## Sunday, May 29, 2016

### Space Elevator Calculations

In this post we use the equations derived in the previous posts and allow the user to plug in some numbers.
In particular, when a cable thickness has been chosen to have the optimal thickness, we determine the mass of satellite required to achieve the given tension at the earth's surface.
We also work out the total mass of the cable.

 Inputs Cable Specific Strength M Yuri Safety factor - Tension at earth's surface Newtons Satellite height above earth km Outputs Total Cable mass - kg Satellite Mass - kg Tension at Satellite - Newtons

## 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.

### 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.

## 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: $$r_e = 6.37 \times 10^6 m$$