $\begingroup$ Not an answer because any calculation is missing but I would say mount an big ass railgun on a small ice moon (titan might be an option, but smaller is better due to gravity) and start shooting bricks of ice towards mars. On titan or other likely moons you might be able to use the methane as power supply or even as propellant. Nice writing by the way. $\endgroup$ – D.J. Klomp Jan 9 '18 at 2:05
  • $\begingroup$ @D.J.Klomp Thank you about the writing. But regarding the question, I'm not looking for a method of delivering atmosphere, I'm looking for a calculation of how much it costs. The energy costs to move so many kg are the same for railgun, atomic rockets, or chariot hitched to swans. $\endgroup$ – kingledion Jan 9 '18 at 2:11
  • $\begingroup$ Actually I don't think so, the differences in efficiency are huge. With a railgun you only need to move the actual kg of atmosphere while with a rocket you also need to move all accompanying fuel and the rocket itself. Our current rockets are I thought for 90% fuel that also needs to be accelerated. Also whether you need to overcome a high or a low gravity has a huge influence on your needed energy consumption. If you can nudge an ice asteroid towards Mars this would require very little energy compared to shooting it from a moon. $\endgroup$ – D.J. Klomp Jan 9 '18 at 2:21
  • $\begingroup$ @D.J.Klomp Thanks for the update, I added a bullet in my question to ignore the cost of rockets and fuels and such. The gravity part is very much important. The cost of exiting Titan's gravity well, for example, is significant. $\endgroup$ – kingledion Jan 9 '18 at 2:25
  • 2
    $\begingroup$ @Renan You do not need a magnetosphere in the short term. See here. $\endgroup$ – kingledion Jan 9 '18 at 13:20
2
$\begingroup$

I found you a moon: Mimas

Since I couldn't sleep I thought I would do some of the math so here it goes:

Radius_Mars = 3390 * 10^3   %in meter
Thickness_Atm = 66 *10^3    %Earth atomosphere is about 100km, took 2/3 for 
mars in meters
Volume_Atm = 4/3*pi*(Radius_Mars+Thickness_Atm)^3 - 4/3*pi*Radius_Mars^3;   
Volume_Atm = 9.7E18 m^3
Air_Density = 1.2 %kg/m^3
Needed_Mass = 0.3*Air_Density*Volume_Atm;   %3.4E18 kg
Needed_Mass = 3.4E18 kg

So you need a mass of 3.4E18 kg to shoot from some moon (according to wikithe astroids in the belt are mainly of C, S and M type so unsuitable), so find an ice moon with the lowest gravity. It would seem that Mimas orbiting Saturnus is mostly ice and has a surface as small as Spain. The mass of Mimas is 3.7E19 kg so only one order of magnitude higher. So instead of having to deal with escape velocity simply move to complete moon to Mars, the it will only become a question of how fast do you want it.

So if you have patience for a 10 years you have to move the intervening distance between Saturn and Mars, being roughly 1.2E9 km. So your average speed, taking into account that you need to decelerate for just as long would need to be

 average velocity = 1.2E12/31E7 = 3834 m/s

So to calculate the energy just use the formula for kinetic energy

 energy = 1/2*m*v^2

So the needed energy becomes:

 energy = 1/2*3.7E19*3834^2 = 2.719E26 J

To put it into prespective if you harness all the solar radiation falling on earth you would need to harvest it at a 100% efficiency for 12.4 years to get the required energy. So I guess we won't be moving planets any time soon :D.

Since it is late it might be good to check the numbers.

$\endgroup$
  • $\begingroup$ Do we really have to worry about deceleration? Lower acceleration and letting Mars' gravity do the rest might save massively on power. I'm not sure, though - I'm not a rocket scientist. $\endgroup$ – Andon Jan 9 '18 at 4:19
  • $\begingroup$ No you really need to decelerate, otherwise all that kinetic energy will be impacted on mars which will be tremendous and more than enough to kill everything on the planet. Although the velocity does not seem that high the mass still is. $\endgroup$ – D.J. Klomp Jan 9 '18 at 6:40
  • $\begingroup$ Isn't it the point? If we are going to terraform Mars, it's much easier to do so when nobody lives there. Just crash your moon / few asteroids into Mars. Few hundred years later you'd the effects will stabilise. $\endgroup$ – Oleg Lobachev Jan 9 '18 at 8:28
  • 4
    $\begingroup$ @Oleg Lobachev: The back story clearly has people living there. $\endgroup$ – D.J. Klomp Jan 9 '18 at 8:32
  • 1
    $\begingroup$ This is good for half the problem, but you don't discuss what ices you are actually getting out of Mimas? We need nitrogen gas on Mars, are there any estimates of how much nitrogen is on Mimas? What do we need to do to turn these nitrogen compounds into atmosphere? $\endgroup$ – kingledion Jan 9 '18 at 13:25
2
$\begingroup$

An important preface: without a stated time period to accomplish this task the energy cost cannot be calculated.

I will make assumptions about time.

Current real Mars data. https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

Atmosphere

Surface pressure: 6.36 mb at mean radius (variable from 4.0 to 8.7 mb depending on season)
[6.9 mb to 9 mb (Viking 1 Lander site)] Total mass of atmosphere: ~2.5 x 1016 kg

  • 1 mb = 0.1 kPa. 6.36 mb = 0.636 Kpa.
  • The proposed Mars from OP has air pressure of 10 kPa. 10 / 0.636 = 15.7. The total mass of atmosphere has increased by 15.7 times.
  • The current total mass of atmosphere is (2.5 x 1016) x 15.7 = 3.92 x 1017 kg.
  • You ask to raise atmospheric pressure 40 kPa more (to a total of 50 kPa). Noted: adding gas mass might not cleanly convert to raising pressure but we will assume.
    You want to add 4 times the current atmospheric mass to the current atmospheric mass.
  • You want 4 x (3.92 x 1017 kg) = 15.7 x 1017 kg or 1.57 x 1018 kg of gas.

For reference, let us consider what it would take to import this much mass from Earth. I know that the conditions of the OP rule out Earth but this will give a sense of scale for these huge numbers.
https://www.space.com/24701-how-long-does-it-take-to-get-to-mars.html We will assume the distance between Earth and Mars is 225 million km. We will allow a leisurely 1 year to traverse this distance.
225,000,000 km/year is 7134 meters/second. It is so cool Google will do calulations like that for you!

Again to get things to scale: this is 25684 km/hour. That is a good clip but the New Horizons probe (from above link) went twice that fast, so OK.

Kinetic energy = 1/2 * mass * velocity ^2, where Kinetic energy is in joules, mass is in kilograms, and velocity is in meters per second

The joules: speed in m/s ^2 = 7134 ^2. = 50893956
Mass moved (from above) = 1.57 x 1018 kg
1.57 x 1018 kg x 50893956 = 7.9903511e+25.

Divided by 2 = 3.9951755e+25 joules to move required mass from Earth to Mars over 1 year.

Of course one must decelerate this mass when it arrives on Mars, unless you have some scheme to decelerate it for free by ramming the mass into the surface. Which could have ramifications, so to speak. The energy you put in to get it up to speed you must then put back in to slow it down: x 2, which fortunately is already done: 7.9903511e+25 joules.

It would be energetically more expensive to do it faster (because one must accelerate to a higher velocity) and less expensive to do it slower. If moving mass from farther away (e.g. Titan) the same holds: it would cost the same energy as moving it from Earth but take more time, or cost more energy to traverse the greater distance at the same speed. Without a stated time period to accomplish this task the energy cost of the task cannot be calculated.


Considering an alternative: this society must have a metric buttload of energy available to consider such an endeavor. From my point of view, this society has unlimited energy. Maybe they tap Casimir forces or are masters of fusion. Moving mass would be a trick with potential for disaster at many steps along the way. The logistics of gathering this much mass at its source, keeping it together en route, and decelerating it into Mars are daunting.

Instead, how about using that free energy to make the mass on site?

How much energy is in that much mass? Sweet online calculators! http://www.wolframalpha.com/widgets/view.jsp?id=b3aa19fe9dc706a3b4cdaa8ddb37d852 1.57 x 1018 kg converts to 1.411 x 1035 joules.

Double checking with this calculator http://www.1728.org/einstein.htm I get 1.348e+35 in that much mass.

For scale, the sun puts out 3.725e+26 joules / second. To make the required mass using the entire energy output of the sun would take 361879194 seconds or 11.4 years. Depending on how your energy source worked, you could set up mass generation plants on Mars and let them chug away.

Of course if you want to get super fussy, these numbers assume that the new Martian atmosphere is the same gas composition and so same kg weight as the existing real Martian atmosphere which is mostly CO2. It is not specified in the OP what the new Martian atmosphere is made of: at 10 kPa and 50% O2 there is 50% something else which must be CO2. (You would be breathing really hard in that atmosphere and it would feel like you burped Coke into your nose).

  • All gases have the same volume.
  • All CO2 atmosphere (current real Mars) would be 44g/mol gas.
  • N2 atmosphere imorted or created would be 28 g/mol gas. That is 0.63 the mass I used above. Those who are very interested can multiply accordingly for new numbers.
  • You could make (but probably not find and import) an atmosphere of neon with 71% of the mass of your diatomic nitrogen atmosphere. That would be 0.45 the mass I used above for calculations. So just 4.5 years to produce with your energy to mass factories!
$\endgroup$
  • $\begingroup$ In markdown you have to press Enter two times to get a paragraph. Or you can add two spaces at the end of a line before pressing Enter once to get a soft linebreak as you would get with <br>. You should try not to mix HTML and markdown if it's not absolutely necessary (and apart from <sup></sup> it's nearly never necessary), that frequently messes with the displaying and causes weird errors. To make a list you have to have one empty line followed by lines that start with a dash and a space "- " to get dots. Then you only need to press Enter once for each new dot. $\endgroup$ – Secespitus Jan 10 '18 at 15:43
  • $\begingroup$ @Secespitus: Thank you very much for cleaning this up and for the formatting advice! $\endgroup$ – Willk Jan 10 '18 at 16:23
2
$\begingroup$

I would seriously consider the main belt asteroids. Delta V from Mars transfer to Ceres transfer orbit is 1.3 km/s. You shouldn't need to go to an asteroid the size of Ceres and you'll be throwing small chunks that will burn up in the atmosphere so you don't need to slow anything down.

Let's say you shoot 1kg chunks from 0 to 1300 m/s in 1 second.

This takes 650 meters (d = 1/2*a*t^2).

Energy is kg* m^2 / s^2 or kg * meters * a

so this is 845000 Joules per kg (you can change the time or acceleration just so you still get that delta v and the energy won't change. Let's call that 1 x 10^6 because our asteroid has some gravity.

Someone smarter than me said you're looking for ~2 x 10 ^18 kg. The total asteroid belt mass is 3 x 10^21 kg so that makes sense to me.

You need 2 x 10^24 Joules or 2 x 10^21 KJ.

"Palo Verde nuclear power plant in Arizona is the largest nuclear power plant in the United states with three reactors and a total electricity generating capacity of about 3,937 MW."

and 1MWh = 3.6 x 10^6 KJ

So 555.5 x 10^12 MWh is 1000 Palo Verdes running at max for 141 million hours (16000 years).

That's a little disappointing but maybe your launcher could use Ceres for a gravity assist. If you could get the delta v to more like 300 m/s that drops it to 7 million hours (800 years).

$\endgroup$
1
$\begingroup$

Unfortunately, the only gases you can safely put in a terraformed body's atmosphere are oxygen and nitrogen.

Earth is out, so the only other feasible source for nitrogen is Titan (unless you have an economic way of instating a partial CNO cycle and have sufficient carbon or carbon dioxide source, and energy, in place).

Nitrogen on Titan - or any significant gravity well outside Mars's - possesses quite a lot more potential energy than it's feasible to unleash unto the red planet. The simplest way of delivering the nitrogen, i.e. supercooling it in frozen missiles and shooting them at Mars using linear accelerators, will result in even more energy. The frozen missiles will burn and disintegrate in the Martian atmosphere, but they will also heat it (and, not too much later, the ground) way too much.

Delivering the goods using reaction drives is unfeasible for the same reason. Each kilogram at Mars ground level would cost too much in either waste heat and/or environmental pollution.

Inertialess grav-drives would of course solve this problem, but they're not current technology; not even in sight.

What follows is also almost certainly not feasible, but due to logistic and economic constraints, not technological limitations.

Stage One: build a base on Titan, install a launcher (actually, many launchers).

The Titan base will send projectiles onto a Hohmann transfer orbit, supplying the first part of the required delta-V. That'll set you back around 11 MJ/kg plus other 3.48 MJ/kg of escape velocity.

The launcher might be installed on a skyhook on Titan to reduce the second part of those costs.

A projectile is just an insulating shell over one ton of supercooled solid nitrogen, plus a solar-powered beacon, and will arrive at Mars after 5 to 18 years depending on planets positions. It has no manoeuvering capabilities.

Stage Two: build the "catcher" (or a forest of them) in orbit around Mars.

The catcher will have to neutralize the incoming projectile's kinetic energy using electromagnetic braking. You will need a minimum of around 28.5 MJ/kg for that. This will also determine an alteration in the catcher's orbit. To balance that, some kind of thruster, maybe a ion engine, will need to be installed. We could perhaps use Phobos as a waystation - but that would limit us to one waystation, restricting both the number of catch windows and volley size.

Another possibility could be to harpoon the incoming projectile - it will be moving at around 7500 m/s - and decelerate it using a dynamo reel. That would allow recovering most of those 28.5 MJ/kg, even if even braking at 10 G will still require more than one minute, and paying out some 290 km of wire. A one-ton projectile falling in from Saturn will then develop a power of around 370 megawatt (28500 MJ in 76 s).

Stage Three: drop the pellets on Mars.

Falling 6000 km from as low as Phobos, even with zero initial speed in respect to the planet (the Phobos launcher will simply neutralize the moon's orbital velocity), the frozen nitrogen pellets would still possess a potential energy of around 7.1 MJ/kg; even if they're close to absolute zero, they would only absorb about 0.3 MJ/kg to achieve the gaseous phase. We'd be left with an excess of more than 28 million calories per kilogram, which is unacceptable.

The only other alternative is a skyhook. The skyhook station would orbit in the same plane as the incoming projectiles, and alternately catch them from either side of the planet, thereby dumping the excess energy straight into Mars. We will require a sizeable series of skyhooks all around the planet - essentially, an artificial ring around Mars.

The gas would then be dropped straight down, neutralizing (and recovering a part of) those 7 MJ/kg of energy through magnetic braking.

The big question is now how long would this take.

Assuming each projectile weighs one round ton, we'd need around three thousand million million projectiles. Most of the required energy is supplied by gravity, but still we need launchers and catchers. With one thousand of each, we require three million million volleys, and at one volley per second (remember we're moving one ton of mass per launch, and incidentally we have to dissipate a lot of energy), that's one hundred thousand years.

$\endgroup$
1
$\begingroup$

So, the other answers have established that it's not easy to move huge amounts of mass around the solar system. Why not try to solve the problem like NASA does?

Making thin air out of rocks

In a reversal of the typical "making stuff appear out of thin air", an ingenious chemical process allows the production of thin air out of rocks. Specifically, a modified version of the Fray-Farthing-Chen (FFC) Process takes in metal oxides such as titanium dioxide, magnesium oxide, or sodium oxide and separates the two elements. This was first proposed in 2000, and immediately earned a Nature paper. Although originally designed for the production of high-purity metals, this process picked up a lot of attention in the late 00's when NASA and others realized that it could be used to produce oxygen out of lunar regolith. This is known as "In Situ Resource Utilization" (ISRU), and NASA has put some pretty serious money towards determining feasibility.

Specifically, the Ilmenox process uses a specially-designed reactor that strips the oxygen from the other elements in lunar or Martian regolith and liberates it into the atmosphere.

To quote from the phys.org article summarizing this development,

Based on experiments with a simulated lunar rock developed by NASA, the researchers calculate that three one-meter-tall reactors could generate one tonne of oxygen per year on the Moon. Each tonne of oxygen would require three tonnes of rock to produce. Fray noted that three reactors would require about 4.5 kilowatts of power, which could be supplied by solar panels or possibly a small nuclear reactor on the Moon. The researchers are also working with the European Space Agency on developing an even larger reactor that could be operated remotely.

So, to summarize: we can use 4.5 kilowatt reactors to produce oxygen (or carbon dioxide) from moon rocks. The return on this is something like 1,000kg of oxygen per year. So our energy conversion is

$$\frac{4,500\ J}{1\ sec}*\frac{31,536,000\ sec}{1\ year}*\frac{1\ year}{1,000\ kg\ O_2} = \frac{142\ MJ}{kg\ O_2}$$

To borrow from WillK's excellent answer, we need something like 1.57 x 1018 kg of gas. To synthesize all of that de novo, we'll need 2.23 x 1026 joules of energy. Curiously, that's in the same ballpark as the other answers - cool! However, I'll argue that my method is still superior for several reasons.

Reason #1: Controlled progress

As many of the other answers point out, half of their energy cost comes from slowing down whatever they're throwing at it. With this method, the process is controlled from beginning to end - ramp up production or slow it down as necessary. To put that number in context, let's check out everyone's favorite Wikipedia page:

  • If this was done with traditional fossil fuels, we can expect to take in 10,000 years if we used all the fossil fuels on Earth every year: fossil fuel reserve as of 2010 = 3.9×1022 J
  • If done with nuclear tech, as your question proposes, we can expect to take more like 1,000 years, if we used all the uranium on Earth every year: global uranium-238 resources = 2.2×1023 J
  • If we covered Mars in solar panels, we can expect to be done in about 100 years: total power received by Mars from the Sun $\approx$ 6.6x1016 W
  • If we cover the Sun in solar panels, we're done in one second: solar output = 3.8×1026 W

Stupid Tumblr post about making cookies with the sun (from https://orteil42.tumblr.com/post/125575676555/blxckbiird-spaghetti-western-wannabe)

Reason #2: We're actually producing a gas

Most of the other answers have thrown something like a small moon or a whole bunch of rocks at Mars and called it a day. That's... not an atmosphere? Nor is it especially friendly to the inhabitants? Instead, with this method, we're producing atmospheric gases in any mix of carbon dioxide and oxygen that we'd like, allowing us to find a good balance between phototrophs and heterotrophs as we continue to terraform the place. Unless the other answers manage to find an extraterrestrial object made largely of nitrogen or neon, against all the traditional extraterrestrial body compositions we've thus far found, all that they'll do is add more rocks to the place - mostly silicates and metal oxides.

Reason #3: Abundant high-purity metals as side-effect

"But wait," you say, "what about all the side products of the FFC process?" Well, I'm really glad you asked. As noted earlier, this process is traditionally used to produce high-purity metals by stripping the oxygen off of metal oxides. Now that we've reversed the roles, these metals are essentially "waste material" - guess we'll have to make really cool buildings out of them instead.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.