Solving the Sky Hook (Beanstalk) Equation - 65 Characters

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In summary: Read moreIn summary, the conversation discusses a problem involving a cable of constant mass per unit length hovering over a point on the Equator. The question is how long the cable must be. The conversation includes discussions on the gravitational force, centrifugal acceleration, stress (tension), and the assumptions made in solving the problem. Ultimately, the tension at the ends of the cable is found to be zero due to the lack of external forces.
  • #1
Feynmanfan
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Dear friends,

<<A cable of constant mass per unit length (p) points straight up hovering over a point on the Equator. How long has to be the cable?>>

I believe there's a mistake in my answer and I need somebody to give me a hint. What I've done is: integrate dF(gravitational force) over the whole string of length L. So the net force equals the centrifugal acceleration times mass, in the centre of mass. Doesn't it?

so i get mw^2(R+L/2)=GMm/(R(R+L)) where R is the Earth radius and w it's angular velocity.

A friend of mine has told me that I'm missing the stress (tension) in this answer. Thanks for your help.
 
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  • #2
Feynmanfan said:
Dear friends,

<<A cable of constant mass per unit length (p) points straight up hovering over a point on the Equator. How long has to be the cable?>>

I believe there's a mistake in my answer and I need somebody to give me a hint. What I've done is: integrate dF(gravitational force) over the whole string of length L. So the net force equals the centrifugal acceleration times mass, in the centre of mass. Doesn't it?

so i get mw^2(R+L/2)=GMm/(R(R+L)) where R is the Earth radius and w it's angular velocity.

A friend of mine has told me that I'm missing the stress (tension) in this answer. Thanks for your help.

There is tension that holds the string together, but the tension is internal force and when you integrate the length of the string the tension integrates out. You can break the problem down to look at the force acting on an infinitesimal length of the string, including the tension gradient for each infinitesimal length, then integrate over the length

[tex]
\int_R^{R + L} {\frac{{GM\rho dr}}{{r^2 }}} - \int_R^{R + L} {\frac{{dT(r)dr}}{{dr}}} = \int_R^{R + L} \rho \omega ^2 rdr} }
[/tex]

The second inegral results in the tension difference between the two ends, which is zero since both ends have zero tension. That leaves the first and last integrals equal and gives your result, to be solved for L
 
  • #3
Feynmanfan said:
Dear friends,

<<A cable of constant mass per unit length (p) points straight up hovering over a point on the Equator. How long has to be the cable?>>

I believe there's a mistake in my answer and I need somebody to give me a hint. What I've done is: integrate dF(gravitational force) over the whole string of length L. So the net force equals the centrifugal acceleration times mass, in the centre of mass. Doesn't it?

so i get mw^2(R+L/2)=GMm/(R(R+L)) where R is the Earth radius and w it's angular velocity.

A friend of mine has told me that I'm missing the stress (tension) in this answer. Thanks for your help.
I am having a little difficulty understanding the problem. What makes the cable stand straight up? What is it connected to? What is holding the fixed end and where is the fixed end?

AM
 
  • #4
This is the infamous "sky hook". Look at the cable from the frame of the rotating earth. A segment of cable at the bottom has greater weight than centrifugal force, but at the top the centrifugal force is greater than the weight. The idea is that if the cable is long (and strong) enough, the centrifugal force will be enough to support the entire weight of the cable.

So... it's not connect to anything and nothing external is holding it up!

Given those assumptions, I'd say that Feynmanfan got it right. As OlderDan points out, the tension is just an internal force.

Interesting problem.
 
  • #5
Thanks a lot for your kind answer. Once that I've determined the length of the cable, I'll try to find a material strong enough to withstand the maximum tensile stress.

Let's see if I get it right.
 
  • #6
Good Luck! You might want to write Arthur Clarke for advice.
 
  • #7
Sorry to bother you again with this thread, but I don't know how to justify that the tension is zero on the ends. Could you give me a good physical reason?
 
  • #8
To have tension in a segment of rope requires a force be exerted at each end of that segment. But at the free end of a rope, there's nothing to exert a force.

Another way to look at it: Imagine an infinitesimal slice of rope at the very end. If there were anything other than an infinitesimally small tension pulling the rope slice, it would accelerate. Thus the tension must be zero at the ends.
 
  • #9
Doc Al said:
To have tension in a segment of rope requires a force be exerted at each end of that segment. But at the free end of a rope, there's nothing to exert a force.

Another way to look at it: Imagine an infinitesimal slice of rope at the very end. If there were anything other than an infinitesimally small tension pulling the rope slice, it would accelerate. Thus the tension must be zero at the ends.
Is the end not accelerating? Will there not be differential gravitational (tidal) forces that will tend to stretch the cable?

AM
 
  • #10
Andrew Mason said:
Is the end not accelerating? Will there not be differential gravitational (tidal) forces that will tend to stretch the cable?
I don't think I made my point very clearly. Viewed from the rotating frame of the earth, there is no acceleration. For a tiny slice at the end of the cable, the forces (weight and centrifugal) tend to zero. If the tension at the ends did not also go to zero, there would be a problem.

Yes, the forces will tend to stretch the cable, so there will be tension throughout the cable, but the tension at the ends will be zero. (Correct me if I'm wrong.)
 
  • #11
Doc Al said:
I don't think I made my point very clearly. Viewed from the rotating frame of the earth, there is no acceleration. For a tiny slice at the end of the cable, the forces (weight and centrifugal) tend to zero. If the tension at the ends did not also go to zero, there would be a problem.

Yes, the forces will tend to stretch the cable, so there will be tension throughout the cable, but the tension at the ends will be zero. (Correct me if I'm wrong.)

You're not wrong about the tension. There can be no tension at the free end of a rope. Newton's third law tells us that. There has to be a reaction force for there to be a force. There could be other forces like the electric force on a charged rope with an excess charge at the surface in an electric field or some such thing.

At the upper end, the weight and the centrifugal force do tend to zero, but their ratio would have to converge to a finite value. For that last bit of mass dm there has to be a tension on the lower side that exceeds the weight by the necessary centripetal force to keep that bit of mass moving in a circle. That does not change the fact that on the other side of dm there is nothing to pull it, so the tension goes to zero.

I didn't try to calculate it, but I expect the tension in the middle region of this thing to be enormous.
 

Related to Solving the Sky Hook (Beanstalk) Equation - 65 Characters

1. What is the Sky Hook (Beanstalk) Equation?

The Sky Hook (Beanstalk) Equation is a mathematical equation that calculates the optimal length and strength of a space elevator, also known as a sky hook or beanstalk, which is a proposed method of transportation between Earth and space using a cable anchored to the Earth's surface and a counterweight in outer space.

2. What is the purpose of solving the Sky Hook (Beanstalk) Equation?

The purpose of solving the Sky Hook (Beanstalk) Equation is to determine the necessary parameters for building a functional and efficient space elevator, which could greatly reduce the cost and energy required for space travel and transportation of goods and materials.

3. How is the Sky Hook (Beanstalk) Equation solved?

The Sky Hook (Beanstalk) Equation is a complex mathematical equation that takes into account various factors such as the Earth's gravity, the cable's material properties, and the counterweight's position and mass. It is typically solved using computer simulations and numerical methods.

4. What are the potential challenges in solving the Sky Hook (Beanstalk) Equation?

Solving the Sky Hook (Beanstalk) Equation can be challenging due to the complexity of the equation and the need to consider multiple variables. Additionally, there may be uncertainties in the input parameters and limitations in our current understanding of materials and their properties.

5. What are the implications of solving the Sky Hook (Beanstalk) Equation?

If the Sky Hook (Beanstalk) Equation is successfully solved, it could lead to the development and implementation of space elevators, which could revolutionize space travel and open up new opportunities for space exploration and commerce. It could also have significant impacts on the environment and economy due to the reduced need for rocket launches and the potential for more sustainable space transportation methods.

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