Recently I was talking to a colleague about physics, and we got onto the subject of gravity in a hollow sphere - don't ask how we got there, I have no idea.

Any way My colleague said that in a Physics class he had learned that there was no gravity in side a hallow sphere. Ever since I was pretty young I have pondered the idea of what happens to gravity at the centre of the earth, or any other spherical object, I had sort of assumed that there would a a point where Gravity would be Zero - because there is an equal ammount of earth each side of you, so the gravitational force exerted by one side would the equal to the force from the other side, so you wouldn't move if you were right in the centre (ignoring all the other problems you would be having) I had assumed that as soon as you move from your place in the centre, you would be closer to one side than the other and the gravity from the closer side would have a stronger pull than that from the further side and you'd end up stuck to the wall.

However according to the

**Gauss-Ostrogradsky theorem**at every point inside a hollow sphere gravity is Zero. there is all sorts on nasty math to "prove" this but I think the general idea is that if you move off your point at the centre, there is more sphere behind you than in front, so the fact that is further away is made up for by there being more of it, so the force exerted by the wall in front and behind you are still the same. I am still not 100% convinced - I think a practical demonstration is called for.

## 2 comments:

It seems that I have had a lot of hits recently by people looking for answers to the "Gravity in a hollow sphere" problem. If got here looking for information of this problem why not leave a comment or start a discussion on this little brain teaser.

I remember doing this as a homework problem in either Physics or Calculus. It's not really THAT hard, though it's been 30 years, so I probably couldn't do it now. All you do is say that for a sphere of radius r (you can ignore the thickness and call it infinitesimally thin) calculate the gravity at point x, y, z (which is a fairly straightforward, though lengthy triple integral). If you assume all the masses to be 1 to make the math easier, the gravity is just G/d-squared where d is the distance between the point and the part of the sphere in question. When you do the math, you find that all the variables "fall out" and the answer is 0. If it's 0 for an infinitesimally thin sphere, you can just "stack up" as many of them as you like to get any thickness, and it will still be 0.

Or, in English, the further-away parts of the sphere are more numerous in just the right proportion to balance out the force from the closer-by parts of the sphere.

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