Pens and heavy boots always do the same thing in any gravitational field, unless they modify it somehow, like by moving the moon. Acceleration due to gravity does not depend on the mass of the accelerated object.
If the moon is small and spinning quickly, a space elevator only needs to be a few feet tall. In this admittedly contrived scenario, the boots will anchor the astronaut because they are going around it more slowly. The pen will float because it is actually in orbit.
To land on this moon you would achieve orbit, and then put your feet down.
I don’t think you’d be landing at all, in any meaningful sense. Any moon massive enough to make walking possible at all is going to be large enough that an extra meter or so at the surface will have a negligible difference in gravitational force, so we’re talking about a body spinning so fast that its equatorial rotational velocity is approximately orbital velocity (and probably about 50% of escape velocity). So for most practical purposes, the boots would be in orbit as well, along with most of the moon’s surface.
Of course, since the centrifugal force at the equator due to rotation would almost exactly counteract weight due to gravity, the only way the thing could hold itself together would be tensile strength; it wouldn’t take much for it to slowly tear itself apart.
My rough calculation says that the density would need to be about a million times greater than Earth’s, around 10^10 kg/m^3. This is too low to be a neutron star, but too high to be anything else I think. It may very well be impossible in this universe.
That’s assuming uniform density though. Of course you could just have a microblackhole with a hard 1-meter-diameter shell encasing it. How you keep the shell centered is … trickier.
Similarly, my quick calculation, given an escape velocity high enough to walk and an object 10 meters in diameter, was about 7 * 10^9. That’s roughly the density of electron-degenerate matter; I’m pretty sure nothing will hold together at that density without substantial outside pressure, and since we’re excluding gravitational compression here I don’t think that’s likely.
Keeping a shell positioned would be easy; just put an electric charge on both it and the black hole. Spinning the shell fast enough might be awkward from an engineering standpoint, though.
Keeping a shell positioned would be easy; just put an electric charge on both it and the black hole.
This won’t work for spherical shells and uniformly distributed charge for the same reason that a spherical shell has no net gravitational force on anything inside it. You’ll need active counterbalancing.
Would it be possible to keep the black hole charged (use an electron gun), then manipulate electric fields to keep it centered? I don’t know enough physics to tell.
Pens and heavy boots always do the same thing in any gravitational field, unless they modify it somehow, like by moving the moon. Acceleration due to gravity does not depend on the mass of the accelerated object.
If the moon is small and spinning quickly, a space elevator only needs to be a few feet tall. In this admittedly contrived scenario, the boots will anchor the astronaut because they are going around it more slowly. The pen will float because it is actually in orbit.
To land on this moon you would achieve orbit, and then put your feet down.
I don’t think you’d be landing at all, in any meaningful sense. Any moon massive enough to make walking possible at all is going to be large enough that an extra meter or so at the surface will have a negligible difference in gravitational force, so we’re talking about a body spinning so fast that its equatorial rotational velocity is approximately orbital velocity (and probably about 50% of escape velocity). So for most practical purposes, the boots would be in orbit as well, along with most of the moon’s surface.
Of course, since the centrifugal force at the equator due to rotation would almost exactly counteract weight due to gravity, the only way the thing could hold itself together would be tensile strength; it wouldn’t take much for it to slowly tear itself apart.
Hmm, I suppose it’s too much handwaving to say it’s only a few meters wide and super dense.
My rough calculation says that the density would need to be about a million times greater than Earth’s, around 10^10 kg/m^3. This is too low to be a neutron star, but too high to be anything else I think. It may very well be impossible in this universe.
That’s assuming uniform density though. Of course you could just have a microblackhole with a hard 1-meter-diameter shell encasing it. How you keep the shell centered is … trickier.
Similarly, my quick calculation, given an escape velocity high enough to walk and an object 10 meters in diameter, was about 7 * 10^9. That’s roughly the density of electron-degenerate matter; I’m pretty sure nothing will hold together at that density without substantial outside pressure, and since we’re excluding gravitational compression here I don’t think that’s likely.
Keeping a shell positioned would be easy; just put an electric charge on both it and the black hole. Spinning the shell fast enough might be awkward from an engineering standpoint, though.
This won’t work for spherical shells and uniformly distributed charge for the same reason that a spherical shell has no net gravitational force on anything inside it. You’ll need active counterbalancing.
Ah, true, I didn’t think of that, or rather didn’t think to generalize the gravitational case.
Amusingly, that makes a nice demonstration of the topic of the post, thus bringing us full circle.
Would it be possible to keep the black hole charged (use an electron gun), then manipulate electric fields to keep it centered? I don’t know enough physics to tell.
Yes, this could work.