SPACE ELEVATORS

1 INTRODUCTION

Since the start of the space age in 1957 all satellites and spacecraft launched into space have been put there by rocket power.

However, well before this various people have proposed the use of a tower or stairway to reach space. Even as far back as 4000 years ago, it has been said that man tried to reach the stars by building a very tall tower. This tower never made it to completion but that has not stopped people speculating on the idea ever since.

The tower of Babel in the plains of Shinar

Current ideas envisage a ribbon of very strong material extending from the surface of the Earth up to and well beyond geosynchronous orbit so that the force on the cable upward from geosynchronous orbit would approximately equal the cable force below geosynchronous orbit. The centre of mass of the cable would thus be at geosynchronous orbit. Not only would the cable be able to support its own weight but it would also be able to support the weight of 'cable cars' attached to the cable that would climb and descend the length of the cable carrying passengers and cargo to and beyond geosynchronous orbit.

2 HISTORY

1895 Tsiolkovsky (Dreams of Earth and Sky)

1957 Artsutov (Cosmic Railway)

1966 Vine, Bradner and Bachus - Sky Hook (paper to Science)

1975 Jerome Pearson - Acta Astronautica (first serious analysis)

1979 Arthur C Clarke - Novel "Fountains of Paradise"

1991 Samio Iijima - Carbon Nanotubes

Late 1990's NASA interest - David Smitherman (advanced project at Marshall SFC)

2000 Bradley Edwards - NASA report

2005 - 2012 Meetings and Competitions

2008 International Space Elevator Consortium [ISEC] formed

2012 Obayashi Corporation - Space Elevator by 2050

2013 International Astronautical Association (IAA) - technical feasibility assessment

2014 Google X prize

2019 IAA publishes "Road to the Space Elevator Era"

3 MATHEMATICS

3.1 The Basics

At its simplest a space elevator consists of a cable extending from the surface of a planet up to well beyond the planetary synchronous orbit. The planet is rotating and carries the cable with it. The cable is thus subject to both the force of gravity (due to the planetary mass) and the centrifugal force (due to the planet's rotation). To keep the cable in relatively stable equilibrium the total force on the cable, acting through its centre of mass must be close to zero.

The above figure shows a cable of total length L extending above a planetary surface. We then need to consider a small part of the cable which has a mass m at a height h above the surface. There are two forces acting on this part of the cable, a gravitational force and a centrifugal force.

3.2 The Gravitational Force

Newton's law of gravitation gives us an expression for the gravitational force:

F_g = G M m / r² = m g

where G is the universal gravitational constant and g is the acceleration due to gravity at the location in question. We can write g as

g = G M / r² = G M / (R + h)²

At the surface of the planet (that is, where h = 0) we write g = g_o which gives us

G M = g_o R²

and thus

g = g_o ( R / r )² and F_g = m g

3.3 The Centrifugal Force

The expression for the centrifugal force is given by Newton's second law F = m a where a is the acceleration of the section of cable under consideration. Now acceleration is a vector quantity where we must consider both the magnitude and direction of the movement. So although the speed of the cable is a constant at any point along the cable (although it increases as we move upwards), the direction is constantly changing because that cable is rotating with the planet. The point in question is moving in a circle, and for circular acceleration it is relatively easy to show that a = r ω². The centrifugal force is thus:

F_c = - m r ω²

The negative sign indicates that the force acts 'upwards', opposite in direction to the gravitational force.

The total force on the specified mass element of the cable is then:

F_T = F_g + F_c = m g_o ( R/r )² - m r ω²

If the planet in question is the Earth then the values that are appropriate for this equation are:

Equatorial radius of the Earth (R) = 6,380 km = 6,380,000 metres

Surface gravity (go) = 9.8 m s^-2

Rotational (angular) velocity = 2π radians / 86,164 seconds = 7.29 x 10^-5 radians/s

The value of 86,164 seconds corresponds to the Earth's sidereal rotation period of 23 hours 56 minutes and 4 seconds. This is the time it takes the Earth to rotate through 360 degrees. This is 3 minutes and 56 seconds short of 24 hours which is the time it takes the Earth to rotate the extra amount (about one degree) to face the Sun again. This is due to the fact that the Earth is moving about the Sun as well as rotating on its own axis.

Although not necessary in our calculation the other two constants we have encountered are:

Mass of the Earth (Me) = 5.98 x 10²⁴ kg

Gravitational Constant (G) = 6.67 x 10(sup>-11 N m² kg^-2

3.4 Planetary Synchronous Orbit

Synchronous orbit for any planet is defined as the altitude at which an orbiting satellite has exactly the same rotational speed (ω) as the planet itself. It thus appears to be stationary as observed from the surface of the planet. At this point the total force acting on a space cable is zero - the centrifugal force is exactly equal to the gravitational force. And because there is no force there is no movement. We denote the altitude of this point as h = h_s and r_s = h_s + R.

F_T = 0 implies that:

g_o ( R/r_s )² = r_s ω²

g_o R² = r_s³ ω²

This gives the value for the synchronous radius as

r_s = ³√ ( g_o R² / ω² )

For the Earth the geosynchronous radius is

r_s = 42,183 km and the geosynchronous altitude (height) is

h_s = r_s - R = 35,803 km

3.5 Forces on the Cable

The force on the cable at synchronous orbit is zero. For the cable to remain stationary with respect to a given point on the rotating Earth the total force on the cable must be zero. This means that the downward forces, acting below synchronous orbit, must exactly equal/balance the upward forces above acting above synchronous orbit.

Let us look at a small portion of the cable at a height h. We consider only a very small portion of height dh.

If the cable is assumed to have a constant areal cross section dA and a uniform density ρ, then the mass of this small portion will be

dm = ρ A dh

The forces acting on this element are, as before

dF = [ g_o (R / (R + h))² - (R + h) &omega² ] ρ A dh

This force acts toward the planet below synchronous orbit and away from the planet above synchronous orbit.

The total force acting on the cable of length L is found by integration:

F = ∫ _h=0^h=L dF = ρ A ∫₀^L [ g_o (R / (R + h))² - (R + h) ω² ] dh

The integration becomes easier if we switch variables from h to r where r = h + R and dh = dr

F = ρ A ∫_R^R+L [ g_o (R / r)² - r ω² ] dr

= ρ A { g_o R² [∫_R^R+L r^-2 dr] - ω² [∫_R^R+L r dr]}

= ρ A { g_o R² [ - r^-1 ]_R^L+R - ω² [ r² / 2]_R^L+R }

and finally

F = ρ A [ g_o L R / (L + R) - ω² L (L + 2R) / 2 ]

which is the force equation for the cable. Sometimes this might be written

F / (ρ A) = L [ g_o R / (L + R) - ω² (L + 2R) / 2 ]

The quantity on either side of the equation is then the force per unit mass, that is the mass per unit length or the lineal mass density (ρ is the volume mass density).

3.6 A Stationary Cable

For a cable in equilibrium (that is, one that will stay vertical as observed from the surface of the rotating planet), the right hand side of the equation must equal zero. That is:

g_o R / (L + R) = ω² (L + 2R) / 2

This expands to a quadratic equation in L:

L² + 3 R L + 2 R ( R - g_o / ω² ) = 0

which has the solution

L_o = R { √ [ 2 g_o / (ω² R) + 0.25 ] - 1.5 }

where the subscript on L designates the length for zero total force on the cable.

For the Earth this equates to a total cable length of L_o = 144,000 km. Of this, about 36,000 km will lie below geostationary orbit and 108,000 km must lie above geostationary orbit. That is, three quarters of the total cable is needed to balance the one quarter below geostationary orbit.

For a space elevator the cable must be in tension. This just occurs when the total force is zero. So L_o is in fact the minimum length of the cable. In practice we must have F < 0. That is, the upward force (negative by definition) must slightly exceed the downward force and the base attachment to the Earth provides the additional force to produce equilibrium and ensure the cable is truly in tension. It would also provide a compensation for the extra mass that would be introduced when 'cars' were attached to the cable to climb to geosynchronous orbit. The force tugging upward on the Earth anchor would thus be variable to compensate for the changing mass below geosynchronous orbit.

3.7 Reducing the Cable Length

Now 108,000 km of cable above geosynchronous orbit is a lot of cable that would be used for nothing else but to provide a counterweight to the useful cable below geosynchronous orbit. To reduce this length we could use an actual concentrated counterweight with a much shorter cable.

Now the counterweight mass, at an orbital radius of rm must provide a force equal (or slightly greater) than the force due to the length of the cable from r_m to r_L.

The force due to the mass M is given by

F_M = M [ g_o (R/r_m)² - r_m ω² ]

The force due to the length of the cable from r_m to r_L is given by

F_cable = ρ A ∫_rm^rL [ g (R/r)² - r ω² ] dr

F_cable = ρ A (r_L - r_m) [ g R² / (r_m r_l) - ω² (r_l + r_m) ]

M = ρ A (r_L - r_m) [ g_o R² / (r_m r_l) - ω² (r_l + r_m) ] / [ g_o (R/r_m)² - r_m ω² ]

Before we can put any numerical values into this equation to try and determine typical values for an actual counterweight mass we need to discuss actual cables and their material properties.

4 CABLES

4.1 Cable Requirements

We assume the properties of the cable material are the same in compression as in tension. We now need only to consider the cable length up to geosynchronous orbit ( h = hs ).

The total force at the base of the cable is:

F = h_s [ (g_o R) / (R + h_s) - ω² (2 R + h_s) / 2 ] ρ_V A

The pressure at this point is:

p = F / A = h_s [ (g_o R) / (R + h_s) - ω² (2 R + h_s) / 2 ] ρ_V

For a ribbon structure about one metre wide this pressure is:

p = 1.346 x 3.6x10⁷ x 1.3x10³
= 6000 MPa = 6 GPa

4.2 Materials

Values of volume density ρ_V and yield stress Y_s for steel and carbon nanotubes (CNT) are given below:

STEEL
ρ_V ~ 8000 kg m^-3 (volume density)
Y_s ~ 200 MPa (yield stress)


CARBON NANOTUBES (CNT)

ρ_V ~ 1300 kg m^-3 (volume density)
Y_s ~ 100 x Y_s for steel ~ 20 GPa (yield stress)
ρ_l ~ 10 kg/km (lineal density)

Comparison of these figures with the requirements for a yield strength of 6000 MPa (6 GPs) show that steel falls short by a factor of 30, whereas CNT exceeds the requirements by a factor of 3.

CNT thus appears able to meet the requirements for s space elevator. One possible configuration is shown below:

4.3 Design

One possible design:

Length ~ 100,000 km Counterweight mass ~ 600 tons Carbon Nano Tubes ρl ~10 kg / km Cable Mass ~ 1000 tons Cable Shape - ribbon ~1m wide Cable Cars ~ 20 ton Speed ~ 200 km/hr Drive - electric motors Power - photovoltaic cells receiving energy from infrared lasers beamed from the ground Base - floating 'oil-rig' in eastern Pacific Ocean (near Galapagos Islands) with +/- 1 km mobility to avoid large space objects/debris

5 PROBLEMS

• Fabricating cable with required strength and length at an affordable cost.
  - 60,000,000 metres of cable at a speculated cost for mass production of $100 per metre implies $6 billion minimum just for the cable

• Cable instabilities, oscillations
  - the devil is in the details

• High power lasers to deliver power to the 'climbers'
  - not yet available at sufficient power
  - motor overheating

• Weather
  - terrestrial and space weather
  - hurricanes in tropical regions
  - tsunamis

• Electromagnetic interactions
  - between the cable and equipment and the geomagnetic field
  - geomagnetic storms

• Space radiation
  - van Allen radiation belts
  - this is particularly concerning for passengers spending a few days traversing this region

• Cable erosion from atomic oxygen

• Deliberate damage
  - sabotage, terrorists

• POLITICAL OPPOSITION
  - COST

6 REFERENCES

WEB

Space Elevator
ISEC


BOOKS

Edwards and Ragan (2000), "Leaving the Planet by Space Elevator"
Michael van Pelt (2009), "Space Tethers and Space Elevators" [Copernicus Books]


FICTION

Arthur C Clarke, "Fountains of Paradise", 1979
Charles Sheffield, "The Web Between the Worlds", 1979
Kim Stanley Robinson, Mars Trilogy (1992+), Space elevators on Mars and Earth
William R Forstchen, "Pillar to the Sky", 2014



Australian Space Academy