- #1

RohitRmB

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i mean to ask that if we make a free body diagram of the space elevator then where and in which direction forces will act on it.

and why that forces will act as they are doing

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- Thread starter RohitRmB
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- #1

RohitRmB

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i mean to ask that if we make a free body diagram of the space elevator then where and in which direction forces will act on it.

and why that forces will act as they are doing

- #2

mfb

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Non-rotating: Downward forces from gravity everywhere, towards the center of earth. An additional part is tension in the cable, which is a bit hard to show in a sketch as it is relevant everywhere. It has its largest value at geostationary orbit. Below, the reduction of tension (with lower altitude) gives a net force upwards for every cable element, and above, the reduction of tension (with higher altitude) gives a net force downwards for every cable element.

Every cable element with mass m gets a total force of mωr^2 inwards, keeping in a circular orbit.

Rotating: Downward forces from gravity everywhere, towards the center of earth. Additional, every mass element gets a force of mωr^2 outwards, and tension as described above. The sum of all forces is 0 for all mass elements.

- #3

A.T.

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- #4

grahamgk

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- #5

A.T.

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It's not that simple. The stress at each point is not just a function of the mass below it, but also of the angular velocity:

http://en.wikipedia.org/wiki/Space_elevator#Cable_section

And above the geosynchronous level, there is stress too.

- #6

grahamgk

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- #7

jbriggs444

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If you assume a motionless cable (in the rotating frame) or a uniformly rotating cable (in the inertial frame) then it is sufficient to integrate the force contributed by each mass element from one end of the cable up to the chosen point.

It does not matter which end you integrate from. The magnitude of the calculated force will be identical either way. The signs of the calculated forces will be opposite, of course.

If you were to integrate both "below and above" and then add the results together you'd be sure to get the wrong answer, either by double-dipping (with a sign error) or cancellation (without the sign error).

The wording of a counterweight "just barely pulling the ribbon taut" seems off. The ribbon is not going to go slack without a counterweight. It is taut under centrifugal force. The counterweight is there so that the ribbon's orbit neither falls from nor flies away from a circular orbit.

- #8

grahamgk

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If you were to integrate both "below and above" and then add the results together you'd be sure to get the wrong answer, either by double-dipping (with a sign error) or cancellation (without the sign error).

The wording of a counterweight "just barely pulling the ribbon taut" seems off. The ribbon is not going to go slack without a counterweight. It is taut under centrifugal force. The counterweight is there so that the ribbon's orbit neither falls from nor flies away from a circular orbit.

If you use the point on the ribbon at the geosynchronous level as an example, the net force on the downward side will be toward Earth (gravitational force will be higher than centrifugal force), and on the upward side will be away from Earth (centrifugal force will be higher than gravitational force). Therefore, the forces will be pulling in opposite directions by an equal quantity. You must, of course, get the sign correct when you add them.

The ribbon is only taut from centrifugal force when there is a counterweight or an incredibly long ribbon past the geosynchronous level. Without the counterweight (or the incredibly long ribbon), the net forces for the entire ribbon will be toward Earth and the ribbon will fall to Earth.

- #9

jbriggs444

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If you use the point on the ribbon at the geosynchronous level as an example, the net force on the downward side will be toward Earth (gravitational force will be higher than centrifugal force), and on the upward side will be away from Earth (centrifugal force will be higher than gravitational force). Therefore, the forces will be pulling in opposite directions by an equal quantity. You must, of course, get the sign correct when you add them.

When you add them, the result should be zero. If your goal was to compute the tension in the ribbon then adding upward force to downward force is just plain wrong.

Suppose that you have two teams playing tug-of-war, each pulling on a rope. Team A is pulling with 1000 pounds-force to the left. Team B is pulling with 1000 pounds-force to the right.

Is the tension on the rope:

A. 2000 pounds-force

B. 1000 pounds-force

C. 0 pounds-force

D. None of the above

My answer is B. It seems that your answer is A.

The ribbon is only taut from centrifugal force when there is a counterweight or an incredibly long ribbon past the geosynchronous level. Without the counterweight (or the incredibly long ribbon), the net forces for the entire ribbon will be toward Earth and the ribbon will fall to Earth.

The ribbon is taut regardless. It is rotating and not massless. To a first approximation, the ribbon will have an elliptical orbit such that [portions of] the ribbon may or may not intersect with the surface of the earth. Whether the ribbon falls "to the Earth" will depend on that detail.

- #10

grahamgk

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When you add them, the result should be zero. If your goal was to compute the tension in the ribbon then adding upward force to downward force is just plain wrong.

Suppose that you have two teams playing tug-of-war, each pulling on a rope. Team A is pulling with 1000 pounds-force to the left. Team B is pulling with 1000 pounds-force to the right.

Is the tension on the rope:

A. 2000 pounds-force

B. 1000 pounds-force

C. 0 pounds-force

D. None of the above

My answer is B. It seems that your answer is A.

I agree with your analogy and that the tension in the rope would be 1000 pounds-force, and, therefore, I agree that it is not correct to add the opposing forces at the geosynchronous point for the space elevator ribbon. The tension on the space elevator ribbon at the geosynchronous level (or any point below that point) will be equal to the net force below that point. The closer to the Earth, the lower the tension.

The ribbon is taut regardless. It is rotating and not massless. To a first approximation, the ribbon will have an elliptical orbit such that [portions of] the ribbon may or may not intersect with the surface of the earth. Whether the ribbon falls "to the Earth" will depend on that detail.

When I used the word "taut," I meant from the point the ribbon is connected to the Earth and upward. If the net forces are downward at the geosynchronous level, the entire ribbon will gradually be pulled toward the Earth until it completely falls to the ground, becoming slack at points close to the Earth as it falls. For example, if the counterweight of a space elevator were released from the ribbon, the net forces at the geosynchronous level would change from being net zero or upward to being downward and the ribbon would be pulled toward the Earth.

- #11

jbriggs444

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When I used the word "taut," I meant from the point the ribbon is connected to the Earth and upward.

Ahh, that makes sense. I was automatically assuming that the tension at the bottom end of the elevator would be zero (and even thinking in terms of an unanchored elevator) and you were carefully arranging things so that the tension at the anchor would be zero.

- #12

grahamgk

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An additional complexity in the analysis for a real space elevator ribbon, which is usually incorporated into conceptual designs, is that the ribbon would be tapered. Since the tension decreases going from the geosynchronous level downward, the ribbon does not need to be as wide either. This reduces the mass, and downward gravitational force, for the entire ribbon, and reduces the width (and mass) all along it. This tapering is also applied to the ribbon section between the geosynchronous level and the counterweight.

- #13

mrspeedybob

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An additional complexity in the analysis for a real space elevator ribbon, which is usually incorporated into conceptual designs, is that the ribbon would be tapered. Since the tension decreases going from the geosynchronous level downward, the ribbon does not need to be as wide either. This reduces the mass, and downward gravitational force, for the entire ribbon, and reduces the width (and mass) all along it. This tapering is also applied to the ribbon section between the geosynchronous level and the counterweight.

I would think in a real space elevator the bottom end would be under compression instead of tension. It would be built like a tower supporting its weight from the ground up. There would be a transition section which would be under minimal tension or compression, and an upper section under tension as you were supposing. Building the elevator this way it is supported from both ends, allowing for a stronger structure.

- #14

mfb

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Tension is way easier to handle - you do not need additional support structures everywhere, and materials can handle more tension than compression forces.

- #15

grahamgk

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However, a Lunar Space Elevator could be built using existing materials, because of the moon's lower gravity. LiftPort Group is currently pursuing this. http://liftport.com/

- #16

mikeph

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If you build something to 1km high then the weight of the top floor will be transferred down to the foundations. But if you build it to 36km, the weight of the top floor is entirely compensated for by the centripetal force keeping it in circular motion. All floors above this floor then exhibit a centrifugal force upwards, which, when transferred down through the building, acts to nullify the weight of one of the floors below 36km. So theoretically could we not keep building until we reach a point where the upward centrifugal forces of all the floors above 36km serve to cumulatively nullify the weights of all the floors below 36km?

If we stop at this point, should the tension at the point of contact with the Earth not be zero? And then the only limiting factor is the tensile strength of the cable at some point (maybe 36km?) where it is maximised?

- #17

jbriggs444

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If you build something to 1km high then the weight of the top floor will be transferred down to the foundations. But if you build it to 36km, the weight of the top floor is entirely compensated for by the centripetal force keeping it in circular motion. All floors above this floor then exhibit a centrifugal force upwards, which, when transferred down through the building, acts to nullify the weight of one of the floors below 36km. So theoretically could we not keep building until we reach a point where the upward centrifugal forces of all the floors above 36km serve to cumulatively nullify the weights of all the floors below 36km?

If we stop at this point, should the tension at the point of contact with the Earth not be zero? And then the only limiting factor is the tensile strength of the cable at some point (maybe 36km?) where it is maximised?

Yes, all of this is correct, except that it's 36,000 km, not 36 km.

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