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Authors: Charles Sheffield

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Note that our old friend, the rocket, satisfies
none
of our ideal system requirements. The Space Shuttle, our first reusable spacecraft, is not suited to anything beyond low earth orbit activities, and is, with all its advantages over its non-reusable predecessors, still a very primitive system.

It may sound improbable, but an ideal space system, satisfying
all
our requirements, could be here in a couple of generations. As we shall see, the technology needed is not far from that already available to us.

It is curious that science fiction, which likes to look beyond today's technology, has remained so infatuated with the idea of rockets. Some people even use them to
define
the field. Look at the 'sf' section in public libraries and you will often see a small drawing of a rocket attached to the spine of each volume. It may be a perverse choice of label for a branch of writing that covers everything from 'Ringworld' to 'Flowers for Algernon,' but you can see how the logic goes; science fiction means space travel, and space travel means rockets—because they are 'the only way of getting up to space and around in space.' After all, there is nothing for any other sort of transportation to 'push against' in space. Right?

Not quite. We will try and dispose of that peculiar viewpoint here. Our preoccupation with rockets for space travel will probably amaze our descendants.

"Why use something as wasteful and noisy as a rocket," they will ask, "when there are simple, clean, efficient alternatives? Why didn't they use Beanstalks?"

The Age of Rockets may look to them like the Age of Dinosaurs. Let's try and see it through their eyes, beginning with the most basic principles.

* * *

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 area 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.)

BOOK: Vectors
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