A Cosmic Funicular

"...where little cable cars climb halfway to the stars...."

"I Left My Heart in San Francisco", sung by Tony Bennett

Of the two main methods proposed to get Humanity off Earth and into space to spread into the Galaxy, the stranger consists of what designers call a space elevator, basically a cable that extends from Earth’s surface to geosynchronous orbit and a car that pulls itself up the cable. But technically speaking, we must say that the device is not truly an elevator: the cable doesn’t lift the car. Instead, the car pulls itself up the cable, which remains stationary, so we must more properly call the device a funicular. Arguments over nomenclature aside, we want to know whether this thing is feasible.

Can we build a cable nearly 36,000 kilometers long and make it strong enough to support itself and a useful load? We will have to make the cable out of a substance that has the largest strength-to-weight ratio that we can find and use as an engineering material. The one material that seems to give us what we want is carbon fiber.

What magnitude of force does the cable have to withstand without weakening to the breaking point? Two forces act directly on any given segment of the cable – the segment’s weight (due to Earth’s gravity) and its centrifugal force (due to its following Earth’s rotation). On a short segment pondering mass dm those forces conform to the statement that

(Eq’n 1)

For the product of Earth’s mass and the gravitational constant we have
MG=3.99x10^{14} meters cubed per second. We also know that Earth rotates
at an angular rate of ù=7.27x10^{-5} radian per second. If the cable has
a uniform linear density L such that dm=Ldr, then we can calculate the total
force exerted on the cable at the Clarke orbit through a straightforward
integration; that is, we have

(Eq’n 2)

We want to calculate the force exerted by a cable extending between r_{0}=6,378
kilometers from Earth’s center to r_{1}=42,251.2 kilometers from Earth’s
center. Equation 2 gives us that number as

(Eq’n 3)

If L=10^{-3} kilograms per meter (one gram per meter), Then F=-48,500
newtons (the minus sign indicating that the force is directed in the negative
radial direction; that is, toward Earth), the equivalent of what 4944 kilograms
weighs lying on Earth’s surface. Can we make a cable that will support that
weight?

Amoco makes a carbon fiber, called Thornel™, that has a density of 1.76 gram per cubic centimeter and a strength of 3.2 GigaPascals; that is, it can endure 3.2 billion newtons per square meter without breaking. If we make our cable out of Thornel™, it must have a cross section of 0.568 square millimeters in order to ponder one gram per meter of length. That cross section will withstand a force of 1817.6 newtons, not enough to support the cable by a factor of 26.68, which fact seems to thwart our project.

Perhaps a different material will help. Experiments have shown that diamond can withstand pressures as high as 60 GigaPascals, 18.75 times stronger than Thornel™. Diamond has a density of 3.5 grams per cubic centimeter, so if we use diamond fiber (call it Atlasite, which can be made by our Omnifex technology) in our cable, then our cable must have a thickness of 0.286 square millimeters to ponder one gram per meter of length. That cross section will withstand a force of 17,137.4 newtons. That’s still not enough, but we’re not done yet.

Near the Clarke orbit centrifugal force becomes strong enough to balance the weight of each segment, so we could make the cable thicker there without adding any significant stress to the structure. We can then make the cable narrower where it comes closer to Earth, where Earth’s gravity becomes stiffer and centrifugal force has nearly faded out. So now we want to calculate the cross section of that tapering cable as a function of altitude above Earth’s surface.

For every million newtons of weight that we want to hang from the cable (the weight of 102 tonnes at Earth’s surface) we must add 16.7 square millimeters of Atlasite to the cross section of the cable. That adds 58.3 grams per meter of length to the mass of the cable. Of course, that adds even more weight to the cable. Near its lower end the cable gains weight rapidly with increasing altitude, but near the geosynchronous orbit each increment added to the cable weighs close to nothing (the true weight being mitigated by centrifugal force), so the cable’s thickness remains more or less constant at that altitude.

At any given point on the cable’s length the increment of cross section dS added to the cable equals the amount required to support the weight of the previously added section. Algebraically we express that statement by writing

(Eq’n 4)

In that equation sigma represents the yield stress (the elastic limit) of the cable’s material in Pascals (newtons per square meter) and rho represents the density of the material (in kilograms per cubic meter). The acceleration of gravity modified by centrifugal force, g=dF/dm, comes from Equation 1. But in Equation 1 the force increases as we descend: we want the force to increase with increasing altitude, so we must reverse the algebraic sign on Equation 1. We can then rewrite Equation 4 as

(Eq’n 5)

and then integrate it to get

(Eq’n 6)

Just above Earth’s surface (r_{0}=6378 kilometers)
the cable has a cross section of S_{0}=16.7 square millimeters. What
cross section does the cable have at r_{1}=42,251.2 kilometers?
Calculating the antilogarithm of Equation 6 tells us that

(Eq’n 7)

For Atlasite the ratio of rho to sigma equals 5.83x10^{-8} kilogram
per newton-meter, so we have

(Eq’n 8)

As small as that is, it will hold 102 tonnes dangling from its lowest point.

One of the more precious aphorisms favored by motivational speakers says that "There are no problems, only opportunities for achievement" (that’s straight from the 1970's, the age of disco and the leisure suit). The cosmic funicular will present us with a number of "opportunities for achievement".

Coriolis force is simply one manifestation of the law of conservation of angular momentum. An object possesses angular momentum by virtue of being part of a rotating system and the Coriolis force comes into play whenever an object moves further away from or closer to the center of rotation. For calculations, the angular momentum of a small body relative to some center of rotation equals the product of multiplying the body’s mass, its distance from the axis of rotation, and its velocity perpendicular to the straight line (the radius vector) connecting it to the axis of rotation. If the body’s mass does not change, its velocity around the axis of rotation will change only in response to a change in its position on the radius vector:

(Eq’n 9)

that is, the ratio of the change in the body’s position on the radius vector to its distance from the axis of rotation equals the negative of the ratio of the change in the velocity around the axis to that velocity itself.

On Earth’s surface at sea level on the equator a body has an eastward speed of 463.68 meters per second and it lies 6,378 kilometers from the axis of Earth’s rotation. On the geosynchronous orbit, 42,251.2 kilometers from Earth’s axis of rotation (35,873.2 kilometers above sea level), the body must have an eastward speed of 3071.66 meters per second. If a body rises from Earth’s surface to the geosynchronous orbit, it must accelerate eastward to make up the almost 2608 meters per second difference between resting speed on Earth and proper orbital speed on the Clarke orbit. If a body rises at a uniform speed, say 160 kilometers per hour (100 miles per hour, making the trip in about 9-1/3 days), it must accelerate eastward at 11.64 meters per second per hour in order to maintain a velocity that will keep it positioned above one point on Earth’s surface. If it comes down at the same speed, it must accelerate westward at the same rate. The magnitude of the acceleration remains the same over the entire distance because v=ùr, so the ratio expressed in Equation 9 yields

(Eq’n 10)

To compensate the Coriolis force and prevent the car from deforming the cable as it ascends and descends, the car must have rocket engines that will thrust eastward or westward as needed. Little thrust will be needed, about 0.0003296 gee’s worth if the car moves at 160 kilometers per hour, so the car can use ion engines. Those will presumably draw power from a rectifying antenna on the car intercepting a microwave beam.

Another "opportunity for achievement" comes to us when we notice that the geosynchronous orbit is not perfectly circular. It’s not that we can’t produce the required velocity to appropriate precision: we can certainly give a satellite the right speed in the right direction to put it into what should be a perfectly circular orbit about Earth. But that satellite will not actually trace out a perfect circle. That’s true to Reality because Earth possesses a relatively massive moon. Just as the moon raises tides in the oceans, so does it produce tidal forces that distort orbits. Our cosmic funicular will bob up and down twice a day and will do so to a degree that will put the Bay of Fundy to shame.

We can calculate a first estimate of how far the funicular
will move up and down and leave the refinement of the figure to others. We base
that first estimate on the assumption that the gravitational potential energy of
the system does not change as the system goes around its orbit; that is, we
assume that the system’s center of mass follows a gravitational isopotential. To
avoid complications, we assume that the moon follows a circular orbit of radius
R=363,300 kilometers and that without the moon’s presence our geosynchronous
orbit has a radius of r_{1}=42,251 kilometers. When the funicular’s
center of mass, lying on the geosynchronous orbit, passes between Earth and the
moon, it rises to a radial distance r_{1}' in accordance with

(Eq’n 11)

The 81 in that equation expresses the fact that the moon ponders 81 times less mass than Earth does. Dividing that equation by MG and rearranging it lets us calculate

(Eq’n 12)

To simplify the algebra leading to that calculation I used the approximation
that r_{1}' equals r_{1}. We thus find that the funicular will
rise 68.594 kilometers above the presumedly circular geosynchronous orbit.

Other effects will change that figure slightly, but for our purpose here is suffices to know that the funicular will rise and fall by about 135 kilometers roughly every 25 hours. That fact tells us that the funicular will not terminate at a ground station. Instead, a pair of hoverjets will fly out of the ground station. Each car on the funicular will be built to carry a cargo pod on its underside. When a car comes to the bottom of the descending cable, the first hoverjet will come up beneath it, take the cargo pod onto its back, and couple to it as the couplings on the cable-car release it. The hoverjet drops away and the car sidles across a bridge to the ascending cable. As the cable-car clamps onto the cable the second hoverjet moves the pod it carries into position and the pod couples to the cable-car as it uncouples from the hoverjet. The second hoverjet drops away as the car begins its nine-day climb to the station on geosynchronous orbit.

It helps to remind ourselves that the cosmic funicular involves a delicate balancing act. We want to maintain the system’s center of mass right on the geosynchronous orbit. That requirement necessitates that we extend the cable beyond the geosynchronous orbit, using the excess centrifugal force to balance the weight of the cable and its cargo below the geosynchronous orbit. In the present example the counterweight must exert 22.176 million newtons just to balance the cable alone. Calculating the length of the counterweight involves solving a cubic equation, which goes well beyond the scope of this essay. Suffice it to say that the counterweight will ponder even more mass than the cable below geosynchronous orbit does.

Calculating the mass of the cable involves solving an integral that has never been solved for finite limits. But, for the purpose of this discussion, we can make an approximation that exploits the fact that the cable’s cross section expands rapidly at low altitude and stops expanding at higher altitude. We simply take the 369.6 square millimeters that we calculated for the maximum cross section and multiply by the length of the cable: the calculation will give us too much mass, but not by much. At that cross section Atlasite ponders 1.2936 kilograms per meter of length, so the 35,873 kilometers of cable ponders a little less than 46,405 tonnes. Add in the counterweight and the infrastructure on geosynchronous orbit and we have a system that easily ponders over 100,000 tonnes.

As noted above, the balance of the funicular is delicate. Any perturbation of the cable’s location is subject to positive feedback. If the cable rises slightly, its weight decreases, the centrifugal force on it increases, and it rises faster. If the cable descends slightly, its weight increases, the centrifugal force on it decreases, and it descends faster. It won’t fly off into space and it won’t fall completely out of orbit, but as the forces acting on it redistribute its energy it will take up a new orbit and, due to the action of the Coriolis force, it will tumble. It will become less than useless: it will become a major hazard to astrogation.

To avoid that situation, the builders will put a heavy traveler on the counterweight and move it up and down the counterweight to compensate shifts in the system’s center of mass due to the motions of the cable cars. As the cable cars move up and down the cable, the system’s center of mass will shift by several kilometers and the traveler will have to move to compensate that shift as it happens, cancelling out the shift in real time.

We also note that we plan to build our funicular in what is essentially a shooting gallery. "Faster than a speeding bullet" is an understatement when we talk about objects that move as fast as eight kilometers per second or faster. Orbital debris moves that fast. A tiny screw can blast through the cable with enough impact to snap it.

Tension in the cable will be released abruptly in the case of a break. Backlash will strike in both directions from the break. Coriolis force will make the cable above the break whip westward as the counterweight, freed from the weight of the cable below the break, rises into space. The part of the cable below the break will whip eastward as it falls toward Earth. If the break occurs near geosynchronous orbit, the lower part of the cable will, in concept, wrap itself almost completely around the Equator. In actuality the uppermost part of that section of the broken cable will hit the atmosphere hard enough to burn up. It would be a spectacular display, certainly, but it’s one that we wish to avoid.

Between Earth’s surface and the altitude of geosynchronous orbit, space consists of 945 trillion cubic kilometers of vacuum. But it’s not empty vacuum. Bits and pieces of spacecraft float about in their own orbits, microsatellites of Earth. For pieces that span more than one millimeter we have something less than ten thousand cubic kilometers per piece around an altitude of one thousand kilometers going up to a bit over ten million cubic kilometers as we approach geosynchronous orbit. It goes back down to one million cubic kilometers per piece on the geosynchronous orbit itself. Even five thousand cubic kilometers per piece (between 1000 and 2000 kilometers of altitude) gives us a mighty big haystack in which to find a millimeter-wide needle.

How often can we expect the cable to get hit by a piece of
debris? At an altitude of 1000 kilometers, where the density of objects reaches
a maximum, each piece, on average, has 5000 cubic kilometers to itself. We can
get an order-of-magnitude estimate of the time between successive hits on the
cable at that altitude by imagining a length of the cable as forming part of the
side of a cylindrical container enclosing 5000 cubic kilometers. Make the
container one hundred kilometers long and make one strip of the wall 37
millimeters wide. That width is twice the minimum circumference of the cable,
which has a cross section of 27.32 square millimeters at 1000 kilometers of
altitude, the double width being necessary to account for the fact that an
object can strike while entering the container or leaving it. The container has
a total surface area of 2606.6 square kilometers, of which 3.7x10^{-3}
square kilometer represents the cable. The probability of a small object hitting
the cable as it passes through that volume is 1:704225. At an altitude of 1000
kilometers an object in a circular or near-circular orbit takes 1.88 hours to
revolve once around Earth, taking two hours to return to the same longitude.
Those data lead us to calculate, somewhat crudely, that something will hit the
cable at that altitude once every 80 years on average. That’s too often, but it
gives us enough time to fix the problem.

If we want to build and operate a successful cosmic funicular, we must clear the junk out of cislunar space. It’s not a needle-in-a-haystack problem: radar sees the junk quite clearly. The problem more resembles an effort to chase down a chicken on several square miles of open prairie. The solution of the problem involves a spacecraft that consists of a solar-power array, ion engines, a wide bucket to capture and hold debris, radar, and an on-board computer with a certain amount of machine intelligence.

Imagine one such robotic scavenger floating on orbit. It seeks out debris with its radar, ignoring live satellites and spacecraft. Among the debris in its radar vision it picks the piece closest to it in terms of delta-vee and adjusts its orbit to pursue and intercept that piece. It draws the piece into its storage compartment and repeats the procedure. When its storage compartment gets too full or it runs low on propellant, it executes a rendevous with a space station on the geosynchronous orbit, where it leaves the debris that it has collected and refills its propellant tanks. The space station uses as much of the debris as it can, puts the remainder into a frangible package, and deorbits that package so that it will burn up in Earth’s atmosphere.

Suppose that each scavenger in the bucket brigade captures one piece of debris per week on average. One hundred scavengers will remove 5200 pieces of debris from cislunar space every year. Currently (as of November 2014) over 600,000 pieces of debris greater than one centimeter wide float on orbits about Earth. If we expand our fleet of scavengers to 1000, they will take a little over 11.5 years to clear that junk out of cislunar space, but that estimate doesn’t count the millions of pieces smaller than on centimeter.

A space-clearing program of some kind will have to operate if we are to have a space-faring civilization, whether we build the funicular or not, so the cost of such a program will not properly be counted in the construction of the funicular.

Another problem that we must address originates in the fact that our cable will dangle through the hottest parts of the Van Allen radiation belts. In the hottest part of the inner belt over two hundred million protons carrying over one hundred thousand electron volts of energy apiece pass through any given square centimeter every second. If every proton were to hit one carbon atom as it passes through a centimeter cube, it would take over two million years for every carbon atom to get hit once, so we don’t expect the radiation to degrade our cable in any span of time less than thousands of years. The people and the cargo riding up and down the cable, on the other hand, will suffer badly.

In the hottest part of the inner belt one liter (1000 cubic centimeters) will have two hundred billion particles, each carrying more than one hundred thousand electron volts, passing through it every second. If that volume were filled with water and each radiation particle, on average, deposited 100,000 electron volts into it on passing through, the water would accumulate one joule of energy in a little over five minutes. Five joules of radiation-borne energy deposited per kilogram of living matter (mostly water) guarantees the death of a person within two weeks.

Of course, I’m overstating the danger in implying that a person would receive a fatal dose of radiation in less than half an hour. The Apollo astronauts spent hours within the Van Allen belts on their ways to and from the moon and they suffer no ill effects attributable to radiation exposure. The Apollo spacecraft provided a certain amount of shielding, just as the cable-cars on the funicular will, but the astronauts spent only hours in the radiation belts; the cable-cars will spend days.

Fortunately there’s a solution to the problem and the cable itself is part of that solution. It’s called radiation remediation and it was devised by Dr. Robert L. Forward and his colleagues. Forward’s design calls for an orbiting tether charged to a high voltage. The resulting electric field deflects approaching particles onto trajectories more nearly parallel to Earth’s magnetic field. Those particles then migrate to the polar regions and get absorbed by the atmosphere. One estimate of the effectiveness of such remediation has the high-energy particle flux in the inner radiation belt reduced by a factor of 100 within two months. In a year that would be a trillion-fold diminution.

The funicular goes the tether concept one feature better. It extends all the way through both radiation belts and we can simply charge up the cable. If the builders of the funicular float the cable at about one million volts, they should be able to clear out the radiation belts, insofar as it’s possible, as the cable grows. When the funicular is ready to begin operation, the voltage should have established an equilibrium between the rate at which particles get deflected into the atmosphere and the rate at which new particles come into Earth’s magnetosphere from the sun. The only special feature that the builders will have to add to the funicular is a means of gently charging and discharging the cargo and passenger pods so that they won’t receive large electric shocks at the ends of their journeys.

It’s a daunting prospect. It would certainly be a challenge. But is it a challenge worth meeting?

It will require moving over 100,000 tonnes of material to geosynchronous orbit and yet it will only move on the order of 2000 tonnes per year. With the exception of the geosynchronous orbit itself, the cable and its counterweight will prevent satellites from occupying orbits lower than the upper end of the counterweight: it will be a major hindrance to astrogation. On first impression, it seems like a good idea, but further analysis makes it seem less desirable. For now, then, rockets are the best means we have for getting into space.

Appendix: Funiculì, Funiculà

The following may be sung to the melody of Funiculì, Funiculà, which Luigi Denza and Peppino Turco composed in 1880 to celebrate the opening of the funicular that ran up Mount Vesuvius.

Rising into the Heavens

we see a cable, we see a cable.

Yes, and we would climb it

if we were able, if we were able.

It is anchored to Clarke’s orbit

to keep it stable, to keep it stable.

It rises like the beanstalk

in the fable, in the fable.

Climbing, climbing, trying not to slip.

Climbing, climbing, such a dizzy trip.

Up the cable, down the cable,

coming home from Mars.

Down the cable, up the cable,

flying to the stars.

efefef