## Monday, May 2, 2016

### Sky Hooks

If we want to make it easy(ish) to get into space, then it would be very helpful to have a sky hook. Once we have that in place we can build an elevator.
Lets look into the physics / maths of how it could be done.
Suppose a small object of mass m is travelling in a cirle round a sphere. The object is a distance R from the centre of the sphere.
In vector notation, write its position as: $\textbf{p}=R\left( \begin{array}{c} \sin(\omega t) \\ \cos(\omega t) \end{array} \right)$ We can then differentiate that twice with respect to time and we find the acceleration: $\textbf{a}=\frac{\partial^2\textbf{p}}{\partial t^2}=-\omega^2 R \left( \begin{array}{c} \sin(\omega t) \\ \cos(\omega t) \end{array} \right) = - \omega^2 \textbf{R}$
So, when an object is moving in a circle, there is an acceleration towards the centre of that circle.
But we know that the force: $\textbf{F}=m \textbf{a}$ So if we want to maintain the object rotating about the centre, then we need apply a force toward the centre of magnitude: $F=m \omega^2 R$ That is called the centrifical force.
For a satellite travelling round a planet, there is a gravitational attraction: $F_g=\frac{G M_e m}{R^2}$ Now, if the gravitational attraction is equal to the centrifical force, then the satellite can travel in a circular orbit: $m \omega^2 R = \frac{G M_e m}{R^2}$ We can solve for $$R$$: $R = \left( \frac{G M_e}{\omega^2} \right)^\frac{1}{3}$
For a geostationary orbit, as the earth spins on its axis, the satellite will remain above the same point on the equator, we set: $\omega_s \ n_d s_d =2 \pi n_r$ where the number of seconds in a day: $$s_d = 24 * 3600$$
the number of days in a year $$n_d\approx365.2422$$
the number of revolutions of the earth about its own axis in a year $$n_r=n_d+1$$
(The +1 there comes from the fact that the earth is also revolving round the sun, for details see the Sidereal time article in Wikipedia).
And we find: $\omega_s \approx 7.29211 \times 10^{-5} \ rad . s^{-1}$ and plugging that into the equation for $$R$$ we find: $R_g \approx 4.2164 \times 10^7 m = 42,164 km$ That's the distance from the centre of the earth to a geostationary satellite.
It is 35,786 km above sea level.

Suppose now we were to consider a different geostationary satellite, this time with a cable attached. The other end of the cable is anchored to the ground.
Now suppose the cable is perfectly vertical and there is a tension $$T(r)$$ in the cable at r. The net force on the satellite should be the centrifical force: $T (R) + \frac{G M_e M_s}{R^2}= M_s \omega_s^2 R \ \ \ \ \ \ \ \ \ \ \ \ (eqn 1)$ Now we consider a small section from $$r$$ to $$r+\delta r$$ above the centre of the earth.
Suppose the cable had constant mass per unit length $$\lambda$$. The mass of the small section is: $\delta m = \lambda \delta r$ The gravitational force on that section is: $\frac{G M_e \lambda \delta r}{r^2}$ We write the net force on the small section from $$r$$ to $$r+\delta r$$ due to the change in tension as: $$\delta T(r)$$.
For the cable section to remain in circular orbit we need the net force to be the centrifical force. Hence: $\frac{G M_e \lambda \delta r}{r^2} - \delta T(r) = \lambda \delta r \omega_s^2 r$ $\delta T(r) = \frac{G M_e \lambda \delta r }{r^2} - \lambda \delta r\omega_s^2 r$ $T(R) - T(r) = \int^R_r dT(r') = G M_e \lambda \left( \frac{1}{r} - \frac{1}{R} \right) + \frac{1}{2} \lambda \omega_s^2 \left( r^2 - R^2 \right)$ But we already have an expression for $$T(R)$$ above (eqn 1). So, after a bit of re-arranging we find the tension in the cable to be: $T(r) = G M_e \left( \frac{\lambda}{R} - \frac{\lambda}{r} -\frac{M_s}{R^2} \right) + \frac{1}{2} \lambda \omega_s^2 \left( R^2 - r^2 \right) +M_s \omega_s^2 R \ \ \ \ \ ( eqn 2)$ Let $$r_m$$ be the position of the maximum tension in the cable. $0=\frac{\partial T}{\partial r}(r_m)$ We find that occurs at the orbit of an untethered geostationary satellite: $r_m = R_g = \left( \frac{G M_e}{\omega_s^2} \right)^\frac{1}{3}$ You can't push with our rope!
The cable has tension but does not offer vertical support.
Mathematically we write that as: $T(r) \geq 0\ \ \ \ \ \ \ \ \forall\ r : r_e \leq r \leq R$ To support the mass of the cable, the satellite must be further out than $$R_g$$
We can use eqn 2 to determine the tension in the cable at the earth's surface $$T(r_e)$$.
That's the maximum force the climber is allowed exert so as not to pull down the satellite.

## Thin in the right places

Suppose the breaking tension of a cable is $$T_b$$ then we define the specific strength $$Y$$ to be: $Y = \frac{T_b}{\lambda}$ We define an optimal non-uniform cable to be one where at every point it is just thick enough to safely withstand the tension at that point.
Mathematically we write that as: $\lambda(T ) = \frac{T\ \alpha}{Y} \ \ \ \ \ \ \ \ \ (eqn 3)$ where our $$\alpha$$ is the safety factor with: $1 \leq \alpha$ With a little algebra we find: $\alpha = \frac{T_b}{T}$ and so we know that $T \leq T_b$ in other words the tension in the cable is at or below the tension at which the cable breaks. A higher $$\alpha$$ is safer.
But now the mass of a small section from $$r \rightarrow r + \delta r$$ is: $\delta m = \delta r \lambda \left( T (r) \right) = \delta r \ \frac{\alpha T(r)}{Y}$ Just as before, to keep the cable section in a circular orbit the net force should equal the centrifical force. The net force is made up of a combination of the change in tension and the gravitational force. $T (r) - T (r + \delta r) + \frac{G M_e \ \delta r \ \alpha T (r)}{Y \ r^2} = \frac{\omega_s^2 r \ \delta r \ \alpha T (r)}{Y}$ We can rearrange that: $\frac{\delta T}{T} =\delta r \left[ \frac{G M_e \alpha}{Y r^2} - \frac{\omega_s^2 r \alpha}{Y} \right]$ We now integrate both sides between $$r_1$$ and $$r_2$$ where $r_e \leq r_1 < r_2 \leq R$ And we find: $ln \left( T(r_2) \right) -ln \left( T(r_1) \right) =G M_e \frac{\alpha}{Y} \left[ \frac{1}{r_1} - \frac{1}{r_2}\right] + \frac{\omega_s^2 \alpha}{2 Y} \left( r_1^2 - r_2^2 \right)$ and we can write that as: $\frac{T(r_2)}{T(r_1)} =exp \left( G M_e \frac{\alpha}{Y} \left[ \frac{1}{r_1} - \frac{1}{r_2}\right] + \frac{\omega_s^2 \alpha}{2 Y} \left( r_1^2 - r_2^2 \right) \right) \ \ \ \ \ (eqn 4)$ Now bear in mind that when deriving that equation we have used an optimally thin cable, (see equation 3). At all points the cable mass per unit length has been chosen to be just thick enough to be safe.
Now if we know the tension at any one spot from ground all the way up to the satellite, then we can use equation 4 to work at the tension at all other locations along the cable. We could specify the tension at the ground anchor and then work out the tension all the way up to the satellite.
We consider 3 variables:
$$M_s$$: the mass of the satellite.
$$R$$: the distance of the satellite from the centre of the earth.
$$T(r_e)$$: the tension at the anchor on the surface of the earth.

Now if we have any two of those we can work out the third.
For example we could specify the distance of the satellite and the tension at the ground anchor and we can work out the mass of the satellite required.
To do that, we would use equation 4 to work out the tension in the cable when it meets the satellite: $$T(R)$$.
But we also know that at the satellite, the centrifical force is equal to the gravity plus the tension, so: $M_s \omega_s^2 R = T(R) + \frac{G M_e M_s}{R^2} \ \ \ \ (eqn 5)$ and so we have a formula for the satellite mass: $M_s = \frac{T(R)R^2}{ \omega_s^2 R^2 - G M_e}$
Alternatively if we know the mass of the satellite ($$M_s$$) and the distance of the satellite from the centre of the earth $$R$$ then we can use equation 5 to evaluate $$T(R)$$ and then equation 4 to find the resultant tension at the ground anchor: $$T(r_e)$$

And if we have $$T(r_e)$$ and $$M_s$$ and we want to evaluate R, then we can use equations 4 and 5 and with the aid of a numerical solver we can evaluate R.

Which ever way we do it, when the tension has been determined, we can return to equation 3 to find the mass per unit length and then integrate to find the total mass of cable required: $M_c = \int^R_{r_e}dm(r) = \frac{\alpha}{Y} \int^R_{r_e}T(r)dr$ That integral can be done numerically.

I've written a calculator using Google sheets and I've published it as a template. To have a look, log-in using a google (gmail) account and click on this link.

Things to consider:
stability to perturbations including wind
could attached kites be helpful in reducing the maximum tension?
collisions: birds, planes, other satellites, space debris.
safety if it snaps. One option would be for the anchor at the earth to release as soon as a break has been detected.

Notes:
In the calculations above we've used the following:
Gravitational Constant: $$G \approx 6.67 \times 10^{-11} \frac{m^3}{s^2 kg}$$
Mass of earth: $$M_e=5.97 \times 10^{24} kg$$
Radius of earth: $$r_e = 6.37 \times 10^6 m$$

#### 1 comment:

Anonymous said...

Love to see work like this. We inside the International Space Elevator Consortium research committee are trying to establish an effort to refine the many different approaches to simulating the dynamics of the long tether for space elevators. There are so many questions to be asked? How long must it be? how stable can it become? how much dynamic motion will occur at the GEO altitude? what is the magnitude of the forces inside the tether [not only just mass and centripetal force, but wind shear, movement of tether climbers, motion of the tether, or tension variation along the tether and elastic factors]. If you are interesting in participating, please go to www.isec.org and participate as a volunteer in this global effort with the future of CIS lunar flight as its future.
Pete Swan, President of ISEC.