Polar Escape Strength Threshold

Suppose we had a cable with mass per unit length $\lambda$.
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.