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) 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.
silly, the Internet doesn't weigh anything! ;-)
That's from 2011. It probably weighs at least one watermelon by now, given Netflix etc.
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.
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
Those are separate from the disconnect I was talking about; I meant a gentle disconnect after which the spacecraft remains in orbit.
The system is not connected to Earth, so it doesn't change the planet's orbit.
Seems like that would negate the lateral problems, but maybe I'm just introducing a twisting one instead...?
A space elevator would reach at least 36,000km to hit 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%20Founta... )
We can get to about 9km using stone by climbing Mount Everest.
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.