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That feels like a perpetual motion machine, because the Earth coming together in the first place released energy. I'm guessing it would take more energy to get the weight to geostationary orbit than you could get back.
Maybe it would work if you lowered an asteroid down, instead. And then you could mine it on arrival.
Edit: Nope, it maths. I think it's down to angular momentum being kind of separate from the gravity well.
So, first off, I'm definitely not arguing this would be a feasible way to get energy in a practical sense in the real world.
But, it wouldn't be a perpetual motion machine. It'd produce less and less energy as the Earth ran out of angular momentum, ultimately approaching zero.
I don't think I'll do the monster math on this, but my gut tells me one could technically and theoretically (not so much in practice) get more energy out of that than it took to get the weight up there. (It might be that the Moon would limit how much energy could be got out of this scheme as well, but I think even with the Moon involved, I think it could still be a net energy gain.) That said, without running the numbers, you might well be right!
Ah shoot, it looks like you posted a minute after my edit, and probably didn't see it.
Orbital mechanics is big-boy stuff, and gets really subtle the moment you're doing anything non-trivial. This is just two-body, so in theory it should be doable, but the tether pulling out energy as it goes along makes it more complicated. It's a bit much tonight, but maybe I'll give it a shot later. One thing that's clear just from the equation for orbit energies is that there's no limit to how much energy can end up inside the weight itself, as it gets faster in proportion with increased height.
For the calculations, I was thinking maybe one could cheese it a bit and get a relatively decent vague idea of the answer if not a more rigorous idea.
My vague idea was that gravity follows an inverse square law while the centrifigul force equasion is linear relative to the length of the tether. We know that gravity pulls toward Earth and the centrifigul force pulls away. So the net force on the weight at any one time is the centrifigul force equasion (a linear equasion) minus the gravity equasion (an inverse square equasion). We also know that the point at which that sum reaches zero is exactly the altitude of a geostationary orbit.
Work equals force times distance. So suppose we just took the area under the curve of that net force equasion from r equals the radius of the Earth to r equals roughly the furthest we vaguely guess we could send the weight before it starts to get sucked into the Moon's gravity well. And then we divide that by the area under the curve from r equals the Earth's radius to r equals the altitude of a geostationary orbit. That should at least give us a figure like "the amount of energy we could get back in theory would be roughly x times what it takes to get the weight past the geostationary orbit altitude threshold."
The mass of the weight would be a term in that net force equasion, but if we just decided the mass was "one unit", that'd make things a bit simpler. If we only care about the ratio of the energy we get back to the energy we put in, the weight should cancel out anyway.
This approach would certainly ignore a lot of things, but if the answer was "A Large Number(tm)", I think it would still be reasonable to handwave the details. (If the result was like 1.1 or something, probably "no, that doesn't even work in theory" is the much safer bet. Let alone if it was less than 1.)
I guess if we wanted to get even more sophisticated, we could take into account things like the weight and tensile strength of carbon nanotubes and see if it would be infeasible to build a tether sufficiently strong without adding a huge amount of weight during the ascent. But I'd be willing to pretend in this thought experiment that we have some material with infinite tensile strength and zero weight at our disposal.
Anyway! Still not trivial math, quite, and definitely not terribly precise or rigorous, but not quite so "big-boy stuff" as modeling the rotational frames and such.
Yeah, that's probably a better approach than using the energy of the orbits.
Okay, so if we set the weight to 1kg, force is rR~e~^2^ - GM~e~/r^2^, where M~e~ is the mass of the Earth and R~e~ is it's rate of rotation, which is a low number in radians per second. The antiderivative along r is then -1/2r^2^R~e~^2^ - GM~e~/r, but you actually don't need that because you just take the derivative again to find extreme points. rR~e~^2^ - GM~e~/r^2^ can be restated as (r^3^R~e~^2^ - GM~e~)/r^2^, and that numerator is a diverging, increasing function as you move away from 0, which means yes, the energy is unlimited.
Welp, I was wrong. I think the trick here must be that the rotation of the Earth didn't actually come from the gravitational collapse itself, but from the pre-existing inhomogeneities of velocity distributions in the early solar system. Even if you could slingshot the mass around the sun, back into the Earth, and collect it again, you would somehow transfer enough of Earth's angular momentum back to the sun to offset the energy gained.