Space elevator

Classical Mechanics Level 1

NASA is planning to stretch a long cable from the equator into space, so that persons and loads can be transported to the orbiting height of the space station, and that the cable would always remain above the same point on Earth. Now, a short rope would fall under its own weight to Earth, but once its length exceeds a certain limit, the rope will start to support itself due to the centrifugal force.

What minimum length must such a rope have, approximately, so that it is just carrying itself?

Details and Assumptions:

The cable is not elastic, has a uniform density, and is tightly stretched without any curvature.
Set up the force balance for incremental rope pieces and integrate them over the length of the rope.
The circumference of Earth is about $2 \pi R \approx 40,000 \,\text{km}.$
Acceleration due to gravity g = $10 \text{ m/s}^2.$

Bonus question: What is the maximum tensile stress along the rope? Is there any known material that can resist these forces?

145,000 km 6,400 km 36,000 km 384,000 km

2 solutions

Markus Michelmann
Sep 20, 2017

Relevant wiki: Newton's Law of Gravity - Problem Solving

An infinitesimal rope piece of length $dr$ has a mass $dm = \rho A dr$ with the cross sectional area $A$ and density $\rho$ . The total force on this rope piece results to $dF = dF_\text{g} + dF_\text{cf} = -GM \frac{dm}{r^2} + \omega^2 r dm = \rho A \left(-GM \frac{1}{r^2} + \omega^2 r \right) dr$ Caculating the integral over the whole rope provides $F_\text{tot} = \rho A \int_R^{r_0} \left(-GM \frac{1}{r^2} + \omega^2 r \right) dr = \rho A \left(-GM \left[ \frac{1}{R} - \frac{1}{r_0} \right] + \frac{1}{2} \omega^2 \left[ r_0^2 - R^2 \right] \right)$ with the radial position $r_0$ of the space station. The force balance implies $F_\text{tot} = 0$ , so that $\begin{aligned} \frac{2 GM}{\omega^2} \left[ \frac{1}{R} - \frac{1}{r_0} \right] &= (r_0^2 - R^2) \\ \frac{2 GM}{\omega^2} \frac{r_0 - R}{R r_0} &= (r_0 - R) (r_0 + R) \\ \frac{2 GM}{\omega^2} &= (r_0 + R) R r_0 \\ 0 &= r_0^2 + R r_0 - \frac{2 GM}{\omega^2 R} \\ r_0 &= - \frac{R}{2} + \sqrt{\frac{R^2}{4} + \frac{2 GM}{\omega^2 R} } \quad \text{(*)} \end{aligned}$ With the earth radius $R \approx 6366 \,\text{km}$ we can caculate the gravitational parameter of the earth $\frac{GM}{R^2} = g \quad \Rightarrow \quad GM = g R^2 \approx 4.05 \cdot 10^{14} \,\frac{\text{m}^3}{\text{s}^2}$ The angular velocity results $\omega = \frac{2\pi}{1 \, \text{day}} = \frac{2 \pi}{24 \cdot 3600 \,\text{s}} = 7.27 \cdot 10^{-5} \,\text{Hz}$ If we plug these values into equation (*), we obtain $r_0 \approx 152,000 \,\text{km} \quad \text{or} \quad l = r_0 - R \approx 145,000 \,\text{km}$ for the length of the rope.

This is one of those problems that can be answered without doing the math.

36,000 km is approximately geosynchronous orbit. A rope that long or less could not be held up by centrifugal force. 384,000 km is approximately lunar orbit. We know that the math of skyhooks have been worked out, and they didn't involve worrying about the Moon. Therefore 145,000 km is the only viable answer.

Jerry Barrington - 3 years, 7 months ago

Laurent Shorts
Oct 8, 2017

Bonus question

The maximum tensile stress occurs where the centrifugal force equals the weight, that is at geostationary orbit at about 36'000 km (radius is $r_g=4.24·10^{7}$ ) altitude. As the rope just carries itself, the force pulling down equals the force pulling up, so that the stress is twice the weight of the rope downward.

$F = 2 \rho A \left(-GM \left[ \frac{1}{R} - \frac{1}{r_g} \right] + \frac{1}{2} \omega^2 \left[ r_g^2 - R^2 \right] \right) = \rho A · 2.8·10^7$

Now, the specific strength is “a material's strength (force per unit area at failure) divided by its density”, that is $\dfrac{F/A}{\rho}=2.8·10^7$ . On the linked page, we can see the best specific strength is lower than 60'000…

Space elevator

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