How would a space elevator be built?

[Kerbart]

Again lets take the sci-fi mentality out of the science.

You realize you're posting in a thread about space elevators, right?

[PB666]

You realize you're posting in a thread about space elevators, right?

That is important because?

[fredinno]

That is important because?

They are (hard) sci-fi.

[Yourself]

Yeah, divide the acceleration and force by 1000. Kilometers to meters and such.

The acceleration is low, but there is still the force to look at. It's not huge, but it's there. And then you have the mass of the whole elevator to take into account.

That's a bit confusing. The math doesn't need to be that complicated. Of course I say that, but I'm about to write a differential equation for it. So, you know, whatever.

So let's look at a tiny section of cable. We'll say the length of this section is ÃŽâ€L and it's a distance r from the center of earth. So let's do the whole free-body diagram thing only because I'm lazy it's not so much a diagram as it is a paragraph. I'm going to work in the rotating reference frame, so the centrifugal force will show up explicitly. Anywho, we have 4 forces acting on our section of cable:

1. Gravity: μ ÃŽâ€m / r²
   
2. Tension towards Earth: T
   
3. Tension away from Earth: T + ÃŽâ€T
   
4. Centrifugal force: ÃŽâ€m ɲ r
   

So I've introduced some extra variables here:

μ = Standard gravitational parameter of Earth

ÃŽâ€m = Mass of our small section of cable

̉ۡ = Angular velocity of Earth

ÃŽâ€T = The difference in cable tension across our cable section

Great, moving on. Since we don't want the cable to be falling to Earth or flying off into space, we kind of want all the pieces of it to be not accelerating, so that means all these forces need to sum to 0:

0 = -( μ ÃŽâ€m / r² + T ) + ( T + ÃŽâ€T + ÃŽâ€m ɲ r )

Let's do some rearranging:

ÃŽâ€m ( μ / r² - ɲ r ) = ÃŽâ€T

Now, let's say the mass per length of the cable is given by some function we'll call î, that means our little piece of cable's mass can be written as its length, ÃŽâ€r, multiplied by this linear density:

î ( μ / R² - ɲ r ) ÃŽâ€r = ÃŽâ€T

We'll divide by ÃŽâ€R and take a nice limit and, bam, we get our ODE:

dT/dr = î ( μ / r² - ɲ r )

This is a first order ODE so we need a single boundary condition to complete this. One thing we know is that the tension in the cable has to go to 0 at the very top (because there's nothing left up there). so we write:

T(R_Earth + L) = 0

But we never said what L was, the cable is free to be as long as it needs to be. So we can actually specify another boundary condition and then figure out what L should be to satisfy the equation above. So, we can pick this one:

T(R_Earth) = 0

So basically we just made the cable tension at the ground 0. In fact, we can choose the tension to be whatever we want at ground level and it'll just change the length of the cable. We have even more freedom because there's still that î function sitting inside there which corresponds to how the thickness of the cable varies along its length. We could even go further and try to find the thickness that minimizes the the average tension over the whole cable. But, it's been a long time since I've done any calculus of variations, I've spent all day playing Fallout 4, and it's also 1 AM and I'm very tired. So I'll just leave it at this: we can act least mathematically create a cable whose tension at ground level is 0 (provided É isn't 0, which for Earth, it's not).

What this equation tells you is also that the tension in the cable will increase with altitude and reach its maximum at geostationary orbit before it starts decreasing again. This is because R_GEO is the solution to this equation:

0 = μ / R_GEO² - ɲ R_GEO

For values of r < R_GEO, that quantity on the right is positive (which means dT/dr is positive --> increasing) and for values of r > R_GEO, the quantity on the right is negative (so dT/dr is negative --> decreasing).

[gamowin]

Don't worry guys, they just discovered that graphene can be rolled up into a nano scale cylinder, resulting in massive tensile strength ratios. Sound familiar? Sorry, It's just that graphene seems to be the super-material flavor of choice these past few years. Graphene can be used to make supa-dupa strong body armor! Use it to build vehicles, its super light and strong! It's also super-amazing at conducting electricity! Supercomputers here we come! If only graphene was a substance that had practical use outside of paper diagrams (as in you are depositing layers of graphene onto paper with a pencil).

[prophet_01]

Can't you 'just' redirect a big asteroid to stationary orbit and use that to set up the required counter weight? That should reduce the required payload quite a lot.

And for the propulsion of the lift, why not use a ground based laser and downward facing 'solar' panels on the capsule that you send up? If you're constructing the laser emitter a couple of km's away from the ground station of the lift, you're also not firing that laser parallel to the cable which should mostly eliminate the threat of heating up the cable.

Also, aren't there concepts for stations that could produce energy with solar panels in orbits and wend it to earth? If I remember correctly that approach uses remote energy transfers as well and I'm pretty sure the major problem here were the costs, not the technology. I have to look it up though.

Sure, it's likely a decade or two down the road until graphene can be mass produced at that quantities, but a space elevator definitely seems more reasonable to me than fusion powered lift vehicles. Edited November 16, 2015 by prophet_01

[magnemoe]

Read an plan about it and it was pretty interesting.
Idea was to use carbon nanotubes, and an minimal elevator first. put in GEO and start to drop end down to earth, you would need an engine to keep position during this process. because of center of balance the drum and system would move outward.
Once down you grab the edge and attach to an ship. You can move the ship to move the wire in LEO.
The initial wire can not lift very much so you send up spinners to build the elevator up. Empty spinners end up as counterweight.

[tomf]

To answer some earlier questions the thickness of the cable varies with the hight. Each section needs to be strong enough to support the weight of itself and all the cable below leading the a function that is close to exponential for most of the height. How fast it grows is related to the tensile strength of the material. For the strongest steel the cable at the top needs to be 10^16 times thicker than the bottom, so if you have a 1mm^2 cable at the bottom it will be approximately the diameter of the earth at the top. Stronger materials can reduce that factor to much more plausible values, theoretically as low as 1.2.

Numbers taken from [url]http://www.zadar.net/space-elevator/[/url]