7
$\begingroup$

Consider an orbital elevator simply modelled as a large mass in geostationary orbit (altitude approx 36,000km) around the Earth, connected to a point on the equator via a strong, light, likely flexible "cable". Assume this cable can withstand some compressive force without buckling, e.g. by pre-stressing.

Now consider a mass (the cargo to be elevated) ascending the elevator cable with a constant speed. Because this mass is moving along a cable which is rotating in sync with the Earth, a Coriolis effect will be applied to the cargo mass.

This is a problem as this will have a bowstring-like effect on the cable, causing it to deflect in a transverse direction, which make cause transverse forces to occur on the orbiting mass, endangering its orbit and risking the collapse of the elevator.

So my problem is this: is the Coriolis effect large enough to affect the elevator as mentioned above? If so, what are the most practical countermeasures to resolve this issue?

For example, causing the cable to be rigid may have been a thought. However, as the cable is so long, it will likely break in one place, creating a worse issue.

$\endgroup$
  • 2
    $\begingroup$ It is a big issue see for instace: newscientist.com/article/… and content.time.com/time/health/article/0,8599,2099830,00.html $\endgroup$ – user66432 Apr 5 '15 at 18:21
  • 1
    $\begingroup$ The solution: An agonizingly slow ascent by the crawlers, or the careful orchestration of multiple crawlers. In fact, scientists are estimating trips lasting almost a month. $\endgroup$ – user66432 Apr 5 '15 at 18:27
  • 1
    $\begingroup$ Keep in mind that a typical design for a beanstalk is under tension over most of it's length. Tensile structure deal with transverse loading a bit better than compressive ones do, which helps. $\endgroup$ – dmckee Apr 14 '17 at 15:33
  • 2
    $\begingroup$ Re, "Assume this cable can withstand some compressive force without buckling." That is, quite franky, unimaginable. A structure that is tens of thousands of miles long is not going to withstand any significant compressive force. Fortunately, it does not have to withstand it because the counterweight is not in geostationary orbit. The counterweight is further out than that, where it experiences significant centrifugal force, and therefore the "cable" is always under tension. $\endgroup$ – Solomon Slow Jun 24 '17 at 16:12
  • 3
    $\begingroup$ @WetSavannaAnimalakaRodVance I think it is more common among science fiction fans than among pros and also that it has been falling out of favor. Which is too bad because I like it. $\endgroup$ – dmckee Sep 4 '17 at 2:44
1
$\begingroup$

The Coriolis acceleration is easily derived from Newton's $F=ma$ by switching to polar coordinates via $x=r\cdot cos(\theta), y=r\cdot sin(\theta)$ and then working out the second time derivatives $\ddot r$ and $\ddot \theta$ under the condition $F=0$ (i.e. assuming no external physical forces). The tangential acceleration turns out to be $r\ddot \theta = -2\dot r\dot \theta$. This is the Coriolis effect (and solving $\ddot r$ in the same way gives us the centrifugal force). The value of $\dot \theta$ for Earth is 0.0000729 radians per second, so a radial speed of 100 m/s will result in a Coriolis effect of about 0.015 m/s².

Compensating for the Coriolis effect will have the same end result regardless of the elevator's radial speed. The end result is that the elevator will be in a geostationary orbit at $r=42164$ km with the speed $r\dot \theta=3074$ m/s. We can compare the elevator's increased kinetic energy with the energy required to lift the elevator in the gravity well: $$E_{kin} = \frac{mv^2}{2} - \frac{mv_0^2}{2} = m\cdot \frac{3074^2-465^2}{2} = m\cdot 4616626$$ $$E_{pot} = GMm\cdot(\frac{1}{r_0}-\frac{1}{r}) = mgr_0\cdot(1-\frac{6378}{42164}) = m \cdot 9.81\cdot 6378\cdot 10^3 \cdot 0.85 = m \cdot53182953$$ $$\frac{E_{kin}}{E_{pot}} = \frac{4616626}{53182953} = 0.087$$

So compensating for the Coriolis effect will require a small but non-negligible part of the total energy budget.

$\endgroup$
0
$\begingroup$

To maximize throughput and hence revenue of a space elevator it will be necessary to actively compensate for the Coriolis force, simply relying on passive damping is not useful.

Quite honestly, the most practical solution I can think of is to have a small throttle-able variable thrust chemical rocket motor to compensate for the force. The thrust required will depend on the velocity of the climber.

The effective delta-vee required is not that much.

For an Earth-space elevator a delta-vee of about 6,000 miles per hour or 2.8 km per second is involved. Considerably less than trying to travel the entire journey without using an elevator.

For a lunar space elevator the delta-vee would be about 150 metres per second.

$\endgroup$
0
$\begingroup$

The definition of the Coriolis Effect is that due to Earth's rotation fluids on Earth have extra movement. This only effects fluids such as air or water. These can in turn effect objects such as rockets. Although to fully answer you question more information is needed.

1- What is the force (In Newtons) that this "cable" can withstand? Newtons per second to meters per second.

2- Can it withstand this force at a constant rate?

From here we can use Earth rotation speed of 460 meters per second and through some math, find out if it can withstand this force.

3- How fast does it go up?

3- What is the cargo weight?

With the cargo weight we can find out and speed we can determine if the cargo can go up without being knocked off the cable. Although I don't have this info, you can use what I have to solve your problem.

$\endgroup$

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.