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I think I have made a mistake somewhere in my design of an untapered steel cable space elevator down to Earth. I was building this as a design reference to figure out how much stronger "stronger" materials needed to be to make one, and have made a mistake, I'm pretty sure.

Some units :

  • Gravitational Constant (G) : $6.67 \times 10^{-11}$
  • Mass of Earth (M) : $5.97 \times 10^{24}$ kg
  • Rotational Period of Earth : 24 hours
  • Density of Steel Cable ($\rho$) : 7,560 ${{kg}\over{m^3}}$
  • Payload : 22,400 tons = 22.4 million kg

Some formulae :

  • Rotational Velocity ($\omega$) : ${2 \pi} \over { 24 {{hr} \over {day}} \times 60 {{min} \over {hr}} \times 60 {{sec} \over {min}}}$ = 0.0000727, for Earth
  • Height of Space Elevator mid-station (d) : $\sqrt[3]{{GM} \over {\omega^2}}$ = 42,233 km, for Earth
  • Height (above ground) of Space Elevator (h) : $d - R_{earth}$ = 35,970 km, for Earth
  • My understanding is that, if a sufficiently heavy counterweight is used, that not much more cable than (h) is required for the full cable run.
  • Relation of cable strength (S) to cable cross-sectional area (A), cable density ($\rho$), payload (P), cable length (h) and gravity (g ~ 9.8, for Earth) : $ SA = (P + Ah\rho)g \rightarrow S = {{(P + Ah\rho)g} \over {A}}$

When I put all these values in for a 6 cm diameter cable, I get a cable strength of 85.8 GPa (gigapascals).

This is well within the boundary of modern steel (around 200 GPa).

The amount of steel cable required would be 1.85 million tons.

While the construction method is still beyond are technology, I was under the impression that the materials were also. Where have I made my mistake?

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  • $\begingroup$ Looks like it’s a common assumption. I found an article questioning the assumption, but haven’t gotten through it yet - scholarsmine.mst.edu/cgi/… $\endgroup$
    – James McLellan
    Commented Dec 5, 2021 at 20:45
  • $\begingroup$ Am I missing something? I don't see the weight of the cable in the load--and that's what prevents elevators on Earth. (And note that you can't just look at mass & gravity as the gravity varies from 1g at the surface to 0g at geosync.) $\endgroup$
    – Loren Pechtel
    Commented Dec 6, 2021 at 5:20

2 Answers 2

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The glaring error is the fantastic value for the tensile strength of steel.

The tensile strength of steel is about 2500 MPa tops, OK let's assume super optimistically 5000 MPa. I have no idea from where you can get steel with a tensile strength of 200,000 MPa.

Ignoring the payload, your formula reduces to $S = h\rho g$.

Plugging in $h =$ 39,970,000 meters, $\rho =$ 7500 kg/m³, and $g =$ 9.81 m/s² I get $S =$ 2,940,793 MPa or 2941 GPa, about 600 times greater than the most optimistic tensile strength of any kind of steel.

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  • $\begingroup$ I must have lost a zero somewhere. Thank you for the sanity check. $\endgroup$
    – James McLellan
    Commented Dec 5, 2021 at 20:53
  • $\begingroup$ Found it! I was not converting cable length from kilometers to meters in the last bit of calculation. Thank you, again. $\endgroup$
    – James McLellan
    Commented Dec 7, 2021 at 11:31
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Redoing your last equation, I got a figure of 2745 GPa (wolfram alpha link), which is quite a bit higher than your number. I'm not sure where the 85.8 GPa mentioned came from

That said, I believe you also need a counterweight. Right now, the centre of mass of your system is pretty far from geostationary orbit because the mass of the cable is a lot more than the mass of the orbital station. As a result, it would not have a stable position relative to the anchor point on Earth.

Including the current mass of the cable into the payload mass to represent the counterweight increases the result of your equation to 9164 GPA, which is even higher. (wolfram alpha link)

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  • $\begingroup$ I had missed a unit conversion in my final calculation. Thank you! Great question about the counterweight. I’m just trying to think through it : in an ideal case, there’s neither tension or compression on the Earth side. In that setup, the whole weight of the cable and payload (but not the counterweight) will be expressed as tension on the counterweight side. Is that wrong? $\endgroup$
    – James McLellan
    Commented Dec 7, 2021 at 11:40

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