7 ms·

I'm curious about space elevator phenomena related to lateral ∆V. At Earth's equator (zero altitude - i.e., on the ground), an object moves spinward at about 0

I'm curious about space elevator phenomena related to lateral ∆V. At Earth's equator (zero altitude - i.e., on the ground), an object moves spinward at about 0.46 km/sec. In geosynchronous orbit, an object has to move at about 3.07 km/sec. A space elevator car ascending from stationary on the equator to a terminal at geo altitude would require a tangential (lateral) ∆V of 2.61 km/sec. This is in addition to the energy needed for ∆ altitude-related potential energy change. Am I thinking about this right? The faster the ascent, the greater the lateral reaction force on the elevator because ∆V/time increases. Has anybody done that model? I didn't find anything in a brief search. The lateral ∆V also suggests it could get windy on ascent because the atmosphere rotates with the earth on a macroscopic basis. Maybe later tonight, I'll calculate lateral wind velocity as F(altitude) and see if I can fold in pressure drop to get wind pressures as F(altitude). I would love to see even a prototype attempt in my lifetime.

WtfRuSerious 8y ago
What about at the poles? Seems like that would negate the lateral problems, but maybe I'm just introducing a twisting one instead...?
- zamadatix 8y ago
  Then you have an enormously large tower held up by nothing but its own strength instead of being naturally pulled into space. Also takes additional delta-v to get satteliges into common orbits post release.
  - angel_j 8y ago
    This is arguably more doable than cables. I've been told that it's completely possible to build steel structures several miles tall (and the main reason we don't is b/c elevators become a problem).
    - chriswarbo 8y ago
      We can build massive steel towers (e.g. https://en.wikipedia.org/wiki/X-Seed_4000 https://en.wikipedia.org/wiki/X-Seed_4000 would be 4km tall), but they would still be nowhere near space (one common definition of reaching space is https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line which is at 100km). A space elevator would reach at least 36,000km to hit geostationary orbit ( https://en.wikipedia.org/wiki/Geostationary_orbit https://en.wikipedia.org/wiki/Geostationary_orbit ). Most designs are even longer, to provide a counterweight. No known material could support a tower that tall (towers need compressive strength; elevators need tensile strength). It would need to rely on 'active support' or 'dynamic structures', like a space fountain ( http://www.orbitalvector.com/Orbital%20Travel/Space%20Fountains/Space%20Fountains.htm http://www.orbitalvector.com/Orbital%20Travel/Space%20Founta... )
      - asib 8y ago
        A friend of mine who studies engineering told me that masonry has infinite compressive strength meaning, in theory, it's possible to build monumental structures out of stone.
        chriswarbo 8y ago
        The base needs to be large enough to support the weight, so you get less of a tower and more of a mound (like the great pyramids). We can get to about 9km using stone by climbing Mount Everest.
- ridgeguy 8y ago
  I think it would be the same ∆V requirement (for geosynchronous orbital altitude). But it's not possible to have a geo orbital plane (so the subsatellite point is stationary on the Earth) that is out of the equatorial plane.
larkeith 8y ago
(In case you were unaware, this is the Coriolis force)
- mirimir 8y ago
  But this is the vertical component of the Coriolis force, which is usually negligible. Yes?
mirimir 8y ago
What if the cable isn't radial? Edit: And bowed, basically in equilibrium with lateral forces.
deleted 8y ago
[deleted]
toxik 8y ago
I’m not sure but I think since angular momentum is conserved, and angular velocity remains constant at all distances to Earth’s center here, then there will not be any lateral forces due to Earth’s rotation about its own axis.
- codethief 8y ago
  While total angular momentum is certainly conserved in any rotationally invariant system (read: Earth + space elevator + objects traveling upwards + atmosphere), the situation only for an object traveling to space in the elevator is different: Angular momentum is given by L = I * ω where I is the moment of inertia and ω is the angular velocity. Let's assume that we put the internet (https://m.youtube.com/watch?v=iDbyYGrswtg https://m.youtube.com/watch?v=iDbyYGrswtg) into the elevator and ship it off to space. Let's also say that the internet has a mass m and is of negligible diameter compared to Earth's radius. Clearly, its angular velocity ω will stay constant, assuming that the space elevator was built perfectly straight and perpendicular to Earth. However, the internet's moment of inertia I = m r^2 will not stay the same but increase quadratically with its radial position r and its angular moment will thus have to increase, as well. If you take the initial radius to be Earth's radius (~6,300km) and the final radius to be the geostationary orbit (~36,000km), the quadratic dependence will give you an enormous difference in angular momentum which will have to be accounted for by some torque that we apply to the internet. It doesn't matter where this torque comes from--for instance we could imagine the internet to have some kind of jet propulsion engine, where the emitted gas's angular momentum would exactly match the internet's change in angular momentum, so that total angular momentum is conserved. Moreover, it could actually be a small torque, provided that we move the internet at a small velocity in the perpendicular direction. (Recall that the change in angular momentum is given by the integral of the torque over time, so the smaller we want the torque to be, the slower we have to move the internet and the longer it will take the internet to reach its final position.) Now what happens if we don't provide said torque? Recall that I assumed earlier that the angular velocity ω of Earth and the elevator (and thus the internet) be constant. If we don't provide the mentioned torque, moving the internet upward will increase the moment of inertia of the system Earth + space elevator + internet due to more mass being located farther away from the axis of rotation. By conservation of angular momentum L = I ω, the entire system's angular velocity then has to decrease, similar to a figure skater stretching out her arms while rotating in order to slow down. All this assumes, however, that the space elevator's cable is completely rigid in the lateral direction and doesn't move under the effective torque it will feel as the internet moves upward. In practice, however, this will not be the case unless further measures are taken which counter the effective torque on the cable and keep the cable perfectly perpendicular to Earth. (Only then Earth's rotation will actually slow down under the backreaction.) This seems somewhat difficult to achieve, though, and I think the better way would indeed be to equip the internet with a jet propulsion engine which provides the necessary angular momentum. [EDIT] Made the part about backreaction on Earth and stability of the space elevator a bit clearer.
jimmcslim 8y ago
Wouldn’t payloads travelling up the space elevator and departing for the rest of the solar system and beyond ultimately deprive the earth of angular momentum and send it hurtling towards the Sun?
- leereeves 8y ago
  Not hurtling toward the sun, but payloads traveling up the space elevator would slow the Earth's rotation (the 24 hour daily rotation about the Earth's axis) in the same way an ice skater spins slower when they spread their arms. When the payload detaches from the elevator, the angular momentum of the Earth about the sun does decrease (some of the AM goes with the payload) but the mass of the Earth also decreases, and the angular velocity of the Earth about the Sun doesn't change.
  - schiffern 8y ago
    >the angular velocity of the Earth about the Sun doesn't change. It does if your spaceship escapes from Earth in a prograde[1] direction (equivalent to Earth throwing a ball forward, causing a retrograde reaction upon the Earth). This type of escape trajectory is necessary to reach any destination farther from the Sun (Mars, asteroids, etc). It's a tiny change obviously, but it's non-zero. Note that returning from Mars/asteroids and aerobraking should add angular momentum, speeding up the Earth and at least partly counteracting this effect. [1] https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion
    - leereeves 8y ago
      You're right, the portion of a rocket's exhaust that is recaptured by the Earth would raise or lower the angular velocity a bit, depending on direction, and gravity between the Earth and the spacecraft would have a tiny effect in the opposite direction. Those are separate from the disconnect I was talking about; I meant a gentle disconnect after which the spacecraft remains in orbit.
- marcosdumay 8y ago

WtfRuSerious 8y ago

What about at the poles? Seems like that would negate the lateral problems, but maybe I'm just introducing a twisting one instead...?

zamadatix 8y ago
Then you have an enormously large tower held up by nothing but its own strength instead of being naturally pulled into space. Also takes additional delta-v to get satteliges into common orbits post release.
- angel_j 8y ago
  This is arguably more doable than cables. I've been told that it's completely possible to build steel structures several miles tall (and the main reason we don't is b/c elevators become a problem).
  - chriswarbo 8y ago
    We can build massive steel towers (e.g. https://en.wikipedia.org/wiki/X-Seed_4000 https://en.wikipedia.org/wiki/X-Seed_4000 would be 4km tall), but they would still be nowhere near space (one common definition of reaching space is https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line which is at 100km). A space elevator would reach at least 36,000km to hit geostationary orbit ( https://en.wikipedia.org/wiki/Geostationary_orbit https://en.wikipedia.org/wiki/Geostationary_orbit ). Most designs are even longer, to provide a counterweight. No known material could support a tower that tall (towers need compressive strength; elevators need tensile strength). It would need to rely on 'active support' or 'dynamic structures', like a space fountain ( http://www.orbitalvector.com/Orbital%20Travel/Space%20Fountains/Space%20Fountains.htm http://www.orbitalvector.com/Orbital%20Travel/Space%20Founta... )
    - asib 8y ago
      A friend of mine who studies engineering told me that masonry has infinite compressive strength meaning, in theory, it's possible to build monumental structures out of stone.
      - chriswarbo 8y ago
        The base needs to be large enough to support the weight, so you get less of a tower and more of a mound (like the great pyramids). We can get to about 9km using stone by climbing Mount Everest.
ridgeguy 8y ago
I think it would be the same ∆V requirement (for geosynchronous orbital altitude). But it's not possible to have a geo orbital plane (so the subsatellite point is stationary on the Earth) that is out of the equatorial plane.

larkeith 8y ago

(In case you were unaware, this is the Coriolis force)

mirimir 8y ago
But this is the vertical component of the Coriolis force, which is usually negligible. Yes?

mirimir 8y ago

What if the cable isn't radial? Edit: And bowed, basically in equilibrium with lateral forces.

deleted 8y ago

[deleted]

toxik 8y ago

I’m not sure but I think since angular momentum is conserved, and angular velocity remains constant at all distances to Earth’s center here, then there will not be any lateral forces due to Earth’s rotation about its own axis.

codethief 8y ago
While total angular momentum is certainly conserved in any rotationally invariant system (read: Earth + space elevator + objects traveling upwards + atmosphere), the situation only for an object traveling to space in the elevator is different: Angular momentum is given by L = I * ω where I is the moment of inertia and ω is the angular velocity. Let's assume that we put the internet (https://m.youtube.com/watch?v=iDbyYGrswtg https://m.youtube.com/watch?v=iDbyYGrswtg) into the elevator and ship it off to space. Let's also say that the internet has a mass m and is of negligible diameter compared to Earth's radius. Clearly, its angular velocity ω will stay constant, assuming that the space elevator was built perfectly straight and perpendicular to Earth. However, the internet's moment of inertia I = m r^2 will not stay the same but increase quadratically with its radial position r and its angular moment will thus have to increase, as well. If you take the initial radius to be Earth's radius (~6,300km) and the final radius to be the geostationary orbit (~36,000km), the quadratic dependence will give you an enormous difference in angular momentum which will have to be accounted for by some torque that we apply to the internet. It doesn't matter where this torque comes from--for instance we could imagine the internet to have some kind of jet propulsion engine, where the emitted gas's angular momentum would exactly match the internet's change in angular momentum, so that total angular momentum is conserved. Moreover, it could actually be a small torque, provided that we move the internet at a small velocity in the perpendicular direction. (Recall that the change in angular momentum is given by the integral of the torque over time, so the smaller we want the torque to be, the slower we have to move the internet and the longer it will take the internet to reach its final position.) Now what happens if we don't provide said torque? Recall that I assumed earlier that the angular velocity ω of Earth and the elevator (and thus the internet) be constant. If we don't provide the mentioned torque, moving the internet upward will increase the moment of inertia of the system Earth + space elevator + internet due to more mass being located farther away from the axis of rotation. By conservation of angular momentum L = I ω, the entire system's angular velocity then has to decrease, similar to a figure skater stretching out her arms while rotating in order to slow down. All this assumes, however, that the space elevator's cable is completely rigid in the lateral direction and doesn't move under the effective torque it will feel as the internet moves upward. In practice, however, this will not be the case unless further measures are taken which counter the effective torque on the cable and keep the cable perfectly perpendicular to Earth. (Only then Earth's rotation will actually slow down under the backreaction.) This seems somewhat difficult to achieve, though, and I think the better way would indeed be to equip the internet with a jet propulsion engine which provides the necessary angular momentum. [EDIT] Made the part about backreaction on Earth and stability of the space elevator a bit clearer.

jimmcslim 8y ago

Wouldn’t payloads travelling up the space elevator and departing for the rest of the solar system and beyond ultimately deprive the earth of angular momentum and send it hurtling towards the Sun?

leereeves 8y ago
Not hurtling toward the sun, but payloads traveling up the space elevator would slow the Earth's rotation (the 24 hour daily rotation about the Earth's axis) in the same way an ice skater spins slower when they spread their arms. When the payload detaches from the elevator, the angular momentum of the Earth about the sun does decrease (some of the AM goes with the payload) but the mass of the Earth also decreases, and the angular velocity of the Earth about the Sun doesn't change.
- schiffern 8y ago
  >the angular velocity of the Earth about the Sun doesn't change. It does if your spaceship escapes from Earth in a prograde[1] direction (equivalent to Earth throwing a ball forward, causing a retrograde reaction upon the Earth). This type of escape trajectory is necessary to reach any destination farther from the Sun (Mars, asteroids, etc). It's a tiny change obviously, but it's non-zero. Note that returning from Mars/asteroids and aerobraking should add angular momentum, speeding up the Earth and at least partly counteracting this effect. [1] https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion https://en.wikipedia.org/wiki/Retrograde_and_prograde_motion
  - leereeves 8y ago
    You're right, the portion of a rocket's exhaust that is recaptured by the Earth would raise or lower the angular velocity a bit, depending on direction, and gravity between the Earth and the spacecraft would have a tiny effect in the opposite direction. Those are separate from the disconnect I was talking about; I meant a gentle disconnect after which the spacecraft remains in orbit.
marcosdumay 8y ago