Saturday, May 7, 2016

Lifting a cable above the equator

Suppose you were at the equator and decided to hold a cable at some height above the earth's surface. The cable dangles vertically down below, touching the earth, but it is not anchored there. As you go higher and higher, you would stay above the same point on the earth. I.e. the top of the cable would be geostationary as the earth spins.
If the cable were not made of a very strong material then as you tried to go higher, it would eventually break.
It turns out that if you have a cable of specific strength 48.5 M Yuri, then you'd be able to reach all the way up to a geo-stationary satellite (35,797km up).
If you reach that height, then you'll be able to go on out to about 143,937 km without requiring a stronger cable.
Something a tad surprising ( to me at least) happens 143,937 km up. You will no longer need to hold the top end. The tension at both the top and the bottom will be zero. The maximum tension will be at the height of a geostationary satellite orbit.
A cable that long, will be able to support itself. The gravitational attraction is exactly the required centrifical force to maintain a circular orbit.
If you then attempted to use a longer cable and go higher, the cable would lift off the ground. In other words the net force would be upwards. So if you want to go out further, then you'd need to anchor one end to the surface of the earth.

I've made lots of statement there, now lets start to justify some of that mathematically.
This post is a follow on from the previous one: Polar Escape Strength Threshold .
We'll use the same variables as defined there.
The cable is vertical and as the earth spins, the top of the cable remains over the same point on the equator. So the cable is geostationary and hence the net force acting on a small section of cable is the centrifical force: \[ \delta m\ \omega_s^2 r = \frac{G M_e \delta m}{r^2} - \delta T \] but we know: \[\delta m = \lambda \ \delta r\] and so: \[ \delta T = \delta r \left( \frac{G M_e \lambda}{r^2} - \lambda \omega_s^2 r\right) \] We can integrate that and use the fact that the tension in the cable at the earth's surface is zero. \(T(r_e) = 0 \).
We find:
\[ T(R) = G M_e \lambda \left( \frac{1}{r_e} - \frac{1}{R} \right) + \frac{\lambda \omega_s^2}{2} \left( r_e^2 - R^2 \right) \ \ \ \ \ \ \ \ \ (eqn 1) \] We define the specific strength Y to be the tension at breaking point divided by the mass per unit length: \[ Y = \frac{T(R_b)}{\lambda} = G M_e \left( \frac{1}{r_e} - \frac{1}{R_b} \right) + \frac{ \omega_s^2}{2} \left( r_e^2 - R_b^2 \right) \]
We now return to eqn 1 above. We can differentiate that with respect to R to find where is the maximum tension: \[ 0=\frac{\partial T}{\partial R}= \frac{G M_e \ \lambda}{R^2}- \lambda \ \omega_s^2 R \] We find that the maximum tension occurs at the geostationary orbit: \[ R_g = \left( \frac{G M_e}{\omega_s^2} \right)^{1/3} \] The threshold specific strength that would enable the cable to reach this max tension is: \[ Y_E = GM_e \left( \frac{1}{r_e} - \frac{1}{R_g}\right) + \frac{\omega_s^2}{2} \left( r_e^2 - R_g^2\right) \approx 48.5 M Yuri \] Returning to eqn 1, we see that \(T(R)=0\) when \(R=r_e\), in other words the tension is zero at the earth's surface. The tension reaches a maximum at the orbit of a geostationary satellite. It then decreases again. But we can find R such that \(T(R)=0\) \[ 0 = G M_e \left( \frac{1}{r_e} - \frac{1}{R_s} \right) + \frac{ \omega_s^2}{2} \left( r_e^2 - R_s^2 \right) \] That implies: \[ 0 = R_s^3 \ \frac{ \omega_s^2} {2} - R_s \left( \frac{ \omega_s^2 \ r_e^2 }{2} + \frac{G M_e }{r_e} \right) + G M_e \] That is cubic in \(R_s\). We solve that numerically and we find that the distance from the centre of the earth to the point on the cable where the tension is zero is: \[R_s \approx 1.5030 \times 10^8 m\] which 143,937 km from the surface of the earth.
A cable which is that long and is placed vertically above a point on the equator has zero tension at both ends. It neither drifts off into space nor falls to the earth. It just remains where it is.

Returning once again to my (current) favorite equation i.e. eqn 1, we see that it implies that for \(R>R_s\) the tension becomes negative. But we can't allow negative tension since we can't push with our cable. Without being held down the cable would rise up. So we would need an anchor at the surface of the earth. With that anchor we no long would have the conditon \(T(r_e) =0 \) . That was used in the derivation of eqn 1. And so eqn 1 would not be valid.
What we find is that as the cable gets longer beyond 143,937 km, the tension would keep increasing. An infintely long cable would require infinite tension at the anchor.

No comments: