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Newton proved in Principia that the total gravity that a sphere exerts over a material point is the same as if all the mass of the sphere were concentrated on its center.

I wonder if it's possible to prove that a sphere / spherical shell is the only shape with this property (or perhaps there are other shapes? I don't know).

We can define center as "center of mass", and I'm only interested in Newtonian physics (as opposed to relativistic physics).

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up vote 6 down vote accepted

It is very easy to construct arbitrary shapes that have the property that the gravitational potential outside is just like all the mass were concentrated at a point.

Start with the gravitational potential for a point:

$$ \phi(x) = - {M\over r}$$

Then take any shape, take two nested cubes for definiteness. Then make $\phi(x)$ be a constant in the interior of the inner cube larger than the supremum of the values outside the cube, and make the potential rise up in a gradually down-curving way to the inner cube's value.

Then $\rho(x) \propto - \nabla^2 \phi$ is a mass distribution which produces this field, and $\nabla^2\phi$ is zero inside the inner cube and outside the outer cube. The only thing you need to check is that the mass density is everywhere positive.

If the positive mass thing doesn't work on the first try, you can always make the potential on the inner cube bigger, or if worst comes to worst, draw an inscribed sphere in the inner cube, and a circumscribed sphere around the outer cube, and fill the region between the two spheres with a uniform positive mass density which is equal to the maximum negative magnitude of the density in the squares alone.

It is just not true that the sphere is the only shape with a pointlike exterior field, not even close.

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Can you explain it a little further? Close to the vertex of the outer cube the potential is much bigger. What is wrong about that reasoning? –  Eduardo Guerras Valera Nov 4 '12 at 1:47
    
By chosing a surface extremely close to one of the corners of your outer cube, the potential will be larger than in the rest of the surface. No density distribution in the cube can compensate that, since you cannot make two hyperbols match if they are not centered in the same zero, e.g. A/x will never equal B/(x-2) for all x. –  Eduardo Guerras Valera Nov 4 '12 at 2:08
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@Eduardo: This is false, just choose any $\phi$ with appropriate asymptotics, and choose the density to be it's Laplacian. –  Ron Maimon Nov 4 '12 at 2:17
    
Oh sh... You are right. I erase my answer. I was thinking in analogy with the electric potential and conducting surfaces... It is too late in Spain now... –  Eduardo Guerras Valera Nov 4 '12 at 2:28
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@Eduardo: It's ok, I was confused too for a minute or two, and it's easy to get lost in the fog of chalk dust while teaching. –  Ron Maimon Nov 4 '12 at 2:34
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