homework and exercises - What properties do you need for building a tower?

When I was a boy I used to daydream about building a tower so tall that the top of it would project into near space.

There would perhaps be a zero gravity area in the penthouse where my friends and I could bounce around and play space versions of various earth-based games and sports in most excellent zero-g conditions.

Much to my continued disappointment and despite all the technological advances of the last thirty or so years, no one has built such a structure.

Can anyone explain the physical limitations/constraints that are preventing someone from realising my fantasy of a 'Space Tower'?

UPDATE: This Kickstarter Project seems to be pretty confident ...

Answer

First off, the limitation is a material that would not collapse under the weight - earth crust is not quite hard enough. Buckling and other instabilities, nope. Generally, forget a tower built on earth. Not a chance, no such material.

Start building from geostationary orbit and extend the "rope" both inside and outside the orbit. The outside may be just heavy counterweights, as the inside will begin to pull towards earth. Make the orbital part thicker to support extra weight, as you extend the lower part, until it reaches earth surface below.

Now the problem is the material. The only material in existence with sufficient weight-strength ratio is buckytubes. These are currently centimeters long at most, extremely expensive and you'd not only need thousands of kilometers of them... the rope to sustain its own weight would have to be about 1km thick in the thickest place (near the geostationary orbit).

Now consider:

• earth carbon supply, I don't think all coal mines combined could mine that much carbon


• construction craft fuel. This all would have to be lifted high enough. A LEO rocket takes many times more fuel than its payload weight. A geostationary orbit rocket - much more. The good news is the fuel can be hydrogen+oxygen which is water, and we have that aplenty. The bad news is you need at least as much energy to separate them as you gain from burning them, so the power consumption for fuel production would exceed whole world's power production.


• environmental impact of that much steam released into atmosphere


• account for micrometeorites that can really rain over your parade. And this thing being that big, collisions WILL happen. Also account for space junk.


• account for winds and storms once you reach the atmosphere. Also, upper atmosphere is pretty hot... not nice work environment, also nanotubes aren't extremely fire-proof.


• cost and impact on economy. Coal becomes super-expensive and we look for alternate sources of carbon.

And when you finally build it, calculate how long a lift travelling some 300km/h would take to reach 37,000km of the 0-gravity orbit...

EDIT:

I can't currently find the article that listed 1km thickness, but let us try to calculate parameters of the tower merely strong enough to sustain itself.

The nanotube tensile strength is $UTS=6422kg/mm^2$ (1)

The density is $\rho=1.4g/cm^3$

The ribbon is said to be 1m wide.

The thickness will vary. For the needed $M_0=20t=20000kg$ capacity it needs $A_1=0.31mm^2$ cross-section at the bottom. At 1000mm width that's 0.00031mm thick.

Now I'm really not in the mood to solve a differential equation of thickness - mass - tensile strength - gravity so let me try a discretization, approximating with $h=1km$ long wedges. At 35000 samples that should give us a decent approximation.

$$ V_n= {A_n+A_{n+1} \over 2}h \\ M_n=\rho V = \rho {A_n+A_{n+1} \over 2}h $$

Now we can't happily assume weight not to vary with altitude. After all, near the orbit it will be zero. It varies with distance from center of Earth. At surface, $r_0=6378km; M_{earth}=5.97 10^{24}kg; G =6.67300 × 10^{-11} {m^3 \over kg s^2}$;

So, the weight function of each segment will be

$$ Fw_n=G{M_n M_{earth} \over r_n^2} \\ r_n=r_0+n[km] $$

And the tensile strength surface $A_{n+1}$ must overcome is

$$ F_{n+1}=F_n + Fw_n \\ F_{n+1} = A_{n+1} UTS $$

We seek $A_{35000}$ which will trivially yield thickness by dividing by 1000mm.

$$ A_{n+1} UTS = F_n + Fw_n \\ A_{n+1} UTS = F_n + G{M_n M_{earth} \over r_n^2} \\ A_{n+1} UTS = F_n + G{\rho {A_n+A_{n+1} \over 2} h M_{earth} \over r_n^2} \\ A_{n+1} = F_n + (A_n+A_{n+1}){ G \rho h M_{earth} \over 2r_n^2 UTS } \\ X := { G \rho h M_{earth} \over 2r_n^2 UTS }\\ A_{n+1} - = F_n + (A_n+A_{n+1})X \\ A_{n+1} = F_n + X A_n + X A_{n+1} \\ A_{n+1} - X A_{n+1} = F_n + X A_n \\ (1-X)A_{n+1} = F_n + X A_n \\ $$ We get our two fundamental equations for numeric computation: (with helper X, which I'm really not in the mood to transform into something nicer.) $$ X = { G \rho h M_{earth} \over 2r_n^2 UTS }\\ A_{n+1} = { F_n + X A_n \over (1-X) } \\ F_{n+1} = A_{n+1} UTS $$

Now excuse me, it's 3AM and I'll finish the calculations at a different time.