## Saturday, May 7, 2016

### 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.