A 24-hour circular orbit does not have to be geostationary. If the plane of its orbit is at an angle to the equator, it will be geosynchronous, with a 24-hour orbital period, but it will move up and down in latitude and oscillate in longitude during the course of one day. The class of geosynchronous orbits includes all geostationary orbits.

All geostationary orbits share the property that the gravitational and centrifugal forces on an orbiting object there are exactly equal. If by some means we could erect a long, thin pole vertically on the equator, stretching all the way to geostationary orbit and beyond, then every part of the pole below the height of 35,770 kilometers would feel a net downward force because it would be moving too slowly for centrifugal acceleration to balance gravitational acceleration. Similarly, every element of the pole higher than 35,770 kilometers would feel a net upward force, since these parts of the pole are traveling fast enough that centrifugal force exceeds gravitational pull.

The higher that a section of the pole is above geostationary height, the greater the total upward pull on it. So if we make the pole just the right length, the total downward pull from all parts of the pole below geostationary height will exactly balance the total upward pull from the parts above that height. The pole will then hang free in space, touching the Earth at the equator but not exerting any downward push on it.

How long does such a pole have to be? If we were to make it of uniform material along its length, and of uniform cross section, it would have to extend upward for 143,700 kilometers, in order for the upward and downward forces to balance exactly. This result does not depend on the cross-sectional area of the pole, nor on the material of which the pole is made. However, it is clear that in practice we should not make the pole of uniform cross section. The downward pull the pole must withstand is far greater up near geosynchronous height than it is near the surface of the Earth. At the higher points, 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. Thus the logical design will be tapered, with the thickest part at geostationary altitude where the pull is greatest, and the thinnest part down at the surface of the Earth.

We now see that “pole” is a poor choice of word. The structure is being pulled, everywhere along its length, and all the forces at work on it are tensions. We ought to think of the structure as a cable, not a pole. It will be of the order of 144,000 kilometers long, and it will form the load-bearing cable of a giant elevator which we will use to send materials to orbit and back.

The structure will hang in static equilibrium, rotating with the Earth. It will be tethered at a point on the equator, and it will form a bridge to space that replaces the old ferry-boat rockets. It will revolutionize traffic between its end points, just as the Golden Gate Bridge and the Brooklyn Bridge have made travel between their end points a daily routine for hundreds of thousands of people.

That is the main concept. Now we have to worry about a number of “engineering details.”

Designing a beanstalk

A number of important questions need to be answered in the process of beanstalk design:

* What is the shape of the load-bearing cable?

* What materials should be used to make it?

* Where will we obtain those materials?

* How will we use the main cable to move loads up and down from Earth?

* Will a beanstalk be stable, against the gravitational pull of the Sun and Moon, against weather, and against natural events here on Earth?

* And finally, when might we be able to build a beanstalk?

The first question is the easiest to answer. The most efficient design is one in which the stress on the material, per unit area, is the same all the way along the beanstalk’s length. 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 equation is: A(r) = A(R).exp(K.f(r/R).d/T.R)

where A(r) is the cross-sectional area of the cable at distance r from the center of the Earth, R is the radius of a geostationary orbit, K is Earth’s gravitational constant, d is the density of the material from which the cable is made, T is its tensile strength per unit area, and f is a function given by:

f(x) = 3/2 — l/x — x²/2

The equation for A(r) shows that the important parameter in beanstalk cable design is not simple tensile strength, but rather T/d, the strength-to-density ratio of the material. The substance from which we will build the beanstalk must be strong, but more than that it must be strong and light.

Second, the equation shows that the tapering shape of the cable is tremendously sensitive to the strength-to-density ratio of the material, and in fact depends exponentially upon it. To see the importance of this, let us define the 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. Suppose that we have some material with a taper factor of 10,000. Then a cable one square meter in area at the bottom would have to be 10,000 square meters in area at geostationary height.

Now suppose that we could double the strength-to-density ratio of the material. Then the taper factor would drop from 10,000 to 100. If we could somehow double the strength-to-density ratio again, the taper factor would reduce from 100 to 10. An infinitely strong material would need no taper at all.

It is clear that we must make the beanstalk of the strongest, lightest material that we can find. What is not obvious is whether any material will allow us to build a beanstalk with a reasonable taper factor. Before we can address that question, we need to know how strong the cable has to be.

The cable must be able to support the downward weight of 35,770 kilometers of itself, since that length hangs down below geostationary height. However, that weight is less than the weight of a similar length of cable down here on Earth, for two reasons. First, the downward gravitational force decreases as the square of the distance from the center of the Earth; and second, 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 tension in a cable of constant cross-section will be equal to the weight of 4,940 kilometers of such a cable, here on Earth. This is in a sense a “worst case” calculation, since we know that the cable will not be of constant cross-section; rather, it will be designed to taper. However, the figure of 4,940 kilometers gives us a useful standard, in terms of which we can calibrate the strength of available materials. Also, we want to hang a transportation system onto the central load-bearing cable, so we need an added margin of strength for that.

Now let us compare our needs with what is available. Let us define the “support length” of a material as the length of itself that a cable of such a material will support, under one Earth gravity, before it breaks under its own weight.

The required support length is 4,940 kms. What are the support lengths of available materials?

TABLE 1 lists the support lengths for a number of different substances. It offers one good reason why

TABLE 1

Strength of materials

Material Density (gm/cc)  | Tensile Strength (kgm/cm²)  | Support Length (km)

Lead 11.4 | 200 | 0.18

Gold 19.3 | 1,400 | 0.73