What is it that will make it possible to spend \$20 billion of your money to put some clown on the moon? Good old American know-how – as provided by good old Americans, like Doctor Wernher von Braun.

Tom Lehrer

# Stairway to the Stars

Arthur Clarke and Charles Shepherd, both scientists and science fiction authors, wrote books, at roughly the same time (the novel of one was published while that of the other was being prepared for publication) in the 1970s, with the idea of a cable-car to geostationary orbit. The idea originally came from a Russian (whose name I do not know) a decade or three earlier.

An object in orbit just above the Earth's atmosphere goes round several times a day (not quite I'll run a circle round the Earth in 40 minutes, but only slightly more than 80 minutes). Orbits further out take longer to complete; the Moon, a little over a third of a million kilometres (i.e. a little under a quarter million miles – over 60 times the low-Earth orbital radius) away takes almost a month to complete one orbit. Part way between, there's inevitably an orbit in which the period takes that intermediate value, one day, which causes an equatorially-orbiting object to remain above the same point on the Earth's equator at all times. An object in such an orbit appears, to an observer on the Earth's surface, not to move; hence the name geostationary. (In which geo means Earth, as in geology.) Because the orbit is synchronised with the Earth's rotation, it's also called geosynchronous and the analogous orbit for an arbitrary planet is generally called synchronous.

In any orbit, the centrifugal force, due to going round and round, exactly balances the gravitational force; since the latter shrinks as radius grows, so must the former. Specifically, radius times the square of orbital velocity is the same for all orbits around a given centre. An object going round the Earth synchronously but in the wrong orbit thus experiences more centrifugal force than gravitational if outside the right orbit; or the other way round inside. At the Earth's surface, gravity is 289 times as strong as the centrifugal force, so it wins (which is why we don't fly off into space).

If an object in synchronous orbit is large enough that parts of it are significantly closer to Earth (where gravity wins, so these parts are pulled inwards) and others are significantly further away (where centrifuge wins, so these parts are pulled outwards), it experiences a tidal stretching which tries to pull each side of the object away from its synchronous orbit. As long as enough of the object is outside to balance the part inside, the object stays in orbit – provided it's stong enough to not get torn apart by the stretching. A cable-car to synchronous orbit would be, at least in principle, just a very long thin object in synchronous orbit – so long and thin that its bottom end reaches the Earth's surface, making it possible for a vehicle to climb up and down; the ultimate Indian rope trick.

In the aftermath of 2001/September/11, I visited my friend TimH who (knowing that the sum in question would keep the poorest fifth of the world's population alive for at least a month) spent some time wandering around muttering, in shocked/awed amazement, twenty billion dollars. [2008/September update: so seven tenths of a teradollar would keep the poorest fifth going for about three years.] This prompted me to think up uncontentious uses for such a sum of money. I've traditionally handled big sums of money by involving (literally) astronomical distances – for instance, when I heard that the U.S.A.'s defence budget for 1984 was half a terradollar, or roughly a quarter of a million million pounds at the time, I rendered it intelligible as a road to the moon on a budget allowing for a million pounds per mile (which is what road planners had to allow for, at the time, when building urban motorways in the U.K.). So I worked out that twenty giga-dollars would provide for a space elevator if it could be built on a budget of order five hundred dollars per metre [and 0.7 T\$ would let us get away with 35 times as much: 17.5 k\$/m]. (2009/May: for another large pot of money to think about, the war on drugs costs of order 25 giga-dollars per year.)

This, in turn, prompted me to consider the mechanical practicalities of building a space elevator. The cable must stretch beyond geostationary orbit, and must carry the greatest tension at that orbit. Outwards from there, its tension holds a counterweigt (which is orbiting faster than gravity can hold it) in; inwards from there, the cable is supporting the (surplus to centrifugal effects) weight of its lower reaches. Thus one may expect the cable to be thickest (so as to be strongest) at geostationary orbit and comparatively thin (so as to be light) at the Earth's surface (whose radius is about 15% of the geostationary orbital radius).

Note that sending things to space using NASA's shuttle costs about \$15 per gramme of payload – feeding the ISS crew is rather expensive ! – and it should be possible for a space elevator to reduce that cost dramatically. Even allowing that more modern rocketry aims to get that \$15/g price tag down by a factor of two or three, a space elevator should be cheaper by a wide margin; if it only saves \$5/g, it'll have saved G\$20 when it's shipped 4 million kg of payload. That may sound like a lot of payload, but if it can manage one (metric) tonne per day it'll meet that target in eleven years. If (as I suspect is likely) it can reduce costs by a factor of ten or better, the larger savings can bring that total payload and time-span down by a factor of about three.

## Theory

So, how must the thickness of and tension in the cable vary with altitude for a space elevator ? I'll describe a general case, going round with angular velocity w, with R as the radius at which this is the orbital angular velocity: thus w.w.R.R.R = G.M with G as Newton's constant, M as the mass of the planet orbited; so w.w.(R.R.R/r/r −r) is the net force per unit mass pulling inwards on the cable at distance r from the centre of rotation.

Now, the fatter our cable is, the more stress it can bear: but equally a fatter cable weighs more heavily on the parts supporting it. So we'll want to ensure that the cable is thick enough to support the tension in it, but no more. Thus we'll have some acceptable stress level, S, which will be some fraction of the ultimate tensile stress of the material of which we make our cable: we'll aim to make the cable fat enough, at each position, to spread out the tension in the cable over enough area that the stress in the cable is no more than S; but we'll make the cable no fatter than that, so as to avoid burdening portions of the cable at radii nearer to R. So the tension, T, in the cable will dictate its cross-sectional area as T/S; if the cable's density is D, this gives a mass T.D/S per unit length of cable.

While one might chose to vary the material of the cable with radius, I'll suppose that D/S is constant, as will arise (for instance) if the cable is made all of the same material. Then the tension, T, in the cable varies with radius, r, (from Earth's centre) as:

• dT/dr = T.(D/S).w.w.(R.R.R/r/r −r), or
• 0 = (S/D).d(log(T))/w/w + R.R.R.d(1/r) +d(r.r)/2, i.e.
• log(T).S/D/w/w +R.R.R/r +r.r/2 is constant.

Now, w.R is a velocity and S/D is the square of another; so introduce v = w.R.√(D/S), which is dimensionless. Likewise, introduce a dimensionless radial coordinate u = r/R to put the above in dimensionless form; this makes our net force per unit mass w.w.R.(1/u/u −u), i.e. v.v.(1/u/u −u).S/D/R towards the planet. Let the cross-sectional area of the cable be A(u) at radius u.R; thus we have T = S.A(u) at this radius and the above reduces to

• log(S.A(u))/v/v + 1/u +u.u/2 is constant, so = log(S.A(1))/v/v +3/2, whence
• log(A(u)/A(1))/v/v = 3/2 −1/u −u.u/2, or
• log(A(u)/A(1)) = v.v.(1−1/u +(1−u.u)/2) = v.v.(1−u).((1+u)/2 −1/u)

with u ranging from a minimum, the planet's surface radius divided by R, to a maximum exceeding 1. Let these bounds be b for bottom and B for Big with b<1<B and u ranging from b to B. Since 1/u +u.u/2 has a minimum at u=1, A has its maximum at u=1 (as is only to be expected) and log(A(u)/A(1)) is negative for any u other than 1. The volume of material in the cable is the integral of its cross sectional area with respect to radius; this is R.integral(: A(u).du ←u; b<u<B :) which our formula gives us as

• R.A(1).integral(: exp(v.v.(3/2 −u.u/2 −1/u)).du ←u; b<u<B :)

and the mass of the cable is D times this. The counter-weight, at u=B, needs to pull outwards with tension S.A(B), requiring a mass M for which −S.A(B) = M.v.v.(S/D/R).(1/B/B −B) whence

M
= S.A(B).R.D/S/v/v/(B −1/B/B)
= (R.D/v/v).A(1).exp(v.v.(1−B).((1+B)/2 −1/B)) / (B −1/B/B)

The load which the cable can bear at its bottom will have weight S.A(b); so (in similar vein to the counterweight) its mass is (R.D/v/v).A(1).exp(v.v.(3/2 −b.b/2 −1/b)) / (1/b/b −b). When this is divided by the mass of the cable, the factor of D.R.A(1) cancels and we are left with a dimensionless answer which depends on S and D only via v.v; it depends on R only via b and B. The argument to exp in the payload can be combined with that in the integral just given, giving the cable / payload mass ratio

• Q = integral(: exp(v.v.(b−u).((b+u)/2 −1/b/u)).du ←u; b<u<B :).(1/b/b −b).v.v

The moment of inertia of the cable will be

I
= D.R.R.R.integral(: A(u).u.u.du <u; b<u<B :)
= D.R.R.R.A(1).integral(: u.u.exp(v.v.(1−u).((1+u)/2 −1/u)).du ←u; b<u<B :)

Note: total energy per unit mass = w.w.r.r/2 −G.M/r = (r.r.r −2.R.R.R).w.w/r/2 is positive for r/R > the cube root of two; so if the cable is cut at u > 1.6, the outer part will escape from the gravitational field of the planet being orbited. (Indeed, if the relevantly-adjusted centre of mass of the portion beyond a cut is at u > 1.6, the same will apply, even if the cut itself is closer in.)

### Angular momentum, tides and stability

The formal solution for a static ladder to space, above, involves a radial structure subject only to radial forces. However, the Earth orbits the Sun and the Moon orbits the Earth; both exert tidal effects which will exert forces on the cable. We must examine the stability of the structure when exposed to such forces: it is vitally important that it be stable. The cable is roughly the same length as the equator: if it were to come tumbling down, it would cause damage across a vast swath of the world.

At the same time, if we are to send any kind of a payload up or down the space elevator, its angular momentum will change by two orders of magnitude: its angular velocity is the same at top as at bottom, but angular momentum is proportional to the square of radius, which varies by an order of magnitude in the transit: this will require the cable to apply a force to the payload at right angles to the direction of the cable. Looked at in the rotating frame of reference of the Earth's crust and the space elevator, the payload is subject to a Coriolis force at right angles to its direction of movement, along the cable: the cable will have to resist this force. From either perspective, this is a force at right angles to the cable, the one direction in which the above simple model provides no mechanism for the cable to exert force.

We do have one factor working in our favour, though: the tidal effects of the Earth's gravitational field will tend to keep the cable correctly oriented. Tidal effects act to stretch radially but compress in the two perpendicular directions. So let us look at the effects of Earth's tidal forces on our structure.

We have a rotating frame of reference, turning at rate w; the circular orbit with this angular velocity has radius R. Gravity's inward force per unit mass, at radius r, is then G.M/r/r = w.w.R.R.R/r/r inwards. This is countered by centrifugal w.w.r outwards and complicated by the Coriolis effect, since we're in a rotating frame of reference. Let W be a vector pointing along the Earth's spin axis, towards Polaris (a.k.a. the North star), with magnitude w (which we've been measuring in radians/second, though I din't mention it). Use co-ordinates whose origin is the Earth's centre (and, crucially, on its spin axis), rotating with the Earth's crust. Then a body with velocity V at (vector) position x experiences forces per unit mass induced by our choice of rotating frame equal to

• 2.V^W +W^(x^W)

where ^ denotes the antisymmetric vector outer product. The first term here is the Coriolis force, the second is the centrifugal force already taken into account. The net force per unit mass, acting on the given body, is thus equal to

• R.R.R.W·W/(x·x) +2.V^W +W^(x^W).

We can take spherical polar co-ordinates, using u = r/R as a dimensionless form of our radial co-ordinate; if we select the equatorial radius through the geosynchronous point on our cable as origin for our angular co-ordinates, our longitude φ (with nominal range −turn<2.φ<turn) and latitude θ (with range −turn<4.θ<turn) will both be small (that is, each will be much less than a radian). Gravity acts radially inwards and centrifuge acts away from the axis of rotation; these are the principal forces on our cable and thus tend to force it towards the θ=zero plane. Provided the cable remains substantially radial, we can treat it as being described as a curve parameterised by u, with θ and φ varying along it.

In our dimensionless variables, consider the piece of cable at radial co-ordinate u = r/R; allow that it is substantially stationary, so subject to force per unit mass R.w.w.(1/u/u −u) inwards along the nominal direction of the cable, balanced by an equal and opposite tension in the cable. To this should be added various corrections due to its small velocity (relative to our rotating frame) and assorted extraneous forces (solar and lunar tides, for example); however, these are very small. By comparison, the forces due to not being at exactly zero values of θ and φ are merely small. First let us consider the forces due to θ being slightly off zero. Gravity acts radially inwards, so has no component affecting θ. The centrifugal force is in the plane of the equator, so exerts a component u.R.w.w.Sin(θ) tending to bring θ back to zero; this will tend to produce a simple harmonic oscillator, for small θ. We should also consider how the tension produces redirected forces as a result of the cable not lying exactly radially; but this will be easier to analyse for the φ-variation first.

We have (to a good approximation) tension R.w.w.(1/u/u −u) per unit mass of the cable acting along the cable at radius r = R.u, at longitude φ(u). At radius R.(u+δu) and longitude φ(u+δu) we have essentially the same tension acting in slightly different directions; this shall produce a sideways force, causing φ(u) to vary.

to be continued …

## Application

The Space Elevator will be built about 50 years after everyone stops laughing.

Arthur C. Clarke

I have a `python` package which implements, as module `study.space.ladder`'s class `Ladder`, the computations (or reasonable approximations thereto) implied by the analysis above.

For my home planet, b is about 0.1512; I have heard discussion of ladders with B at about 1.3. A(b) is the cable's cross-section at ground level; with b=0.1512 we get log(A(b)/A(1))/v/v = 3/2 −1/b −b.b/2 = −5.63 whence A(1) = A(0.1512).exp(5.63 v.v). Note that exp(5.63) is 278, so v=1 will give a cable 278 times as fat at orbit as at ground; this ratio climbs to the fourth power of 278, namely 5.9e9, at v=2 so we'll be needing a small value of v if we're to have any hope of success. Indeed, a sphere with radius Earth's synchronous orbital radius has surface area 22e15 square metres; a cable with cross-section less than a square millimetre would be impractical even if strong enough (it'd get lost ;^) so it'd be ridiculous to have A(1)/A(b) even as big as about 22e15/1e−6 = 22e21; this requires log(A(1)/A(b)) < 51.46, whence v = √(log(A(1)/A(b))/5.63) must be less than 3 (and, in practice, even 2 is probably impractical; 3 would be preposterous).

Now, w.R is roughly 3 km/s, so we need w.R/v to exceed 1 km/s (and, in practice, probably also 1.5 km/s); its square is S/D and S is necessarily less than the ultimate tensile stress of our building material. (For want of a better name) I describe the square root of the ratio of ultimate tensile stress over density as the densile speed of the material; we need a material whose densile speed exceeds 1 km/s (by a good factor, almost certainly at least 2). As it happens, the densile speeds of various materials, e.g. metal alloys, are large fractions of (but less than) 1 km/s and one or two forms of carbon fibre actually cross the 1 km/s barrier (albeit I suspect the data refer to individual fibres; building one thousands of miles long might present some practical difficulties). My sources vary on Kevlar®; DuPont's data, if I've understood it right, puts its densile speed just over 1 km/s; but other data puts it closer to half a km/s.

Thus materials, suitable for building a cable capable of lifting us out of our planet's gravity well, are tantalisingly almost practical. The best carbon fibre for which I have data manages 1.83 km/s, yielding v = 1.683; the cable's cross-section at orbit will need to be 2 million times that at the ground; a cable 1 mm across at ground level would need to be 1.43 metres across at orbit. Who knows, maybe we'll manage something impressive with carbon nanotubes (a.k.a. bucky tubes or Buckminster Fullerenes) … I've seen (most likely in New Scientist) 130 GPa claimed as their strength and their density can't be much above 3 g/cc; with those values, it'd have a densile speed of 6.6 km/s, pushing v below 0.5, and yield A(1)/A(b) only slightly greater than 3 ;^>

Since it's the best on offer, and the only material for which I have data indicating a densile speed better than 1.7 km/s, I'll work with the nominated (heat-treated) carbon fibre; it has D = 2.1 gram/cc, S = 7 GPa. An elevator out to radius B = 1.3 times geostationary would, for each square metre of orbital cross section, mass 55 million tonnes and be able to lift a third of a tonne. A material with the same density but twice as strong would drive v down to 1.19, have only 1430 times the cross-section at orbit as at ground (so 38 times in diameter); with B still at 1.3, each square metre at orbit would require 65.4 million tonnes and enable us to lift a thousand tonnes. Modest improvements in material properties are worth huge improvements in performance.

### Constraints, given a 20 gigadollar budget

As noted above, the mass ratio cable/payload depends on the the material used only via its densile speed; the ratio will also depend on b and B. We can use this ratio, given a required payload of, say, 1 tonne (a plausible minimal lift for one person plus supplies) to infer the total mass of the escalator; for a given budget, e.g. our original G\$20, this implies a maximum price per unit mass for affordable building materials (the price cieling below; multiply it by 35 for the corresponding threshold if your budget is T\$ 0.7). The following tables give values of Q and this price cieling for illustrative values of √(S/D) and B, presuming b = .1512 (i.e. Earth's surface radius divided by geostationary radius). For the sake of a sense of scale on prices, the price of Kevlar® (as at 2002/February) ranges from \$8/lb = \$17.6/kg to \$50/lb = \$110/kg.

First, here's how Q varies with B at fixed √(S/D):

√(S/D) = 2 km/s√(S/D) = 3 km/s
BQ/millionprice cielingQ/thousandprice cieling
1.211\$1.90/kg6.9\$2.91/gram
1.26611.6\$1.72/kg7.5\$2.66/gram
1.312.2\$1.60/kg7.8\$2.50/gram
1.515\$1.37/kg9.6\$2.09/gram
1.7516.3\$1.23/kg11.2\$1.78/gram
216.7\$1.20/kg12.1\$1.66/gram

Now vary √(S/D) with fixed B = 1.266 [the ton used in the small-speed cases is the US ton, just over .907 tonne; M = million, k = thousand]:

√(S/D).s/kmQprice cieling
118e2210 cents/teraton
1.191e1719 cents/megaton
1.52.2e118.4 cents/ton
1.6832200 M\$8.3/ton
1.8230 M\$78/ton
1.946 M\$0.43/kg = \$390/ton
211.6 M\$1.72/kg
2.13.5 M\$5.7/kg
2.21.23 M\$16.2/kg
2.20851.14 M\$17.6/kg = \$8/lb
2.3490 k\$41/kg
2.4219 k\$91/kg
2.425181 k\$110/kg = \$50/lb
2.5106 k\$188/kg
2.656 k\$358/kg
2.731.3 k\$640/kg
2.818.6 k\$1.08/gram
2.911.6 k\$1.73/gram
37500\$2.66/gram
3.15100\$3.9/gram
3.51410\$14.1/gram
4450\$44/gram
5104.4\$192/gram
6.07340.0\$500/gram
722.4\$891/gram
115.42\$3690/gram

Note that the last price I heard for bucky tubes (late 2001) was \$500/gram, so they'd need a densile speed over 6 km/s; which is compatible with the estimate I get, reading their 130 GPa strength as an ultimate tensile stress, unless they're denser than diamond. Of course, a budget of G\$20 might be hard to come by, and there are other things for this budget to cover aside from the building materials, but (as TimM pointed out) a little expenditure on discovering how to make bucky tubes cheaper could go a long way … all the way to the stars, in fact ;^)

### Other Limitations

You still have to lift mass out of the Earth's gravity well; and that takes 57.8 MJ/kg. A 100 kg adult thus costs 5.78 GJ (a.k.a. 1600 kW hours). If we had some way to collect the total solar energy incident on the Earth (about 179 peta Watt, several orders of magnitude higher than total power output from human industrial activities; total electrical power consumption is presently around 2 terawatts) and use it to power lifting payload out to synchronous orbit, this would let us shift 3 mega tonnes per second from bottom to top; in practice, total power available to us is unlikely to be so great. We might be able to use the energy from payloads coming down the ladder to lift payloads up the ladder, but efficiency is always going to be an issue – at least until we've got superconducting materials at the temperatures to be experienced by our tapes, light enough to be built into the tape and cheap enough that we can afford mega-metres of them.

When it comes to emigration, we have to consider each human to take up at least of order a tonne of payload – of order ten times as much as the person – to allow for supplies of air, water and food for the journey (taking of order a week) plus pressure vessel to hold it all in and probably some personal belongings they want to take with them. The wildly optimistic 3 mega tonnes per second would thus translate to three million emigrants per second, but I doubt we could even deliver one millionth of that power supply to all our space elevators put together; even emigration rates of order three per second seem optimistic. Given that the global birth rate is of order four per second, we can't rely on emigration to halt population growth on Earth – we still need to reduce birth rates if we're to avoid over-population.

## Pay-back

Getting payload to the Moon would cost several tens of thousands of dollars per kilogram (Charles Chafer of Celestis, in New Scientist, 2002/Feb/16th edition) by rocketry. Getting payload to low Earth orbit costs NASA about \$15/g, which is comparable, though smaller. I presume that most of that cost is in getting the payload out of Earth's gravity well: a space elevator would make it possible to cut costs by of order \$10 per gram; so a G\$20 ladder will pay for itself once it's lifted of order 2 thousand tonnes of payload out to orbit. After that, the price of lifting mass into orbit can be hugely reduced.

Two thousand tonnes is a lot of payload; but the price of building materials should come down – especially when one customer puts in a bid for several thousand miles of the material, empowering the producer to achieve economies of scale; albeit producing the cable in space, from space-sourced materials, would probably make most sense. If nothing else, the usual candidate counter-weight is a small asteroid; the project calls for resource-gathering in the asteroid belt, so why not source the cable material there as well …

### Other planets, etc.

Of course one can also build space elevators for other heavenly bodies. As for Earth, the critical parameters are synchronous orbital speed and ratio of surface radius to synchronous orbital radius. Let's be optimistic and assume we have a material with densile speed 10 km/s; then (with the possibly unreliable data available to me, and taking B=1.3 for the Q calculation) we get

NameR/Mmvbb.b/2 +1/b −3/2A(1)/A(b)1/Q
Sun252807.2450.027534.81e7940
Mercury250.0310.0101971.100.077
Venus1600.0470.0042501.720.0033
Earth42.14.30730.15125.1261.6230.1383
Moon93.0230.018652.21.02810.50
Mars20.5.14540.16634.531.10041.068
Jupiter1602.830.4480.838000.080
Saturn1101.90.5460.485.30.034
Uranus78.870.3331.563.30.057
Neptune93.880.272.305.90.019
Pluto20.0230.07312.211.006428.23

Note that the Sun's synchronous orbit is almost half way to Mercury's orbit; unless the Sun's ladder had B < 2.29 it'll have to be careful to avoid bumping into Mercury – but, in any case, we'd need a much stronger material for the Sun's ladder; 1e794 is very huge ! Even Jupiter is probably infeasible. The Moon's ladder needs B < 4.13 to avoid hitting the Earth; but if Earth also has a ladder, we need to be sure they don't get tangled; each can have B up to 2.8 before we have to worry about that, though. If the moon has no ladder, Earth's can have B up to 9 before worrying about collision.

For cables with B=1.3, to get Q down to 1000, so the payload is 0.001 of the cable mass, one needs these values of √(S/D):

Name√(S/D).s/kmvA(1)/A(b)
Sun2820.257.1008
Mercury2.360.130.19
Venus7.070.066.34
Earth3.650.842.0263
Moon1.180.195.14
Mars1.600.91.0240
Jupiter12.32.29.012
Saturn6.073.1.011
Uranus5.271.7.014
Neptune6.561.34.016
Pluto0.4580.5.0473

so we could build a viable elevator for Pluto with existing materials; and Kevlar® would certainly suffice for the Moon and Mars. For the Sun, and probably for Jupiter, the best we can realistically hope for (and the most we're ever likely to want in any case) is a non-synchronous escalator that grazes its outer atmosphere. Such a structure, for Jupiter, would provide the means to collect Helium 3 from its clouds, which would be a big boost for fusion technology. I don't have data on the moons of the outer planets, but several of these are Earth-scale, so may be of interest.