Wizards, Aliens, and Starships: Physics and Math in Fantasy and Science Fiction

by Adler, Charles L.

  where L is the length of the structure and A its cross-sectional area. The parameter geff is the effective acceleration of gravity acting on the structure. For a short cable, this parameter would simply be the acceleration of gravity at the Earth’s surface. However, the cable is so long that (as the quotation from Tsiolkovsky shows) the effective value of g will decrease as you go up the tower, reversing in direction once you pass the geostationary point. To calculate this value would be difficult, but it must lie between 0 and the actual acceleration of gravity on Earth’s surface, so as a guesstimate let’s choose a value about one-tenth of the actual acceleration—say, 1 m/s2.

  The parameter of interest is not the tension itself but the stress in the cable (Y), which is the tension divided by its cross-sectional area:

  Putting in numbers, if we estimate the cable’s length as 140,000 km (see below) and the density of steel as 8, 000 kg/m3, the maximum stress is about 1012 N/m2. This estimate is three times the value of the stress indicated by an accurate calculation, but it is ballpark correct [24]. Unfortunately, the breaking stress of steel is only about 2.5 × 108 N/m2, about two orders of magnitude too small. Steel clearly won’t work. Is there anything that will?

  8.5 STRESSES AND STRAINS: CARBON NANOTUBES

  The magnitude of the problem can be seen by examining the height parameter of various materials. We can define a “breaking height” for a material by the formula

  where Ymax is the maximum stress the material can take before breaking (or, more realistically, significantly deforming—what is known as the “plastic limit” for the material). Effectively, if we made a cable out of this material and hung it onto some sort of skyhook, this is the maximum length before the cable tore itself apart under its own weight. (I’m using the real value for g here, not the effective value.) For steel, h is about 3 km, or some 50,000 times less than the length the skyhook would have to be.

  The ideal material to build such a structure is light (to minimize its weight) and strong. The best candidates right now are materials built from carbon nanotube fibers [207]. These are carbon compounds whose bonds are arranged in long tubes and, pound for pound, are over a hundred times stronger than steel is. Using values in the paper and a density of approximately 1 g/cm3, we calculate a height parameter h = 1,000 km. This is still not strong enough if we were to try to build a simple cable for the space elevator, but other designs make the building problems much easier.

  The best design anyone has come up with for the space elevator using these fibers calls for a tapered aspect: thickest at geosynchronous orbit and thinning out both above and below [24][128][190]. These tapered structures are designed so that the stress is the same everywhere throughout the structure. The height of the structure is a whopping 150,000 km, or roughly halfway to the Moon. The advantage of this structure is obvious: the strains are minimum at geosynchronous orbit, so it makes sense to put most of the mass there. The shape of the structure is determined by the taper parameter, τ, the ratio of the thickness of the structure at the geosynchronous point to the ratio of the structure on the ground. This can be found to be [190]

  Here, R is the radius of the Earth. Using R = 6,400 km and h = 1,000 km, the ratio is e6.4 = 600, meaning that the middle would need to be 600 times wider than its ends. This is not exactly impossible, but the paper I am citing uses a value of h for these nanotubes of 2,100 km, leading to a 10:1 taper ratio. Under this design, because of the lightweight nature of the carbon nanotubes, the entire structure has a mass of only 150,000 kg. This design is meant to allow a 10,000 kg payload to climb up the tower.

  In principle, spacecraft launched from the top of the tower would have speeds that would let them reach Mercury or Jupiter without any additional rockets, which is another appealing aspect of the space elevator.

  One issue with these carbon nanotube fibers is that as of yet, none has ever been made that is longer than a few centimeters. Stringing them into a strand 150,000 km long is a formidable engineering task, and there is a very good possibility that if this is done, the overall strength of the cable will be much weaker than the yield stress of the individual fibers.

  8.6 ENERGY, “CLIMBERS,” LASERS, AND PROPULSION

  Another tricky issue with the tower is getting stuff up it once it is built. Current designs call for “climbers,” which are essentially robots that pull themselves up along the cable. The power requirements for these climbers is pretty high, as the following estimate shows.

  As far as gravity is concerned, you can think of anything sitting on the surface of the Earth as sitting at the bottom of a deep well. It takes a lot of effort to climb out of the well. The gravitational potential energy of any payload of mass m at some distance r from the center of the Earth is given by the formula

  The law of conservation of energy tells us that to get a climber of mass m from the surface of the Earth to geosynchronous orbit requires us to supply the difference in the potential energy. This is

  We don’t need to worry about the kinetic energy of our payload as the Earth’s rotation will provide that through the tower; if we have a 10,000 kg payload the total energy we need to provide is 5.3 × 1011 J. To provide this power, a 2001 proposal for a space elevator suggested using a free electron laser (FEL) to beam power to optical cells attached to the climber [76]. Other ideas have been to use solar power or electrical power run up the cable. All these ideas have their drawbacks, in particular the beamed power idea. However, assuming that it can be done, let’s look at the costs involved.

  The first thing to realize is that there will always be some waste involved in this process. Consider the FEL idea: first, one must convert electrical power into laser radiation, which can be done with an efficiency of about 10%, give or take. Then it must be beamed to the climber, absorbed by the solar cells, and transformed again into electrical energy. The broad-wavelength solar cells that are commercially available today have a best efficiency of about 20%; because the laser operates at a single wavelength, the proper choice of materials can raise the overall conversion efficiency. The 2001 report estimated a 90% conversion efficiency, but the estimate is based on a personal communication from the developer of these cells [76]. A personal communication is not the same as a peer-reviewed publication and makes one suspicious that the quoted figure was exaggerated. Certainly there has been nothing in the scientific literature to justify this extremely high conversion efficiency. Being generous, I will assume a 60% conversion efficiency, and another 60% conversion efficiency of the electrical power into mechanical (i.e., climbing) power. This gives an overall efficiency e of

  Therefore, the total energy required to bring this 10,000 kg load into orbit is Etot = 5.3 × 1011/.036 = 1.5 × 1013 J. Assuming a cost of $0.1 per kW-hr for the electrical energy required, this is a cost of roughly $400,000, or about $40 per kilogram. This is about 500 times less than the current $20,000 per kilogram via rockets. (The report gives an overall cost of about $500 per kilogram, but this includes all costs, not just energy.) Is this too good to be true?

  8.7 HOW LIKELY IS IT?

  The trickiest part of building the elevator is that it is simply so incredibly long. Here the history of architecture is instructive. From ancient times until the late nineteenth century, the height of tall buildings was constrained to be less than 150 m because of the materials used (usually stone.) In 1889 the Eiffel Tower effectively doubled the maximum achievable height, and buildings have been getting taller and taller since then, with iron, steel, and concrete used in their construction. At the time of this writing, the tallest building in the world is the Burj Khalifa tower in the United Arab Emirates, which stands a whopping 818 meters (2,680 feet) high. The space elevator would be more than 150,000 times taller.

  Maybe this is the wrong way to look at it. Since it’s a structure in tension, maybe we should consider it as being a suspension bridge. The longest suspension bridges in the world are only about 2 km long, or 75,000 times shorter than the space elevator would be. The space ele
vator is so long that it could be wrapped around the world nearly four times.

  Let’s go through a list of things that need to be done to make the space elevator:

  • Develop a climber system. This should be the easy part; space elevator enthusiasts have sponsored a $0.9 million prize for a climber to ascend a 1 km cable at a speed of 2 m/s. To date, the best entry has climbed 100 m at an average speed of 1.8 m/s.

  • Develop the laser system for beamed power. Bringing a 10,000 kg climber up to geosynchronous orbit in one week requires an average power output of roughly 2.4 MW from the laser (based on the estimates provided in section 4.6). This greatly exceeds the maximum power available from any FEL built to date. By comparison, the Jefferson Labs Free Electron Laser has a maximum output power of 14 kW although there is no reason a more powerful FEL can’t be built.

  • Develop the carbon nanotube fibers. The strength-to-density ratio of the best fibers developed to date is still an order of magnitude less than what is needed. There are several papers indicating that long composites built from these fibers cannot achieve the strength needed because of defects. To build the cable within the next 20 years would require an average 50% increase in the yield strength per year.

  • Deploy and construct the cable. The 2001 report gives a detailed analysis of how the cable could be built. First, a strand strong enough to support a climber would be deployed from a satellite initially in geosynchronous orbit; as the strand was lowered, the satellite’s orbit would be boosted. Then climbers would be sent up the strand to weave in additional strands until, ultimately, it would be thick enough to support a 104 kg payload [76].

  None of these would be impossible to achieve except perhaps developing the composite materials. The question is how much money would be needed, and how long it would take to pay off the initial costs. The big costs are the materials costs plus the deployment costs; the costs for R&D for the materials and satellite deployment system are fairly small compared to the materials and deployment costs. Both Google and the country of Japan have put money and effort into developing this idea; in 2008, Japan was willing to commit about $8 billion in funds to develop the project. Let’s assume that Bradley Edward’s estimate is correct and that it will take a total expenditure of $30 billion to build the entire thing. I am going to assume that the space elevator would take 20 years to build, which is what its most enthusiastic supporters contend, and that it would then have a 20-year operating lifetime. This isn’t what its supporters contend, but it is reasonable given the harsh environment of space. Let’s assume that the money is financed at a 7% interest rate, which is reasonable in light of the high risk of the venture. A mortgage calculator on the web tells me that financing this venture over the 40-year period would cost roughly $90 billion.

  At the end of his phase 2 report, Edwards states that the total capacity of the system would be putting 1,000 tons of payload into space per year, or about 106 kg [77]. This strikes me as a very optimistic estimate: because each payload is about 104 kg, this means running 100 trips per year, or once every three days. I’ll assume an order of magnitude less than this: ten payloads per year, or 105 kg into space per year. In a 20-year working lifetime the space elevator could lift about 2 × 106 kg (roughly five million pounds) of payload into space at a cost of roughly $1 billion above construction and development costs, assuming the canonical $250 per pound payload costs of the elevator. The total investment in the space elevator would therefore be about $91 billion, including interest, or the equivalent costs of lifting about two million pounds of payload into space by conventional means. Over the allotted time period we get a roughly 2.5-fold savings in cost in launching satellites into geosynchronous orbit compared to conventional means, if this estimate, which was a very favorable one, is correct. This is decent, but not the 99% savings that some enthusiasts offer. Again, this figure is reached by assuming a much lower capacity than the authors of the study do.

  Bottom line: the space elevator is a very risky investment that offers moderate savings over conventional means of space transportation. Lowering infrastructure costs for conventional space systems might make them even more competitive with the space elevator. I’ve read the statement that people will build the space elevator 20 years after everyone stops laughing at it. Well, no one is laughing anymore, but I think the case has still to be made for it.

  8.8 THE UNAPPROXIMATED ELEVATOR

  This section is for the math and physics wonks who want to repeat the calculations in this chapter in a more exact way. The work here is based on P. K. Aravind’s paper in the American Journal of Physics [24].

  The analysis of this structure is based on the technique of free-body diagrams, which can be found in any standard elementary physics textbook, such as Fundamentals of Physics by Halliday, Resnick, and Walker. In a free-body analysis, we analyze and resolve the forces acting on one element of the structure independent of the rest of it; because the structure is in equilibrium, the net force acting on any independent element must be zero.

  Let’s consider a section of the elevator at distance r from the center of the Earth and of length dr. The tension in the structure at distance r is T(r). There are four forces acting on the structure: the tension at the top and bottom, tending to pull the structure apart; the weight of the structure, pulling it down; and a “centrifugal force” due to the rotation of the Earth, tending to push it outward. We can write this as

  Here, A is the cross-sectional area of the structure (which may depend on height), ρ is the density of the structure, M is the mass of the Earth, G is the universal gravitational constant, and Ω is the angular frequency of the Earth’s rotation (= 2π/day = 7.3 × 10−5/s). If we make dr infinitesimal in length, we can rewrite the equation as

  We can think about solving this in two different ways:

  1. By specifying the cross-sectional area as a function of r. The simplest is cross-section area A is constant.

  2. By specifying some condition on the stress of the elevator, for example, making the stress throughout the structure constant.

  I’ll solve the first case, specifying the cross-sectional area as constant, here. I leave the second problem as an online exercise.

  The stress, Y, is the tension divided by the cross-sectional area: Y = T/A. Therefore, we can rewrite the equation

  The condition at the ground is that Y(R) = 0, where R is the radius of the Earth. The equation can be solved by integration:

  The maximum value of the tension occurs at the radius of the geosynchronous orbit:

  The maximum value of the stress is

  Using parameters appropriate for the Earth and a density of 8,000 kg/m3 (appropriate for steel), then the maximum stress is 1.2 × 1013 N/m2.

  CHAPTER NINE

  MANNED INTERPLANETARY TRAVEL

  9.1 IT’S NOT AN OCEAN VOYAGE OR A PLANE RIDE

  In Robert Heinlein’s novel Podkayne of Mars, the eponymous heroine takes a voyage from her home on Mars colony along the first leg of a projected three-legged voyage to Venus, Earth, and back home again [118]. The ship she travels on, the Tricorn, is essentially an ocean liner moved into space; it has fancy dining halls, a ballroom, and first-class, second-class, and steerage compartments. The ship is described in detail. As is customary in a Heinlein novel, the science is pretty good: the one place where it departs most from reality is the “torch” drive, which is essentially direct matter-energy conversion, allowing the ship to travel under its own acceleration for a good part of the voyage. Even so, the trip to Venus alone takes several months because of the huge distances involved.

  Other novels are filled with similar voyages. Many of them depict long voyages in cramped quarters, and in some the plots are driven by the ennui that results. In Arthur C. Clarke’s 2001: A Space Odyssey, three of the five crewmen aboard the Discovery are placed in suspended animation because the limited life support aboard the ship couldn’t handle more than two crewmen for the multiyear voyage (to Saturn’s moons in the novel, Jupit
er’s in the movie). So: trips to low Earth orbit take a few hours; a trip to the Moon takes a few days; and trips to even the nearer planets take months to years, even in science fiction novels, at least in scientifically accurate ones. Why is that?

  Well, this is a combination of two related factors, distance and gravity. At its closest approach, Venus, the nearest planet to Earth, is about 40 million km from the Earth, or 110 times farther away than the Moon. Mars is even farther away: at the closest approach, it is 78 million km from Earth. The Apollo astronauts traveled to the Moon in three days; at the same average speed, it would have taken them almost a year to reach Venus.

  The other tricky part, gravity, is due to the fact that all of the planets in the Solar System are in orbit around the Sun; they must follow the laws of gravity as they orbit it, as must any spacecraft launched from Earth to another planet. While the spacecraft is traveling from Earth to Mars, it is essentially a tiny planet in orbit around the Sun. This means it can’t simply be pointed at Mars and launched off: Mars is moving in its orbit. Even if the craft could somehow travel on a straight line, when it got there Mars would be somewhere else. And the rocket doesn’t move on a straight line: as I said, it’s essentially a little planet too.

  9.2 KEPLER’S THREE LAWS