Vectors
Page 18
* * *
A spacecraft, orbiting Earth around the equator just high enough to avoid the main effects of atmospheric drag, makes a complete revolution in about an hour and a half. If the Earth had no atmosphere, a spacecraft in a 'grazing orbit' would skim around just above the surface in 84.9 minutes. At the end of that time, it would not be above the same point on the Earth where it started. The Earth is rotating, too, and if the spacecraft revolves in the same direction as Earth it must go farther—about 2,370 kilometers, the distance that a point on the equator rotates in 84.9 minutes—before it again passes over the point where it began.
Now keep the spacecraft in a circular path above the equator, but instead of a grazing orbit, imagine that it travels 1,000 kilometers above the surface. Then the orbital period will be greater. It will now be about 106 minutes: the higher the orbit, the longer the period of revolution.
When the height of the spacecraft is 35,770 kilometers, the orbital period is 1,436 minutes, or one sidereal day (a solar day, the time that a point on the Earth takes to return to point exactly at the Sun, is 1,440 minutes). In other words, the spacecraft now takes just as long as the Earth to make one full revolution in space. Since the spacecraft is moving around at the same rate as the Earth, it seems to hover always above the same point on the equator.
Such a specialized orbit is called geostationary, because the satellite does not move relative to the Earth's surface. It is a splendid orbit for a communications satellite. There is no need for ground receiving antennae to track the satellite at all—it remains in one place in the sky. The term 'Clarkian orbit' has been proposed as an alternate to the cumbersome 'geostationary orbit,' in recognition of Arthur Clarke's original suggestion in 1945 that such orbits had unique potential for use in worldwide communications. Note, by the way, that a 24-hour period orbit does not have to be geostationary. An orbit whose plane is at an angle to the equator can be geosynchronous, with 24-hour period, but it moves up and down in latitude and oscillates in longitude during one day. The class of geosynchronous orbits includes all geostationary orbits.
A geosynchronous orbit has some other unusual features. It is at the distance from the Earth where gravitational and centrifugal accelerations on an orbiting object balance. To see what this means, suppose that you could erect a thin pole vertically on the equator. A long pole, and I do mean long. Suppose that you could extend it upwards over a hundred thousand kilometers, and it was strong and rigid enough that you could make it remain vertical. Then every part of the pole below the height of a geostationary orbit would feel a net downward force, because it is travelling too slowly for centrifugal acceleration to balance gravitational acceleration. On the other hand, every element of the pole beyond geostationary altitude would feel a net outward force. Those elements are travelling so fast that centrifugal force exceeds gravitational pull.
(Every mechanics textbook will point out that there are no such things as 'centrifugal forces'; there is only the gravitational force, curving the path of the orbiting body from its natural inclination to continue in a straight line. The centrifugal forces are fictitious forces, arising only as a consequence of the use of a rotating reference frame for calculations. But centrifugal forces are so convenient that everyone uses them, even if they don't exist! And when you move to an Einsteinian viewpoint you find that centrifugal forces now appear as real as any others. So much for theories.)
The higher that a section of the pole is above geostationary height, the greater the total outward pull on it. So if we make the pole just the right length, the total inward pull from all parts of the pole below geostationary height will exactly balance the outward pull from the higher sections above that height. Our pole will hang there, touching the Earth at the equator but not exerting any downward force on it. If you like, we can think of the pole as an enormously long satellite, in a geostationary orbit.
How long would such a pole have to be? If we were to make the pole of uniform cross-section, it would have to extend upwards a distance of about 143,700 kilometers. This result does not depend on the cross-sectional area of the pole, nor on the material from which it is made. It should be clear that in practice we would not choose to make a pole of uniform cross-section, since the downward pull that it must withstand is far greater up near geosynchronous height than it is near the Earth. At the higher point, the pole must support the weight of more than 35,000 kilometers of itself, whereas near Earth it supports only the weight hanging below it. From this, we would expect that the best design will be a tapered pole, with its thickest part at geostationary altitude where the pull on it is greatest.
The idea of a rigid pole is also misleading. We have seen that the only forces at work are tensions. It is thus more logical to think of the structure as a cable than a pole.
We now have the major features of our 'basic Beanstalk.' It will be a long, strong cable, extending from the surface of the Earth on the equator, out to beyond the geostationary orbit. It will be of the order of 144,000 kilometers long. We will use it as the load-bearing cable of a giant elevator, to send materials up to orbit and back. The structure will hang there in static equilibrium, revolving with the Earth. It is a bridge to space, replacing the old ferry-boat rockets.
That's the main concept. What could be simpler? We have—perhaps an understatement—left out a number of 'engineering details,' but we will look at those next.
DESIGNING THE BEANSTALK.
Let us list some of the questions that we must answer before we have a satisfactory Beanstalk design. The most important ones are as follows:
• What shape should the load-bearing cable have?
• What materials will it be made from?
• Where will we get those materials?
• Where will we build the Beanstalk, where will we attach it to Earth, and how will we get it installed?
• How will we use the main cable to move materials up and down from Earth?
• Will a Beanstalk be stable, against the gravitational forces from the Sun and Moon, against weather, and against natural events here on Earth?
• What are the advantages of a Beanstalk over rockets ?
• If we can get satisfactory answers to all these questions, when should we be able to build a Beanstalk?
We can offer definite answers to some of these questions; other answers can only be conjectures. Let's begin with the first, which is also the easiest.
Suppose that the load-bearing cable is made of a single material. Then the most efficient design is one in which the stress on the material, per unit area, is the same all the way along it. This means there is no wasted strength. With such an assumption, it is a simple exercise in statics to derive an equation for the cross-sectional area of the cable as a function of distance from the center of the Earth.
The result has the form:
(1) A(r)=A(R).exp(K.f(r/R).d/T.R)
(Note to Editor Baen. I know you told me not to use equations, but surely I'm entitled to at least one.)
In this equation, A(r) is the cross-sectional area of the cable at distance r from the center of the Earth, A(R), is the area at the distance R of a geostationary orbit, K is the gravitational constant for the Earth, d is the density of the material from which the cable is made, T is the tensile strength of the cable per unit area, and f(r/R) = 3/2 — R/r — (r/R)2/2.
Equation (1) tells us a great deal. First, we note that the variation of the cross-sectional area with distance does not depend on the tensile strength T directly, but only on the ratio T/d, which is the strength-to-weight ratio for the material. The substance from which we will build the Beanstalk must be strong, but more than that it should be strong and light.
Second, we can see that the shape of the cable is tremendously sensitive to the strength-to-weight ratio of the material, because this quantity occurs in the exponential of equation (1). To take a simple example, suppose that we have a material with a taper factor of 10,000. We define taper factor as the cross-sectional ar
ea of the cable at geostationary height, divided by the cross-sectional area at the surface of the Earth. So, for example, a cable that was one square meter in area at the bottom end would in this case be 10,000 square meters in area at geostationary height.
Now suppose that we could double the strength-to-weight ratio of the material we use for the cable. The taper ratio would drop from 10,000 to 100. If we could double the strength-to-weight ratio again, the taper ratio would reduce from 100 to 10.
It is clear that we should make the Beanstalk of the strongest possible material. Note than an infinitely strong material would need no taper at all.
Two other points are worth noting about the shape of the cable. It is easy to show that the function f(r/R) has its maximum value at r = R. This confirms the intuitive result, that the cable must be thickest at geostationary height, where the load is greatest since the cable must support all the downward weight between that height and the surface of the Earth. Second, a look at the change of f(r/R) with increasing r shows that the cross-sectional area decreases slowly above geostationary height. This is why we need a cable with a length that is much more than twice the distance to that height.
MATERIALS FOR THE BEANSTALK.
The cable that we need must be able to withstand a tension at least equal to its own weight from a height of 35,000 kilometers down to the surface. In practice, it must be a good deal stronger than that. We will certainly want to build in a reasonable safety factor, and we will want to hang other structures on the cable all the way down, to make it into a usable transportation system. So we expect that we will need to work with a very strong material, one with an unusually high strength-to-weight ratio. Of course, if one does not have a material that is quite as strong as needed, one can try and compensate by increasing the taper factor, but we have seen that this would be a very inefficient way to go. Halving the strength of materials would square the taper factor. The incentive to work with the strongest possible materials is very large.
The tension in the cable at a height of 35,770 kilometers, where upward and downward forces exactly balance, is less than the weight of a similar length of 35,770 kilometers of cable down here on Earth, for two reasons. The downward gravitational force decreases as the square of the distance from the center of the Earth, and the upward centrifugal force increases linearly with that distance. Both these effects tend to decrease the tension that the cable must support. A straightforward calculation shows that the maximum tension in a cable of constant cross-section will be equal to the weight of 4,940 kilometers of such cable, here on Earth. This is in a sense a 'worst case' calculation, since we know that the cable will be designed to taper. However, the need for a safety factor means that we need to be conservative, and the figure of 4,940 kilometers gives us a useful standard in terms of which we can calibrate the strength of available materials.
The definition we have chosen of a cable's strength, namely, how much length of its own substance it must support under Earth's gravity, is used quite widely. For a particular material, the length of itself the cable will support is called the 'support length' or 'characteristic length.' It is particularly handy because of the way in which the strength of materials is usually described, in terms of the tons weight per square centimeter (or per square inch) that they will support. (It would be more desirable, scientifically speaking, to give strength in dynes per square centimeter, or in newtons per square meter. These measures are independent of the Earth's surface gravity. But historically, pounds per square inch and kilos per square centimeter came first and things are still given that way in most of the handbooks. Note also that we are concerned only with tensile strength—how strong the material is when you pull it. Compressive and shear strengths are quite different, and a material may be very strong in compression and weak in tension. A building brick is a good example of this.)
Against our requirement of a support length of 4,940 kms, how well do the substances that we have available today measure up?
Not too well. Now we see why no one has yet built a Beanstalk. Table 1 shows the strengths of currently available materials, their densities, and their support lengths. (The physical data that I am using here is drawn, wherever possible, from the "Handbook of Chemistry and Physics," 57th Edition. It is one of the most widely available reference texts and should be in any reasonable library.)
TABLE 1
STRENGTH OF MATERIALS
Material
Tensile
Strength
(kg/cm2)
Density
(gm/cm3)
Support
Length
(km)
Lead
200
11.4
0.18
Gold
1,400
19.3
0.73
Aluminum
2,000
2.7
7.4
Cast iron
3,500
7.8
4.5
Carbon steel
7,000
7.8
9.0
Manganese steel
16,000
7.8
21.
Drawn tungsten
35,000
19.3
18.
Drawn steel wire
42,000
7.8
54.
Iron whisker
126,000
7.8
161.
Silicon whisker(SiC)
210,000
3.2
660.
Graphite whisker
210,000
2.0
1,050
Fictionite
2,000,000
2.0
10.000
Not surprisingly, we won't be trying to make a Beanstalk support cable from lead. As we can see from the table, even the best steel wire that we can find has a support length only one hundredth of what we need. The last entry in the table, Fictionite, would be perfect but for one drawback: it doesn't exist yet. The strongest materials that we have today, graphite and silicon carbide whiskers, still fall badly short of our requirements (for Earth, that is. A Mars Beanstalk has a minimum support length of only 973 kms. We could make one of those nicely using graphite whiskers).
Does this mean that we have a hopeless situation? It depends what confidence you have in the advance of technology. Table 2 lists the strength of materials that have been available at different dates in human history. There is some inevitable arbitrariness in making a table like this, since no one really knows when the Hittites began to smelt iron, and there must have been poor control of times, temperatures and purity of raw materials in the Bronze Age and early Iron Age. All these factors have a big effect on the tensile strength of the products.
It is tempting to try and fit some kind of function to the values in the table, and see when we will have a material available with a support length of 5,000 kms. or better. It is also very dangerous to even think of such a thing. For example, consider a fit to the data of the form: Strength = B/(t - T), where B and T are to be determined by the data, and t is time in years before the year 2000 A.D. This fits the data fairly well if we choose B = 525,000 and T = 17.5 years. Unfortunately, such a form becomes infinite when t = T. If we were to believe such a fit, we would expect to have infinitely strong materials available to us some time in 1982!
TABLE 2
PROGRESS IN STRENGTH OF MATERIALS
AS A FUNCTION OF TIME
Year
Available material
Tensile
Strength
(kg/cm2)
1500 B.C.
Bronze
1,00
1850
Iron
3,500
1950
Special steels
16,000
1970
Drawn steel
42,000
1980
Graphite and silicon whiskers
210,000
Note: Years given indicate the dates when the materials could first be reliably produced in production quantities.
Not surprisingly, extrapolation of a trend without using physical models can lead us to ridiculous results. A much more plausible way of predicting the potential strength of materials is available to us, based on the known structure of the atom. In chemical reactions, only the outermost electrons of the atom participate, and it is the coupling of these outer electrons that decides the strength of chemical bonds. These bonds in turn set bounds on the possible strength of a material. Thus, so far as we are concerned the nucleus of the atom—which is where almost all the atomic mass resides—contributes nothing: strength of coupling, and hence material strength, comes only from those outer electrons.
In Table 3 we give the strengths of the chemical bonds for different pairs of atoms. These strengths, divided by the molecular weight of the appropriate element pair, decide the ultimate strength-to-weight ratio for a material entirely composed of that pair of elements. The final column of the table shows the support length that this strength-to-weight ratio implies, using the carbon-carbon bond of the graphite whisker as the reference case.
TABLE 3
POTENTIAL STRENGTH OF MATERIALS BASED
ON THE STRENGTH OF CHEMICAL BONDS
Element pairs
Molecular
weight *
Chemical
bond strength
(kcal/mole)
Support
length
(km)
Silicon-carbon
40
104
455
Carbon-carbon
24
145
1,050
Fluorine-hdrogen
20
136
1,190
Boron-hydrogen
11
80.7
1,278