1
$\begingroup$

I had a discussion with my wife about physics behind a Space Elevator. Imagine a cable rising 60k into the firmament for its elevator to climb. The cable would have to rotate in unison with the earth, maintaining a straight line. How?

She believes that you need to rocket or a mass of some size to keep the cable rigged; otherwise the earth would roll up the cable like a spinning spool rolling up its thread.

As a model, I mentioned that someone could spin a rope of some mass around himself with the centripetal force keeping he rope taut. However this would not work with a small-mass thread.

So....

How much mass would we need to keep the space-elevator cable rigid?

$\endgroup$
4
$\begingroup$

Your wife could be right about the rocket.

Whenever we say something is small physically, we need to be sure what it is small with respect to. In the case of a thread, the drag force exerted by air beats the centripetal force keeping it taut. Because centripetal forces scale with mass, a denser thread will work.

The issue facing a space elevator is different. For a rigid cable, we need the terminus to be above geostationary orbit. The height of the atmosphere is about 100km from sea level, while geostationary orbit is about 36000 km, so very little of the cable will be exposed to atmospheric drag. What net drag there is will be mostly from the prevailing winds, I suspect. The effect of this will be small but persistent. If it is never corrected, over time the elevator will drift away from vertical. Whether this effect is large enough to matter over, say, the lifetime of a civilization, I don't know.

For the other part of your question - what mass would we need to keep it taut? - I think your intuition about the rope is good. Here our concern isn't wrapping, but whether the rope will collapse. Imagine the rope is made of little beads of mass d$m$ connected by little massless cables of length d$h$. In the rotating frame, a bead at height $h$ in the cable experiences a total force $\mathrm{d}F = -\mathrm{d}m\frac{M_eG}{(h+R_e)^2}+T(h+\mathrm{d}h) - T(h) + \mathrm{d}m \, \omega^2(h+R_e) = 0$ for a system in equilibrium, where $T$ is the tension. If we call the mass per unit length of the cable $\lambda$, then we can turn this into a differential equation:

$$\frac{\mathrm{d}}{\mathrm{d}h} T(h) = \lambda \left ( \frac{M_eG}{(h+R_e)^2} - \omega ^2 (h+R_e) \right )$$ Integrating: $$ T(h) = \lambda \left ( M_eG \left( \frac{1}{R_e} - \frac{1}{R_e+h} \right ) - \frac{1}{2}\omega^2 (h^2 + 2 h R_e) \right) +C $$ We find the integration constant $C$ we look at the base of the cable. The tension there must be sufficient to keep the whole cable in place. The force the whole cable exerts at that point is

$$T(0)=C=\int_0^L \lambda \left ( -\frac{M_eG}{(h+R_e)^2} + \omega ^2 (h+R_e) \right ) \mathrm{d}h$$

where $L$ is the total length of the cable. We can see right away that this integral becomes more and more positive as $L$ increases, so even without finishing the integral we know there is some value of $L$ that will make this base tension positive. The condition that the cable stay taut is the condition $T(h) > 0 \;\forall\; h < L$ - that is, there is tension in the cable everywhere but the very end. Plug in the value of $C$ and you will find that if $C$ is positive, this is the case.

So all we technically need is a long cable. But it might be more efficient to have a counterweight at the end.

$\endgroup$
  • $\begingroup$ This should not be the accepted answer. The jet stream does not come anywhere near the equator, it neglects that the center of gravity must necessarily be at geosynchronous altitude, and it assumes a uniform density per unit length, which makes for a very bad design. $\endgroup$ – David Hammen Oct 14 '14 at 21:22
  • $\begingroup$ Thank you for pointing out that the jet stream doesn't come close to the cable. I fixed it. A uniform density may be a bad design, but the question didn't ask for a design - it asked if a counterweight was necessary. Finally, your statement about the center of gravity is wrong. It corresponds to the condition $C=0$, i.e. a cable that hangs suspended in space. For an actual elevator to be useful, $C$ must be greater. You are perhaps thinking that if the center of gravity is not at this point, the cable will necessarily drift east or west. This is not the case. $\endgroup$ – user27118 Oct 15 '14 at 20:41
1
$\begingroup$

A space elevator would need to extend well beyond the altitude of geosynchronous orbit. The center of gravity (not center of mass) of the space elevator system would need to be at geosynchronous altitude.

The most common architecture is to attach the outer end of the elevator to a counterweight. If the cable broke, this counterweight would orbit elliptically, with perigee at the counterweight's pre-break distance from the Earth. The cable prevents the counterweight from following this natural motion if the cable remains intact. This in turn makes the counterweight exert an outward tension on the cable, which in turn makes the cable system orbit the Earth as if it were a point mass located at the cable's center of gravity. No rockets are needed, at least ideally.

$\endgroup$
  • $\begingroup$ Any idea on the required mass of the counterweight? $\endgroup$ – Frederick C. Lee Oct 14 '14 at 19:49
  • $\begingroup$ There's no clearcut answer. The reason for having a counterweight is to reduce the length of the cable. The cable would need to be considerably longer without that counterweight. The optimal length and counterweight depends on the material used, the variations in thickness of the elevator, on a number of safety and economic concerns. $\endgroup$ – David Hammen Oct 14 '14 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.