So the mass of a small section of length \(l\) is: \[m \ = \lambda \ l \] If the cable is hanging vertically in a constant gravitational field \(g\), then the tension at the top will be the \[T = m \ g = \lambda \ l \ g \] We define the

*specific strength*to be the tension at breaking point divided by the mass per unit length: \[Y=\frac{T_b }{\lambda} \] This is measured in units of: \[[Yuri] = \frac{[Newton] [ meter]} {[kilogram]^2} \] With constant \(g\), a cable with specific strength \(Y\) will be able to support up to its breaking length: \[l_b =\frac{T_b }{\lambda \ g} = \frac{Y}{g}\] Now how about the case when the gravitational follows a one over r squared law. Where r is the distance from the centre of the earth. Lets start by considering a vertical cable that isn't moving: A small section from \(r\) to \((r + \delta r)\) has forces acting up and down. If it is not accelerating, then the forces are balanced: \[F_{up}=F_{down}\] \[T(r + \ \delta r )= T(r)+ \frac{G M_e \delta m }{r^2} \] But we know the mass: \[\delta m = \lambda \ \delta r\] So \[\delta T = \frac{G M_e \ \lambda \ \delta r }{r^2} \] We can integrate both sides of that for a cable that goes from the earth's surface \(r=r_e\) to \(r=R=r_e + h \) a height h above the earth.

We note that the end that touches the surface is not anchored. So the tension: \[T(r_e)=0\] So \[T(R)=G M_e \ \lambda \left( \frac{1}{r_e} - \frac{1}{R} \right)\] Suppose the cable breaks at \(R_b\), returning to the definition of the specific strength: \[Y=\frac{T(R_b)}{\lambda} \] Thus: \[Y=G M_e \ \left( \frac{1}{r_e} - \frac{1}{R_b} \right)\] rearranging we find: \[R_b = \left( \frac{1}{r_e} - \frac{Y}{G M_e}\right)^{-1}\]

So the breaking height above the earth's surface: \[h_b = R_b - r_e = \left( \frac{1}{r_e} - \frac{Y}{G M_e}\right)^{-1} - r_e\] Now when: \[\frac{1}{r_e} - \frac{Y}{G M_e} = 0 \] we can reach an infinite height! In other words the cable can completely escape the earth's gravitational field.

The polar escape strength threshold is: \[Y_p = \frac{G M_e}{r_e} \approx 62.6 M Yuri \] We use the word

*polar*here since is most relevant for a cable above either the north or south pole. If we were to have a geostationary cable above the pole, then we could ignore the spin of the earth. On the other hand for a geostationary cable above the equator, the centrifical acceleration should not be ignored. I'll show the details of how to deal with that in my next post.

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