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The seas wouldn't even be flat to the "surface" of the cube, right? They'd deform into a spheroid. So looking at it from a peak (or from space) it would appea

by blahedo 15y ago

The seas wouldn't even be flat to the "surface" of the cube, right? They'd deform into a spheroid. So looking at it from a peak (or from space) it would appear to be a liquid sphere intersected with a solid cube.

CWuestefeld 15y ago
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 equilibrium between those forces sits, nor do I know how to calculate it.
- tvon 15y ago
  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.
  - CWuestefeld 15y ago
    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.
    - T-hawk 15y ago
      Easier to figure using the radius rather than the diameter. Volume of the sphere is 4pi/3 * r^3, volume of the cube is (2r)^3. Factor out r^3 and you get (4pi/3)/8, or 52% of the cube's volume is within the sphere.
    - masterzora 15y ago
      The math on that is easy. The volume of a cube is X^3, the volume of a sphere is 1/6 * pi * X^3, so the difference is a little under 1/2 X^3. Put another way, the cube is (roughly) twice the volume of the sphere.
  - hammock 15y ago
    That's right. If you are in the center of a face, the center of an edge, or on one of the vertices the gravity vectors all point towards the core. An imagined sphere inside the cube holds 52% of the mass and so the remaining mass is divided among eight vertices, about 6% for each. If you are off a ways from one of the points listed above, the sphere part (52%) will always pull towards the center. A lot of the force from the corners will be mitigated/cancelled by symmetry, and whatever is left over, again, it's only 6% maximum per corner.
  - 7952 15y ago
    You are assuming that every part of the cube has the same density. Perhaps the density increases as you get to the corners.
    - dalke 15y ago
      For all we know, anyone making a square planet would be able to put all the mass in the corners, with only 0.00001% for the super-material used to make the faces of the cube. The math is relatively easy with uniform density. Otherwise we have to start conjecturing on the density distribution, and without real-world feedback it degenerates to wondering if an infinite number of angels can dance on the head of a pin.
- arethuza 15y ago
  One place to start would be the Schiehallion Experiment of 1774 where a symmetric and isolated mountain in Scotland was used as the basis of an experiment to measure how much the gravitational attraction of the mountain pulled a pendulum away from the vertical. The goal being to estimate the mass of the Earth - with what turns out to have been a reasonably accurate result. During the course of the experiment they had to measure the shape of the mountain quite carefully and they had the idea to join points of similar height together with a line on the map - inventing contour lines. http://en.wikipedia.org/wiki/Schiehallion_experiment http://en.wikipedia.org/wiki/Schiehallion_experiment
- muyyatin 15y ago
  Sorry I don't have a reference, but as long as you are outside of an object, its gravitational force on you is the same as if it were a point-mass (at least for classical physics).
  - xsmasher 15y ago
    Which points to the oceans "bunching up," not laying flat.
  - bzbarsky 15y ago
    This is only true for objects that have a spherically symmetrical mass distribution and for a point test object. That is, the field of a spherically symmetric object is the same as the field of a point mass of its mass located at its center. As soon as your test object is non-point you get tidal forces due to the field typically being nonuniform. As soon as your object is not spherically symmetric you get a field that looks nothing like that of a point mass.
    - starwed 15y ago
      It's still the case that if you're sufficiently far away from a nonspherical object, it'll effectively act like a point source. I'd guess this is what the OP was recalling.
      - bzbarsky 15y ago
        Well, sure. If you're far enough away it _looks_ like a point source! That doesn't apply to being on the face of a cube, though; that's not nearly far enough away, obviously. :)

by blahedo 15y ago

CWuestefeld 15y ago

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 equilibrium between those forces sits, nor do I know how to calculate it.

tvon 15y ago
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.
- CWuestefeld 15y ago
  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.
  - T-hawk 15y ago
    Easier to figure using the radius rather than the diameter. Volume of the sphere is 4pi/3 * r^3, volume of the cube is (2r)^3. Factor out r^3 and you get (4pi/3)/8, or 52% of the cube's volume is within the sphere.
  - masterzora 15y ago
    The math on that is easy. The volume of a cube is X^3, the volume of a sphere is 1/6 * pi * X^3, so the difference is a little under 1/2 X^3. Put another way, the cube is (roughly) twice the volume of the sphere.
- hammock 15y ago
  That's right. If you are in the center of a face, the center of an edge, or on one of the vertices the gravity vectors all point towards the core. An imagined sphere inside the cube holds 52% of the mass and so the remaining mass is divided among eight vertices, about 6% for each. If you are off a ways from one of the points listed above, the sphere part (52%) will always pull towards the center. A lot of the force from the corners will be mitigated/cancelled by symmetry, and whatever is left over, again, it's only 6% maximum per corner.
- 7952 15y ago
  You are assuming that every part of the cube has the same density. Perhaps the density increases as you get to the corners.
  - dalke 15y ago
    For all we know, anyone making a square planet would be able to put all the mass in the corners, with only 0.00001% for the super-material used to make the faces of the cube. The math is relatively easy with uniform density. Otherwise we have to start conjecturing on the density distribution, and without real-world feedback it degenerates to wondering if an infinite number of angels can dance on the head of a pin.
arethuza 15y ago
One place to start would be the Schiehallion Experiment of 1774 where a symmetric and isolated mountain in Scotland was used as the basis of an experiment to measure how much the gravitational attraction of the mountain pulled a pendulum away from the vertical. The goal being to estimate the mass of the Earth - with what turns out to have been a reasonably accurate result. During the course of the experiment they had to measure the shape of the mountain quite carefully and they had the idea to join points of similar height together with a line on the map - inventing contour lines. http://en.wikipedia.org/wiki/Schiehallion_experiment http://en.wikipedia.org/wiki/Schiehallion_experiment
muyyatin 15y ago
Sorry I don't have a reference, but as long as you are outside of an object, its gravitational force on you is the same as if it were a point-mass (at least for classical physics).
- xsmasher 15y ago
  Which points to the oceans "bunching up," not laying flat.
- bzbarsky 15y ago
  This is only true for objects that have a spherically symmetrical mass distribution and for a point test object. That is, the field of a spherically symmetric object is the same as the field of a point mass of its mass located at its center. As soon as your test object is non-point you get tidal forces due to the field typically being nonuniform. As soon as your object is not spherically symmetric you get a field that looks nothing like that of a point mass.
  - starwed 15y ago
    It's still the case that if you're sufficiently far away from a nonspherical object, it'll effectively act like a point source. I'd guess this is what the OP was recalling.
    - bzbarsky 15y ago
      Well, sure. If you're far enough away it _looks_ like a point source! That doesn't apply to being on the face of a cube, though; that's not nearly far enough away, obviously. :)