Calculating the Length of a Space Elevator Rope

[Fibo112]

Homework Statement

The following task causes me problems:

The science fiction writer R.A. Heinlein describes in the novel "Friday" a satellite ("space elevator"), which consists of a long rope, placed directly above the equator. The rope is aligned along the Earth's radius. It moves so that it appears to the Earth observer that it is suspended at a fixed point above the equator. The lower end of the rope hangs just above the Earth's surface. Suppose that the linear mass density of the rope ρl is constant. The radius of the Earth is R 6370 km and the Earth mass ME = 5.97 x 1024 kg.

a) How long should such a rope be?

Homework Equations

Fg=Gm1m2/r^2, Fcentripedal=mv^2/r

The Attempt at a Solution

My only idea would be to set the Zentripedal force equal to the gravitational force. Then I would have a length of 144'000km. Would the rope actually behave as described? But then the tension in the rope would have to be exactly right at every point. What do you think? How would you solve the task?

 

[.Scott]

Don't worry about the tension. An implicit part of the problem statement is that the rope will not break.
The part of the rope within the geosynchronous orbit radius will be pulling down.
The part above that radius will be pulling up.

All you have to do is make the total pull zero.
So integrate both the G-pull and the centrifugal force from 6370km to your answer - and make sure it is zero.

 

[Fibo112]

Don't worry about the tension. An implicit part of the problem statement is that the rope will not break.
The part of the rope within the geosynchronous orbit radius will be pulling down.
The part above that radius will be pulling up.

All you have to do is make the total pull zero.
So integrate both the G-pull and the centrifugal force from 6370km to your answer - and make sure it is zero.

Thanks for the reply. If I didnt make a mistake for 144000km the centrifugal force is equal to the gravitational force. What I am not sure about is how stable this is. Will the rope remain straight?

 

[.Scott]

Thanks for the reply. If I didnt make a mistake for 144000km the centrifugal force is equal to the gravitational force. What I am not sure about is how stable this is. Will the rope remain straight?

There may be problems there. I wouldn't get too involved in the practicalities, but yes, it should stay straight and if it begins moving up or down, it would just oscillate over the course of its orbit.

 

[Fibo112]

There may be problems there. I wouldn't get too involved in the practicalities, but yes, it should stay straight and it begins moving up or down, it would just oscillate over the course of its orbit.

why will it begin moving up or down?

 

[.Scott]

why will it begin moving up or down?

Ignoring effects from the atmosphere and the moon (and other such things), no. In reality, yes. If one of these was ever built, you would probably tie it to the ground with a shock absorber to dampen anything that started it moving.

 

[jbriggs444]

why will it begin moving up or down?

An anchored "beanstalk" will not move up and down. Do not worry about that part. We're not concerned with practicalities like perterbations and resonances.

The problem is that if the top end of the stalk is at geosynchronous orbit (which is not at 144,000 km). the whole stalk will fall down. Every point on the rope except for the top would be pulled to the Earth more strongly by gravity than centrifugal force can counter-balance.

 

[Fibo112]

An anchored "beanstalk" will not move up and down. Do not worry about that part. We're not concerned with practicalities like perterbations and resonances.

The problem is that if the top end of the stalk is at geosynchronous orbit (which is not at 144,000 km). the whole stalk will fall down. Every point on the rope except for the top would be pulled to the Earth more strongly by gravity than centrifugal force can counter-balance.

yes, 144000km is the length so that the centrifugal force and the gravitational force will be equal.

 

[litup]

The thing bothering me is with a 'rope' 144Km long, the orbital velocity will be a variable all the way up. So how does the rope maintain straightness from bottom to top, outside of small pertubutions?

 

[jbriggs444]

yes, 144000km is the length so that the centrifugal force and the gravitational force will be equal.

Can you show your work? 144,000 km is not correct. [In addition to not answering the stated problem]

 

[jbriggs444]

The thing bothering me is with a 'rope' 144Km long, the orbital velocity will be a variable all the way up. So how does the rope maintain straightness from bottom to top, outside of small pertubutions?

The individual pieces of rope are not in self-stable orbits. They are held in place by the rest of the rope.

Imagine swinging a rock around your head at the end of a rope. What holds the pieces of rope in place? Now remove the rock.

 

[Fibo112]

I posted an extension to this question here https://www.physicsforums.com/threads/motion-predicted-by-Newtons-laws.933794/.

 

[Fibo112]

I have now realized that the motion as described in the problem will not work even if we assume the string to be rigid since the center of mass will not move as it must. Does this question even make sense?

 

[jbriggs444]

I have now realized that the motion as described in the problem will not work even if we assume the string to be rigid since the center of mass will not move as it must. Does this question even make sense?

Sure it does. Why would the center of mass not move as it must? Can you provide your reasoning?

As long as a rope is thick enough that air resistance is negligible, I can swing one around my head successfully. All pieces move in circular paths, so it is clear that the center of mass does so as well (though the center of mass is irrelevant).

 

[Fibo112]

The center of mass will move as if all the mass and force of the body is concentrated there right? So I calculated the total force and the centripedal force needed for a particle at the position of the center of mass and they weren't equal..

 

[jbriggs444]

The center of mass will move as if all the mass and force of the body is concentrated there right?

For a rigid object, the center of mass will move based on the sum of all forces and torques on all portions of the body.

Add up all the forces, divide by the mass and you know how the center of mass must move.
Add up all the torques, divide by the moment of inertia and you know how the body must rotate.

But you cannot determine the forces and torques by only looking at the position of the center of mass.

Gravity is inverse square and affects different portions of an extended body with different intensities depending on their distance from the center of the gravitating body.

Centrifugal force is proportional to radius and affects different portions of an extended body with different intensities depending on their distance from the center of the gravitating body.

[If the rigid object is not vertical, you have to consider direction as well as intensity]

 

[jbriggs444]

So I calculated the total force and the centripedal force needed for a particle at the position of the center of mass and they weren't equal..

Yes! That is true. They will not be equal. But that does not prevent you from solving the problem. You have to solve for the rod/rope length that yields a total force that keeps the center of mass in circular motion at the prescribed geosynchronous rate.

There is no requirement that the rate must match the rate at which the center of mass would orbit on its own if the mass of the rod/rope were concentrated there.

 

[Fibo112]

but if the ropes motion were as the problem described it the center of mass would undergo uniform circular motion. However since the quantities I stated above are not equal the center of mass cannot undergo uniform circular motion...right?

 

[jbriggs444]

The center of mass can and does perform uniform circular motion. It's just that it's not a circle at the same altitude where a compact mass would be in geosynchronous orbit.