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> On spherical earth the horizon on average is a little over three miles away. I read (long time ago) that the horizon is 29km on a seashore. In other words if you see a ship disappear on a horizon, it was 29km away. And so his 3mi vs my 29km is a bit of a discrepancy. Can anyone set things straight here?
Wikipedia says 3.1 miles.
> On spherical earth the horizon on average is a little over three miles away. I read (long time ago) that the horizon is 29km on a seashore. In other words if you see a ship disappear on a horizon, it was 29km away. And so his 3mi vs my 29km is a bit of a discrepancy. Can anyone set things straight here?
Earlier quoted context omitted.
Interesting point, but I'm not so sure. We take the shortcut of assuming that the earth's gravity is drawing everything toward a point at the center of the earth. But that's not really correct; every bit of matter is drawing everything else toward it. So those giant mountains that comprise the vertexes of the cube would be pulling the oceans toward them. That works against your point. I don't know where the equilibri…
A sphere with some gravitational influence from the points, but with most of the mass towards the center of the cube, it would still be mostly sphere-like, I think.
Of course, this doesn't begin to account for the vectors, and it assumes that the density is uniform. But it should give us a sense of how much of the gravitational pull differs from our simplistic model.
I imagine walking on a cube-shaped planet will be not different that doing it on a sphere shaped one.
Earlier quoted context omitted.
A sphere with some gravitational influence from the points, but with most of the mass towards the center of the cube, it would still be mostly sphere-like, I think.
For a cube, what portion of its volume lies within an inscribed sphere? In other words, given a cube of edge=X and a sphere of diameter=X, what's the difference in volume? Of course, this doesn't begin to account for the vectors, and it assumes that the density is uniform. But it should give us a sense of how much of the gravitational pull differs from our simplistic model.
Earlier quoted context omitted.
A sphere with some gravitational influence from the points, but with most of the mass towards the center of the cube, it would still be mostly sphere-like, I think.
For a cube, what portion of its volume lies within an inscribed sphere? In other words, given a cube of edge=X and a sphere of diameter=X, what's the difference in volume? Of course, this doesn't begin to account for the vectors, and it assumes that the density is uniform. But it should give us a sense of how much of the gravitational pull differs from our simplistic model.
I imagine walking on a cube-shaped planet will be not different that doing it on a sphere shaped one.
Why not? The most important thing about a round planet in this case (and, in fact, the reason planets tend towards roundness) is that it means gravity is (roughly) equal at every point on the surface (and, in fact, at every point on any planet-shaped shell inside it, as well). On a cube, this is entirely gone. The centers of the faces would have the strongest gravities; the corners would have the weakest. For localis…
> On spherical earth the horizon on average is a little over three miles away. I read (long time ago) that the horizon is 29km on a seashore. In other words if you see a ship disappear on a horizon, it was 29km away. And so his 3mi vs my 29km is a bit of a discrepancy. Can anyone set things straight here?
http://en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon Wikipedia says 3.1 miles.
I imagine walking on a cube-shaped planet will be not different that doing it on a sphere shaped one.
Why not? The most important thing about a round planet in this case (and, in fact, the reason planets tend towards roundness) is that it means gravity is (roughly) equal at every point on the surface (and, in fact, at every point on any planet-shaped shell inside it, as well). On a cube, this is entirely gone. The centers of the faces would have the strongest gravities; the corners would have the weakest. For localis…
We walk on a sphere shaped planet, yet we do not "slide down" if we stand on the North Pole, yes? Same with a cubic one, as I imagine, the 90 degrees angle will be crossed without even realizing is there. I imagine we would perceive it as a flat planet.
As you can see, I did not take the whole thing too seriously. And that, of course, diminishes karma. So it goes. Cheers!