Hollow sphere gravity question

In summary, the conversation is discussing the hypothetical scenario of a small object inside a large hollow sphere and whether it would be accelerated towards the inner surface. The consensus is that in both Newtonian mechanics and general relativity, the object would not be accelerated, but would remain in equilibrium at the center due to the canceling out of gravitational forces in all directions. This is true regardless of whether the sphere is rotating or not. Additionally, the conversation touches on the concept of gravity as a vector quantity and how it cancels out inside the sphere, as well as the equivalence of the gravitational field outside the sphere to a point mass at its center.
  • #1
GENIERE
I seem to have managed to confuse myself again. If I assume the only matter in the universe was an object of very small mass and size located inside a hollow sphere of very large mass, would the small object be accelerated toward the inner surface?

Thanks
 
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  • #2
Not in Newtonian mechanics. I seem to remember that Feynman derives this result in his lectures.

I can't remember whether or not it is true in GR. Although I recall using this result (that it is not accelerated to the inner side of the sphere) in cosmology so I suspect it is true in GR too.

Can anyone verify me here?

Matt
 
  • #3
If you started exactly at the center, then symmetry will require that you stay there - independent of Newtonian or GR based approach. I also believe this may be true if you are not at the center, but am not sure if GR predicts otherwise.
 
  • #4
How about if you were not at the center, and the shell was rotating? In Newtonian gravity the rotation would not matter, but in GR it probably would do something. (I'll gloss over the issue of whether it makes sense to talk of a rotating shell when essentially all the matter of the universe is located in that shell!)
 
  • #5
Well, according to gravity and geometry, no. any position inside the hollow sphere would have the same gravitational pulls form all sides, so gravity would not cause any changes. Think about a circle on a graph. say point A is at the center. the circle around it is equal in all directions. now point a moves up the y-axis say 1/2 the radius. now it is closer to the upper limit, right? but, as it gets closer to that upper limit, more circle line has gotten below it. since gravity depends on distance and mass, it will always be the case that inside a perfect sphere at any point gravitational forces will cancel out.

I'm 90% sure.
 
  • #6
If one could excavate a spherical hole, say, 10-foot diameter at the exact mass center of the earth, and a marble was held then released in the center of the hole, what would happen?
The answer to this question is the answer to your question
 
  • #7
GENIERE said:
I seem to have managed to confuse myself again. If I assume the only matter in the universe was an object of very small mass and size located inside a hollow sphere of very large mass, would the small object be accelerated toward the inner surface?

Thanks
Seems to me that if the sphere of matter were at the centre of the surrounding shell it would stay there. This is because the gravitational attractiveness of the sphere of matter would be concentrated at the centre. Whichever way a hypothetical person looked or measured the gravitational attraction of the shell from this point, they would find it to be equal.
The matter would be in equilibrium with the surrounding shell and would not, could not move, unless something changed and upset the balance.
 
  • #8
You have the right analysis but the wrong conclusion. Gravitational force is a vector quantity; it has direction as well as amount. So somebody at the center of the sphere looks in any direction there's an amount of gravity directed toward him from that direction. And in the opposite direction there's an equal amount pointed that way. So the two of them cancel each other out. And that works for any direction, so all the gravity cancels out. Thus the person at the center of the hollow sphere experiences no gravity at all.

Furthermore with a little geometry you can see that this happens not only at the center but at any point inside the sphere.
 
  • #9
berty said:
Seems to me that if the sphere of matter were at the centre of the surrounding shell it would stay there. This is because the gravitational attractiveness of the sphere of matter would be concentrated at the centre.
From the viewpoint of ordinary Newtonian gravity, the field is zero everywhere within a uniform spherical shell. Not just at the center.

The gravitational field outside the spherical shell is equivalent to an equal mass concentrated at its center.
 
  • #10
Thanks folks!
 

1. What is a hollow sphere gravity question?

A hollow sphere gravity question refers to a hypothetical scenario where a hollow sphere of uniform density and thickness is placed in a gravitational field. The question typically involves determining the gravitational force acting on a point inside or outside the sphere.

2. How is the gravitational force calculated for a point inside a hollow sphere?

The gravitational force acting on a point inside a hollow sphere is calculated using the same formula as for any other point mass: F = GmM/r^2, where G is the gravitational constant, m is the mass of the point, M is the mass of the sphere, and r is the distance between the point and the center of the sphere. However, for a hollow sphere, the value of M is only the mass of the portion of the sphere that is closer to the center than the point in question.

3. What happens to the gravitational force if the point is moved closer to the center of the hollow sphere?

If the point is moved closer to the center of the hollow sphere, the gravitational force acting on it decreases. This is because the distance between the point and the center of the sphere decreases, resulting in a smaller value for r in the gravitational force formula. As a result, the force decreases with the inverse square of the distance.

4. What is the gravitational force on a point outside a hollow sphere?

The gravitational force on a point outside a hollow sphere is calculated using the same formula as for any other point mass: F = GmM/r^2, where G is the gravitational constant, m is the mass of the point, M is the mass of the sphere, and r is the distance between the point and the center of the sphere. However, for a hollow sphere, the value of M is only the mass of the portion of the sphere that is closer to the center than the point in question. This means that as the point moves farther away from the center of the sphere, the gravitational force acting on it decreases.

5. How does the thickness of the hollow sphere affect the gravitational force?

The thickness of the hollow sphere does not affect the gravitational force acting on a point inside or outside the sphere as long as the density remains uniform. This is because the mass of the sphere is determined by the total volume, not the thickness of the material. However, if the density is not uniform, the gravitational force may vary at different points on the surface of the sphere depending on the thickness of the material at that point.

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