A classical mechanics problem by Tapas Mazumdar

Consider a cube with constant density.

What is the maximum number of points on the surface of this cube such that all of these points experience the same gravitational force exerted by the cube?

0 2 4 8 \infty

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1 solution

Tapas Mazumdar
Sep 26, 2016

Consider a circular region on a face with its center lying directly at the geometrical center of the square face (point of intersection of its diagonals). Each point on the perimeter of this circular region will subtend the same angle (with one arm taken towards the center of the circle and the other towards the center of mass of Earth) at the center of mass of the Earth (which is at the geometrical center of the cube).

So, from here we conclude that all these points will feel the same amount of Force due to gravity, thus same value of acceleration due to gravity. It doesn't matter how many faces this polyhedron has, as long as it is a regular polyhedron , the number of points we will obtain will be \infty .

I doubt your claim of "circular region (centered at the center)" would work. While I agree that the "North, South, East, West" points of this circle would experience the same force, I am doubtful that any other points of this perfect circle would experience this same force, due to the lack of symmetry of this cube.

I agree that there is a "slightly deformed circular region" that would experience the same force. This is independent of being a regular polyhedron, and is just dependent on how level curves will cut a surface.

Calvin Lin Staff - 4 years, 7 months ago

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