Ask Science
Ask a science question, get a science answer.
Community Rules
Rule 1: Be respectful and inclusive.
Treat others with respect, and maintain a positive atmosphere.
Rule 2: No harassment, hate speech, bigotry, or trolling.
Avoid any form of harassment, hate speech, bigotry, or offensive behavior.
Rule 3: Engage in constructive discussions.
Contribute to meaningful and constructive discussions that enhance scientific understanding.
Rule 4: No AI-generated answers.
Strictly prohibit the use of AI-generated answers. Providing answers generated by AI systems is not allowed and may result in a ban.
Rule 5: Follow guidelines and moderators' instructions.
Adhere to community guidelines and comply with instructions given by moderators.
Rule 6: Use appropriate language and tone.
Communicate using suitable language and maintain a professional and respectful tone.
Rule 7: Report violations.
Report any violations of the community rules to the moderators for appropriate action.
Rule 8: Foster a continuous learning environment.
Encourage a continuous learning environment where members can share knowledge and engage in scientific discussions.
Rule 9: Source required for answers.
Provide credible sources for answers. Failure to include a source may result in the removal of the answer to ensure information reliability.
By adhering to these rules, we create a welcoming and informative environment where science-related questions receive accurate and credible answers. Thank you for your cooperation in making the Ask Science community a valuable resource for scientific knowledge.
We retain the discretion to modify the rules as we deem necessary.
view the rest of the comments
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.