# cubical Earth

**URL:** <https://boards.straightdope.com/t/cubical-earth/594166>\
**Category:** Cecil's Columns/Staff Reports\
**Created:** [August 26, 2011, 12:24pm UTC](https://boards.straightdope.com/t/cubical-earth/594166 "2011-08-26T12:24:54Z")\
**Posts on this page:** 2\
**Page:** 4

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**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\
**Post date:** [September 27, 2011, 5:44am UTC](https://boards.straightdope.com/t/cubical-earth/594166/61 "2011-09-27T05:44:27Z")

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> [@ZenBeam](#):
>
> Almost certainly that could be done for at least a portion of the surface. It would be more difficult as you approached the edges of the surface. You’d need higher and higher density.
> 
> Also, the height above the surface where the pull could be perpendicular would be smaller near the edge. For example, if you wanted perpendicular pull for at least 10 feet above the surface, it might be impossible to achieve that within 10 feet of the edge.
> 
> ETA: Actually, you should be able to do this on all six surfaces simultaneously, again, with restrictions near the edges.

What am I missing here? At the edge, you have all of the mass of one side directly below you, all the rest of the mass of the rest of the cube on one side, and nothing on the other side. How do you end up with a pull that is straight down?

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**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\
**Post date:** [September 27, 2011, 11:33pm UTC](https://boards.straightdope.com/t/cubical-earth/594166/62 "2011-09-27T23:33:21Z")

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Very high density mass right at the edges. If you had a charged conducting metal cube, the electric field tangential to the surface of the cube must be zero, and since it’s charged, it has a non-zero field pointing away from the surface. So if you had a mass density that matched that charge density (a hollow, cubical Earth, combining the best of two current threads!) you’d have a gravitational field perpendicular to the surface everywhere on the surface. The gravitational field would also be zero everywhere in the interior.

ETA: It would rapidly become non-perpendicular near the edges as you moved away from the surface, but would change more slowly in the surface interiors.

In that case, the mass would be concentrated in the twelve edges, and especially at the eight corners, but some mass everywhere on the surface. I’d imagine the charge becomes an line charge along the edges, with point charges at the corner tips.

That’s the most extreme case, but if you only wanted a perpendicular gravitational field over the middle part of the surfaces, it would be possible with a less severe distribution.

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