= 27d567bc2d48d9f40e9ccc20ea370b15)
RolfAndreassen comments on Questions for shminux - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (32)
For a cylinder of zero thickness, you are correct. The trick is to have nonzero thickness. This doesn't (to first order) affect the gravitational attraction at the endpoints, but it limits how close you can get to the axis. Even if you allow a particle to enter the cylinder, it will then no longer feel the attraction from the outer parts. It therefore seems to me that you can choose a density and radius such that there is never an event horizon as you go perpendicularly in, but a length such that there is one at the endpoints.
I have no answer for the argument that the escape velocity is independent of direction; but is it possible that this result was derived for a sphere? At any rate it may be that, if I got around to playing around with numbers, I would find that there was no solution as I described above.
Imagine a long cylinder cut into segments. Instead of moving along the line, you merely have to take the segment at one end and move it over to the other end. Because of the superposition principle, the change in potential is merely the change cause by moving the segment from one end to the other.
Now, when does moving the segment increase the potential, and when does moving the segment decrease the potential? Can we identify a maximum?
That's a good approach. If the segment is closer to the point of interest after it is moved, the potential well is deeper. Which tells you that it is deepest at the center.
I sit corrected; my argument that as you enter the cylinder you can ignore the outer parts breaks down.
This is quite general for isolated potential systems. Total energy of a particle: E=KE+PE. E=0 at infinity. KE=1/2mv^2, PE=mP where P is the gravitational potential (volume integral of -G*rho/r dV). Escape velocity corresponds to E=0, so we get v=sqrt(-2P), regardless of anything else.