A few weeks back, I made the case that relying on space to provide an infinite resource base into which we grow/expand forever is misguided. Not only is it much harder than many people appreciate, but it represents a distraction to the message that growth cannot continue on Earth and we should get busy planning a transition to a non-growth-based, truly sustainable existence. To prove what a distraction it is, I will distract myself again this week with another space post. This time, true to the brand, I will do the math on why the infinite resources of space appear to be of questionable use to our human enterprise.
Part of my motivation comes from the bruised, and bruising comments in reaction to the Why Not Space? post. The faith is strong that technologies are already in hand and that we just need NASA to get out of the way so the commercial bounty of the sky will open up and we’ll finally be off to the races. I myself refrained from ruling out such a future, but the mere suggestion that we may fail to expand into space was clearly considered by many to be ridiculous—as if such a fate is predestined: as sure as the sun will tomorrow. Sociological impulses tugged at my physicist bones, tempting me to study exactly how such an unshakable faith has been implanted in so many obviously smart people. For these folks, the arc of the future is as sure as the historical progression from the Dark Ages until now. Wait? Was there something before the Dark Ages? Something grand? Alas, my history fails me.
Leaving the sociology aside—but before we get busy with the math—I’ll share the story that during the comment firestorm, an individual contacted me from NASA headquarters (not to revoke my funding, thankfully), offering thoughtful perspectives on space policy. The part I can’t shake is the statement that it takes decades of serious research to answer two simple questions: “Can humans live and work in space for the long term?” and “Can an economically viable activity be found in space?” Opinions aside, these are open questions, and have been for some time. We have no proof—or even firm expectation—that either is practical or possible.
Around the time of the final U.S. Space Shuttle flight, a NASA official was asked in a radio interview to explain what was left to inspire young kids about space. The answer was that mining asteroids and the Moon offered a new grand challenge to inspire our kidlets. Granted, space mining probably is a bit more inspiring than off-shore drilling or coal mining as a career choice. It’s got space in it. But are we really serious about getting materials from other bodies within the solar system?
The “easy” pickings are the asteroids, because they are small enough not to require much fuel for a surface landing. The Moon and Mars, by contrast, need lots of fuel to safely touch down and later lift off. Most asteroids are basically rock. We have plenty of that stuff already. The prizes are the metal-rich iron-nickel asteroids, comprising 10% of the total asteroid belt mass, or about 3×1020 kg (1/20,000 of an Earth mass). By comparison, the world has an estimated nickel resource of 130 million tons (≈1011 kg). The Earth contains far more nickel than this, but most of it in the inaccessible core (as opposed to inaccessible space). Nonetheless, one iron-nickel asteroid 1 km in diameter would increase our current nickel resource more than ten-fold. It is indisputable that asteroids contain a trove of metals.
Likewise, our hydrocarbon endowment on Earth cannot compare to the enormous amount of methane contained in Jupiter. And there are lakes of methane on Titan—a moon of Saturn—that dwarf the hydrocarbon resources on Earth. Nobody guards them. Oxygen sold separately.
Finally, folks speak of the helium-3 (³He) resource on the Moon at a trace concentration of about 10 parts per billion by mass. Such fuel could be useful in fusion reactors of the future. I will not address this option here, except to compare the energy density to that of fossil fuels. One gram of lunar soil has about 10−8 g of ³He, or about 3×10−9 moles, becoming 2×1015 particles. Reacting with deuterium, this process releases 18.6 MeV of energy, becoming 1.4 kcal/g. Compare this to coal at 4–7 kcal/g, oil at 10 kcal/g, and natural gas (methane) at 13 kcal/g. The energy density is not ridiculously small, but the resource is very diffuse, expensive to collect and deliver, and with no current application for its use. That’s why I’ll say no more here.
How easy is it to get at these various extraterrestrial resources? Nothing but empty space stands between us. Once in space, how easy is it to zip about from one place to the other? It doesn’t seem too bad, judging from movies.
A popular misconception is that once in space, we escape the grip of gravity. This is a natural intuitive leap if you have watched video of weightless astronauts in low Earth orbit (LEO). It turns out that gravity from Earth in LEO is still 90% what it is on the ground. The trick is that the orbiting spacecraft has tremendous sideways velocity, so that as it falls to Earth (yes, it is always falling!), the ground curves out from under it so that it continuously falls around the Earth. It takes a velocity of 7.7 kilometers per second (17,000 m.p.h.) to pull this off. Because the astronauts fall at the same acceleration as the spacecraft (equivalence principle), they appear to be weightless—much as an unlucky person would be in a falling elevator on Earth.
Low Earth orbit is said to be “halfway to anywhere” in the solar system. Even though a spacecraft in LEO is still deep within the grip of Earth’s gravity, there is some truth to the statement. The 7.7 km/s velocity imparted to the spacecraft gives it some of the kinetic energy required to climb further out of the well, representing 70% of the velocity (and half the energy) needed to escape. A more serious problem is that Earth itself is deep within the well of solar gravity. To go outward means climbing a hill. To go inward means first reducing velocity (takes rocket juice) to start falling toward the Sun, and then slowing down again (more juice) at the destination after picking up speed from the “downhill” ride.
Let’s look at the gravitational potential energy well in the solar system to get a feel for the size of the hills involved.
The blue line is dominated by the solar potential (goes down to −1.0 at the solar surface; not shown), but has little pockets made by the planets. Each pocket goes down to the respective planet’s surface. Mercury is off the plot to the lower-left, but Venus and Earth are visible (Earth at 1.0 astronomical units, or A.U., from the Sun), and Mars barely shows up. Jupiter’s well is the most dramatic, followed by Saturn’s. We live on the steep part of the hill, and Earth’s little dip is dwarfed by the climbs/descents to our neighbors.
But the potential energy graph by itself is misleading, being only part of the story. For instance, we travel around the Sun at a velocity of 30 km/s, while Mars sails at a more sedate 24 km/s. So to meet up with Mars, we have 6 km/s of extra velocity to burn, helping us up the hill. We speak of this as a Δv (delta-vee) adjustment to trajectory. In just the same way, a spacecraft in LEO has in the bank 7.7 km/s of the 11.2 km/s needed to escape Earth for good. Then it’s on to fight the Sun.
The next plot puts this in perspective, albeit only in simplified, approximate terms. The bottom of the plot represents the Earth’s ground. It takes 7.7 km/s of velocity to get to LEO (actually, it takes the equivalent of about 9.5 km/s because much effort is expended just climbing out, in addition to establishing the orbital speed). At 11.2 km/s, we’re free to take on the solar system. The plot is based on minimum-energy Hohmann transfer orbits.
Each planet is represented by three dots: the top one being outside the planet’s grip in an identical solar orbit, the next one down at low-planet orbit (akin to LEO), and the lowest represents being at rest on the surface. For Saturn and Jupiter, these surface points are off the chart—so taxing is this requirement. And for these two, there’s no “there” there anyway to land on. Crudely speaking, we must have the means to accomplish all vertical traverses in order to make a trip. For instance, landing on Mars from Earth requires about 17 km/s of climb, followed by a controlled 5 km/s of deceleration for the descent. Thus it takes something like 20 km/s of capability to land on Mars, and LEO is approximately half the battle. Similarly, it takes about 20 km/s to get to the asteroid belt at 2–3 A.U. LEO is a good deal less than halfway to Mercury, Jupiter, Saturn, and beyond. This is especially true if one wants to enter orbits around the giants.
Why all this talk of delta-v? Isn’t energy a more appropriate metric, since this relates to fuel requirements? Yes. I’ll leave the derivation for the Appendix, but the energy expended comes out to be the mass of fuel exhausted times ½v²exhaust, where vexhaust is the rocket exhaust velocity (usually about 3–5 km/s) in the frame of the rocket. Meanwhile, the velocity acquired by the spacecraft is
where the final piece is the natural logarithm of the ratio of initial spacecraft-plus-fuel mass to the final mass of the spacecraft after the burn. For small burns, the velocity acquired is proportional to the mass of spent fuel. Since the total energy expended is also proportional to the same, energy expended scales with delta-v, making delta-v a useful measure.
For larger burns, though, the non-linearity of the logarithm comes into play. Somewhat counter-intuitively, it is possible to get the rocket moving faster than the exhaust velocity once the fuel mass exceeds 63% of the total initial mass. In order to get delta-v values in the 20 km/s range when the exhaust velocity is less than 5 km/s requires almost nothing but fuel. See the Appendix for more on this.
Since the large delta-v’s required to get around the solar system require a lot of fuel, and we have to work hard to lift all that fuel from the Earth’s surface, could we just grab hydrocarbons from Jupiter or Titan and be on our way?
Let’s say you arrived in Jupiter orbit running on fumes, relying on the gassy giant to restock your coffers. In order to get close enough to Jupiter, you’ll be skimming the cloud-tops at a minimum of 42 km/s. Getting 1 kg of fuel on board will require you to accelerate the fuel to the speed of your spacecraft, at a kinetic energy cost of 885 MJ. The energy content of methane is 13 kcal/g, or 54 MJ/kg. Oops. Not even enough to pay for itself, energetically. Get used to Jupiter. And I have completely ignored the fact that you need marry two O2 molecules to each molecule of methane, meaning you actually get only 11 MJ per kilogram of total fuel. Utterly hopeless.
We could come to a full stop to gather giant amounts of fuel at no kinetic energy cost, and then power out. But Jupiter’s potential well is so deep that the efficiency of energy delivery suffers mightily. Well—what about Titan? It’s a smaller body, and somewhat far from the grip of Saturn. From the surface of Titan, it takes 2.6 km/s to escape the moon, then another 2.3 km/s to escape from Saturn. So about 5 km/s total to end up in a Solar orbit at 9.5 A.U. Add another 15 km/s to get the fuel back to the inner solar system, if that’s part of the plan.
As we saw above, the methane/oxygen mix delivers about 11 MJ/kg. If we could perfectly turn the energy of combustion into rear-directed exhaust of the reacted fuel, we would calculate the exhaust velocity by equating the thermal energy available with kinetic energy of the exhaust: E = ½mv² to find that each kilogram of fuel can attain an exhaust velocity of 4.7 km/s. Note that this ideal rocket engine would be cool to the touch if all the energy was transferred to exhaust kinetic energy. We’ll take 3 km/s as a more realistic number (this may still be high: good rocket fuels deliver 3–5 km/s, and methane is not a top choice for rocket fuel). Using the logarithmic rocket equation, we find we would need an initial-to-final mass ratio of 5.3:1 to reach 5 km/s, or 750:1 to deliver the fuel back to the inner solar system.
In the first case, we necessarily deliver less than 20% of the tank capacity to the environment outside Saturn’s grip. In the second case, the tank arrives about 0.1% full. If you sent a 6,000 gallon fuel tanker truck to the refinery and it came back with 8 gallons, you might be pretty steamed and consider the trip a wasted effort. If you wanted the Titan Tanker to be able to get back to the surface of Titan for another refill, you need to leave enough fuel for another 5 km/s expenditure. Now the tanker can only offer 3% of its capacity to the waiting customer—who was until this unwelcome news enjoying a view of Saturn’s rings.
This scenario pre-supposes that we can find oxygen on Titan, so we can arrive on the surface light and empty. If we had to carry oxygen to Titan from the inner solar system, the energy cost balloons to absurdity—as if it was not there already.
Now on to getting materials from asteroids. I will assume that we want the resources to end up somewhere else in the solar system, so that we must move the mass across the potential well.
The asteroid belt is over 20 km/s away in terms of velocity impulse. If the goal is to use the raw materials for production on Earth or in Earth orbit, we have to supply about 10 km/s of impulse. We would probably try to get lucky and find a nickel-metal asteroid in an unusual orbit requiring substantially less energy to reel it in. So let’s say we can find something requiring only 5 km/s of delta-v. Our imagined prize will be a cube 1 km on a side, having a mass around 1013 kg. This is very small for an asteroid, but we need to moderate our ambitions. From a resource point of view, it’s still a lot.
To get this asteroid moving at 5 km/s with conventional rocket fuel (or any “fuel” that involves spitting the mass elements/ions out at high speed) would require a mass of fuel approximately twice that of the asteroid. As an example, using methane and oxygen, (4 kg of O2 for every 1 kg of CH4), we would require two years’ of global natural gas production to be delivered to the asteroid (now multiply this by a large factor for the fuel to actually deliver it from Earth’s potential well). The point is that we would be crazy to elect to push the asteroid our way with conventional rockets.
So let’s use a solar sail! A big sheet of aluminized mylar would deflect solar photons and provide propulsive force. It works out that the momentum flux (pressure) from 1370 W per square meter (solar flux at Earth) is just 1370/c Newtons per m², where c is the speed of light. A Newton is about the weight of an apple. In our case, this calculates to 4.6 micro-Newtons per square meter. Right: you’ve never been knocked over by the sun. Even squirrels are safe. Solar wind pressure is orders of magnitude weaker still.
Big mass, small force per unit area. We’re going to need a big sail. Let’s try to make it the area of the Earth! I’m not kidding around here. 1014 square meters, baby! At a thickness of 25 microns, the mylar would have a mass of about 1012 kg. Hmmm. If we had a way to get this much mass off the Earth, we wouldn’t need to mess around with solar sails in the first place. So let’s tone it down a bit: I’ll go with an Egypt-size sail 1012 m² in size—which is still 10,000 times the launch mass of a fully-loaded Saturn V rocket, to humble us.
The acceleration of our 1013 kg asteroid, using Newton’s F = ma comes to about 50 nano-g (g ≈ 10 m/s²) It takes 350 years to reach 5 km/s. At least it can do it, given great patience. But there are big problems. Even if we could conceive of an Egypt-sized solar sail (difficult for me to swallow), orbital maneuvers don’t generally have the luxury of time. As the asteroid approaches Earth, it needs to be slowed down during the fly-by, otherwise it will just, well, fly by.
I may not have exhausted the possibilities, but I’m personally exhausted. I had never worked through these computations before, and took my typical approach of estimating obvious, brute-force solutions to a problem. When such things do not work out, it sends up a red flag that maybe this thing isn’t so simple. The ideas sound good, but are easier said than done. It looks to me as if the resources of space are effectively stranded in place.
I have stayed away from the commercial aspects of the problem. If the energetics don’t work, economics can do nothing to save it. No materials of intrinsic value (other than value from rarity) has been brought back from space—and certainly not in any economically advantageous way. It is noteworthy that NASA, ever existing under budgetary pressure, has not developed a viable commercial draw into the solar system. It is not for lack of talent, incentive, and technical capability that such a transition has not happened.
I still support the economics of space-enabled science. Some of our most precious knowledge of the nature of the Universe comes from space missions and observations. Keep that train rolling, by all means! But such endeavors are unlikely to fall outside of the public funding model.
The original post on space was not as quantitative as most Do the Math posts are, so I felt I had some unfinished business. I conclude with a reminder that the space angle is a distraction to the resource problems we will face on Earth this century—chiefly in the energy domain. I enjoyed doing the calculations, but at the same time feel that I could have spent my time on more meaningful and productive pursuits, rather than evaluating fantastical schemes.
For the brave/curious, I include here a mathematical description of the ideal rocket. Not part of the main post, so stop here if you’ve seen enough.
I won’t go through a full derivation, but will hit some of the key points. The basic approach is conservation of momentum. A rocket ejects some small nugget of mass (a molecule, for instance) at high speed. We will describe the total mass of the rocket before the ejection as M + m, ejecting mass m so the remaining rocket has mass M. The rocket travels at velocity V before the ejection, and the nugget is ejected at velocity v relative to the rocket (so V − v in the global frame: the minus sign because the nugget is directed opposite the forward direction of the rocket). After the ejection, the rocket will be traveling at V + ΔV.
The momentum before the ejection is (M + m)V, and after the ejection the combined momentum is M(V + ΔV) + m(V − v). Setting these equal to each other (conservation of momentum), we find that ΔV = (m/M)v.
Integrating this process step by step and starting the rocket from rest leads to the logarithmic result that V = v·ln(Mi/Mf), relating to initial and final masses of the rocket-fuel combination. When Mf is close to Mi (i.e., not much fuel spent), the logarithmic term just turns into the ratio of fuel spent to the total mass. But that’s basically just ΔV = (m/M)v again.
Looking at energy in our ejection scenario, the energy before ejection is Ei = ½(M + m)V². After ejection, we have two pieces to track, and the total energy is Ef = ½M(V + ΔV)² + ½m(V − v)². Working through the algebra, the change in energy is ΔE = ½mv²(1 +m/M), independent of the velocity, V, of the rocket. In other words, the energy changes by the kinetic energy of the ejected particle in the frame of the rocket times a small correction that we can ignore since m/M can be made arbitrarily small (like a molecule compared to the entire rocket). This is the basis for saying in the main post that the energy expended in a burn is essentially the kinetic energy of the spent fuel mass.
In detail, how much of the energy goes to the ejected fuel and how much to the rocket depends on the rocket velocity. At first, when V ≈ 0, almost all the energy in Ef goes to the second term (ejected piece). At the moment when V = v (when Mi/Mf = 2.71), all of the energy goes into the rocket kinetic energy, and the ejected lump ends up stationary (no kinetic energy). Later, both the rocket and the fuel share the kinetic energy pie as the ejected fuel moves in the same direction as the rocket after expulsion.
As an interesting aside, the fraction of energy expended by the rocket that actually goes into kinetic energy of the rocket—thus a measure of efficiency—starts out very small, and reaches a maximum value of 65%. The fueled spacecraft initially has total mass Mi, and after the burn it has mass Mf, meaning that the amount of fuel burned is ΔM = Mi − Mf. The final kinetic energy of the spacecraft is ½MfV², which can also be represented as ½Mfv²ln²(Mi/Mf) using our rocket equation. Meanwhile the amount of energy expended in the burn is ½ΔMv² (just summing the individual ½mv² contributions derived above). Taking the ratio of the two (note the cancellation of v²), we find an efficiency of delivering kinetic energy to the spacecraft that looks like ln²x/(x − 1), where x is the ratio Mi/Mf. For a while, the efficiency tracks ΔM/Mi, but reaches a peak of 0.65 (65%) when ΔM/Mi gets to 0.8 (Mi/Mf ≈ 4.92). To be clear, this has nothing to do with the efficiency of the rocket engine itself. Even a 100% efficient rocket engine that could eject spent fuel carrying away the entire heat of combustion (rocket engine/nozzle cool to the touch) would incur this efficiency hit in terms of transferring propulsive energy into kinetic energy of the spacecraft. I found this to be academically interesting.