Space Elevator

Space elevator or Space tower

Leaving Earth’s gravitational pull requires tremendous energy not to mention the cost associated with it. At present we use rockets to propel object into space. But there is an alternative floating around–space elevator or space tower. The proposal is to build a very tall rising tower from Earth’s equator terminating geostationary orbit.

The idea of the space elevator was first proposed by the Russian physicist Konstantin Tsiolkovsky and again by the Leningrad engineer, Yuri Artsutanov To the cosmos by electric train. It was then popularized by Arthur Clarke’s novel Fountains of Paradise. Yet until the discovery of carbon nanotubes the realization of space tower was considered as farfetched idea. Carbon nanotubes (CNTs) are tubes made of carbon with diameters typically measured in nanometers. Carbon nanotube is the best candidate so far, that is being proposed as a construction material; this is so because it has high tensile strength along with low density.

How to construct the elevator? the proposal is to build a free standing tower of uniform density and constant cross sectional area at Earth’s equator. But, what is free standing tower? a free standing tower is one whose weight is counter-balanced by the outward centrifugal force on it. This means the tower won’t exert force on the ground beneath it. Such type of tower is in tension along its entire length. Each element of the tower is in equilibrium under the action of gravitational, centrifugal, and tension forces.

The forces acting on a small element of the tower are: an upward force \(\mathrm{F}_U\) due to the portion of tower above the element, a downward force \(\mathrm{F}_D\) due to the portion of tower below, a downward force \(\mathrm{W}\) due to the weight of the element, and a fictitious upward centrifugal force \(\mathrm{F}_C\) on the element due to its presence on the rotating Earth. The vector sum of these forces must vanish if the element is in equilibrium.

The weight and the centrifugal forces balances each other at the geostationary height i.e. \(\mathrm{F}_C = \mathrm{W}\) . This implies the tension forces at the two ends should also be equal, i.e. \(\mathrm{F}_U = \mathrm{F}_D\). For an element below the weight force exceeds the centrifugal force, thereby \(\mathrm{F}_U > \mathrm{F}_D\) for an equilibrium. In contrary for an element above the geostationary height, the centrifugal force exceeds the weight and thus \(\mathrm{F}_U < \mathrm{F}_D\) for an equilibrium.

The physics of space elevator

Let us denote mass, radius and rotational angular velocity of the Earth by \(\mathrm{M}, \mathrm{R}\) and \(\omega\)

\[\begin{aligned} -\frac{GMm}{R_g^2} &= -m\omega^2R_g\\ R_g\ &= \Big(\frac{GM}{\omega^2}\Big)^\frac{1}{3} \label{eq:one}. \end{aligned}\]

Where \(\mathrm{G}\) is the gravitational constant \(G=6.674\times10^{-11} \frac{\mathrm{m}^3}{ \mathrm{kg s}^2}\). Let us assume that the standing tower has constant mass density $\rho$ and cross sectional area \(\mathrm{A}\). Suppose a small element of tower of length \(dr\) whose lower end is a distance \(r\) from the center of the Earth.. Equilibrium condition demands that: \(\mathrm{F}_U - \mathrm{F}_D +\mathrm{F}_C - \mathrm{W} =0\). Moreover if the tensile stress $\mathrm{T}$ is defined as force per unit area we write \(\mathrm{F}_U - \mathrm{F}_D = AdT\). The equilibrium condition can be put as

\[\begin{aligned} AdT &= \frac{GM(Adr\rho)}{r^2}-(Adr\rho)\omega^2r, \\ \frac{dT}{dr}&=GM\rho\Big[\frac{1}{r^2}-\frac{r}{R_g^3}\Big] \label{eq:two}. \end{aligned}\]

upon integrating Eq. ([eq:two]) from \(r=R\rightarrow R_g\), with \(\mathrm{T(R)}=0\), we get the tensile stress at geostationary height \(R_g\)

\[\begin{aligned} \mathrm{T(R_g)} =GM\rho\Big[\frac{1}{R}-\frac{3}{2R_g}+\frac{R^2}{2R_g^3}\Big] \label{eq:three}. \end{aligned}\]

Let \(\mathrm{H}\) denote the distance of the top of the tower from Earth’s center. We can determine \(\mathrm{H}\) by integrating Eq. ([eq:three]) from \(r=R_g\rightarrow \mathrm{H}\) subject to the boundary condition \(\mathrm{T(H)}=0\), which expresses the fact the tension drops to zero at the upper end of the tower. In this way we find

\[\begin{aligned} \mathrm{T(R_g)} =GM\rho\Big[\frac{1}{H}-\frac{3}{2R_g}+\frac{H^2}{2R_g^3}\Big] \label{eq:four}. \end{aligned}\]

If we equate the RHS of Eq. ([eq:three]) and Eq. ([eq:four]) and note that \(\mathrm{H}=\mathrm{R}\) is a solution of the resulting in cubic $\mathrm{H}$, we can reduce the cubic to the quadratic equation

\[\begin{aligned} RH^2+R^2H-2R_g^3 &= 0\\ H&=\frac{R}{2}\Big[\sqrt{1+8\Big(\frac{R_g}{R}\Big)^3}-1\Big]=150,000~\mathrm{km} \label{eq:five}. \end{aligned}\]

the height of the top of tower above Earth’s surface is therefore \(\mathrm{H}-\mathrm{R}=140,000~\mathrm{km}\)

Center of Mass of the Tower?

Another point worth looking at is the question of center of mass of the tower based on Kepler’s law of motion. Kepler’s law is valid for objects whose masses are small in comparison to the variations of the gravity field the objects are moving in. In the case of the space tower, gravity decreases by a factor greater than 300. Also recall that the space elevator is not a point mass.

Take for example two identical masses attached by a length \(2L\) of massless cable. Assume one of the masses is at a distance \(R+L\) from the center of the Earth and the other at \(R-L\). If the the masses are in circular orbit of angular velocity $\omega$ in such a way that the cable always goes through the center of the Earth, the centrifugal acceleration of the masses is

\[\begin{aligned} a_{cL} &=(R-L)\omega^2\\ a_{cH} &=(R+L)\omega^2\\ \label{eq:six}. \end{aligned}\]

Where \(L\) and \(H\) in subscripts indicate Low and High. In like manner the acceleration due to gravity on the masses is expressebile as

\[\begin{aligned} g_{cL} &=\frac{GM_e}{(R-L)^2}\\ g_{cH} &=\frac{GM_e}{(R+L)^2}\\ \label{eq:seven}. \end{aligned}\]

At equilibrium we demand that these accelerations cancel each other out

\[\begin{aligned} (R-L)\omega^2 + (R+L)\omega^2 &= \frac{GM_e}{(R-L)} + \frac{GM_e}{(R+L)}\\ \omega^2 &= \frac{GM_e}{2R}\Big(\frac{1}{(R-L)^2}+\frac{1}{(R+L)^2}\Big) \label{eq:eight}. \end{aligned}\]

For \(L=0\), we obtain \(\omega^2 =\frac{GM_e}{R^3}\), which is Kepler’s third law. If L is small then

\[\begin{aligned} \frac{1}{(R+L)^2}\equiv\frac{1}{R^2}-\frac{2L}{R^3} \label{eq:nine}. \end{aligned}\]

Which also is Kepler’s third law. For large values of \(L\) we note that \(\omega\) is greater than Kepler’s third law would predict. The reason for this is because the increase in gravity for the low mass is greater than the decrease for the high mass.

In conclusion for an orbit that extends over a great range of altitudes, the orbit will be faster than the orbit at the center of mass altitude. Thus, even for an elevator not attached to the Earth, the center of mass should be above geosynchronous orbit (GEO).