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.
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