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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 th

by 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.

by 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.