# Space elevator physics for undergraduates

A space elevator is conventionally conceived of as a cable fixed to the (equator of the) earth that reaches all the way to space. At the top of the cable a counterweight is attached, so that the centrifugal force on the cable and the counterweight together counteract the gravitational force on both, fixing the cable upright. What forces are involved?

Assume the cable is made of uniform material of density $$\rho$$, and that at height $$r$$ it has cross-sectional area $$A( r)$$. Then, the gravitational force on the infinitesimal cable segment at height $$r$$ equals

$$d F_g = - G \rho M_{\oplus} \frac{A( r) dr}{r^2},$$

and the centrifugal force on this infinitesimal segment would equal

$$d F_c = \omega^2 \rho A( r) \cdot r dr.$$

The two forces are equal at the altitude of geostationary orbit, the point where $$\frac{G M_{\oplus}}{r^2} = \omega^2 r$$, that is, $$r_s = (\frac{G M_{\oplus}}{\omega^2})^{\frac13}$$ . The centrifugal force dominates above this altitude, and the gravitational force below it.

The change in the tensile force $$dF_T$$ on the cable as we move from height $$r$$ to height $$r + dr$$ must equal the change in gravitational and centrifugal forces: $$dF_T = dF_g + dF_c$$.

Now, we want design the cable for constant tensile stress. In other words, we want to use as little material as possible while preventing the cable from breaking. This means fixing $$dF_T = T d A$$ so that we obtain $$T dA = A \rho (G M_{\oplus} / r^2 + \omega^2 r) dr$$. In terms of the altitude $$r_s$$ of geostationary orbit and the constant $$g = G M_{\oplus} / r_0^2$$ of gravitational acceleration at the surface, this implies that

$$\frac{d(\ln A)}{dr} = \rho g r_0^2 / T (1/r^2 - r / r_s^3)$$

We can conclude here that the cross-sectional area of the cable increases from the surface level on until $$r = r_s$$, at which point it decreases (exponentially). The formula for the optimal cross-sectional area distribution $$A( r)$$ is then found by integrating over $$r$$:

$$A( r) = A(r_0) \exp{ \frac{\rho g r_0^2}{T} \left( \frac1{r_0} - \frac1{r} + \frac{r_0^2 - r^2}{2 r_s^3} \right)}$$

As noted above, the maximum is reached at $$r = r_s$$. The so-called taper ratio $$A(r_s) / A(r_0)$$ is therefore a function of the ratio $$\rho / T$$ alone:

$$\frac{A(r_s)}{A(r_0)} = \exp{ c \rho / T },$$

where $$c \sim 48.5$$ if we denote the specific strength $$\tau = T / \rho$$ in units of MPa/(kg/m3). Using the strongest commercially available materials such as Toray T1100G would lead to a rather large taper ratio of roughly $$2.4e5$$, corresponding to a ratio of roughly 500 between the radii of the cable at these respective altitudes.

A very rough estimate using the trapezoidal rule shows that the mass of the cable will at least have to equal $$(r_s - r_0) \rho ( A( r) + A(r_0) )/2$$. For a cross-sectional area at surface level of 1 square millimeter (meaning the cable would measure approximately 14 square meter at its thickest), that would amount to a not-too-shocking-but-still-prohibitively-large number of 46 million kilograms.

For fixed tensile strength at ground level, the mass estimate scales with specific strength as $$m \propto \tau^{-1} (\exp{48.5 \tau^{-1}}+1)$$. We see that the specific strength is the crucial factor determining whether a space elevator is feasible to build. For comparison: the specific strength of T1100G is in the order of 3.9 MPa/(kg/m3), whereas that of carbon (nano)tubes is in the 40+ MPa/(kg/m3) range. The latter would most definitely suffice.

Calculating the mass of the cable and counterweight (depending on the total cable length) exactly and accomodating an actual elevator along the cable in the design greatly complicate the differential equations. This is the point where the ‘for undergratuates’ label would no longer apply and where a computer might be employed profitably, so we will end the discussion here for now.

This post was loosely based on the excellent introduction ‘The physics of the space elevator’ by P.K. Aravind.