|MadSci Network: Physics|
The answer is, I'm afraid, relatively uninteresting. The gravitational field on the surface of the cube would be nearly the same as that of a sphere of similar size and mass. In other words, the local gravitational force at some location would nearly point towards the center of the object -- the center of the cube.
If you were at the center of a face of the cube, in the middle of a large, flat surface, this would seem perfectly natural. Gravity would pull you straight "down", perpendicular to the surface. If you dropped a ball, it would fall, bounce off the surface, bounce straight back up, gradually slow down, fall again to the same spot, and bounce up and down in place; just as it would on Earth.
But if you were standing on one of the edges of the cube, straddling it with one foot on each face, then things wouldn't be so simple. Gravity would still pull you straight towards the center of the cube, but the surfaces would now be tilted by 45 degrees to this gravitational pull. If you dropped a ball here, EXACTLY centered on the sharp edge, then the ball would (in theory) bounce up and down on the edge. But if you dropped it a bit to one side -- say, to the left of the edge, near you own left foot -- then the ball would not bounce straight "up". Instead, it would bounce away from the surface ... but also away from you, heading further to the left. In fact, it would continue to move away from you with every bounce. When it stopped bouncing, it would roll towards the center of that face of the cube.
It would be nearly the same as if you took a big box here on Earth, tilted it up on one edge, and then stood at the top, with one foot on each side of the edge.
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