Perfect spheres, gravity and charges.

[William Gaatjes]

I was watching a Walter Lewin, MIT video course and i finally got the answer why they would always use these conducting spheres.

Perfect spheres and gravity.
Gravity seems to be zero at the center of a perfect sphere. Every rotating mass by nature is not a perfect sphere. When thinking of planets and stars, this makes a lot of sense. That makes me think that there is no point inside for example the earth or the sun where gravity is absolutely zero. That makes me wonder, the idea is proposed by some people in the field that the core of our sun is a very small neutron only core. But this core would also not be a perfect sphere or would it ?
How about neutron stars , would these be perfect spheres ?
Or a black hole ? Would a black hole be so massive that it could form a perfect sphere ? If we assume it would be, we would have a very large mass, with a strong gravitational field, with exactly inside that outwards "radiating gravity field" (I do not know how to write it otherwise) there is a point of zero gravity. That huge change would make a difference. That makes me think of an extreme dy/dx in the aether or what many people like to call space time or quantum fluid.


When it comes to electric charge and perfect spheres which are good electrical conductors : A perfect sphere has no electrical polarization when it has charge and no electrical field or other charge is present.
If we would take the earth, although it is not a good electrical conductor,
i wonder if we could say it is electrically polarized. Because it is a flattened sphere. Add the sun as well. Which is also not a perfect sphere. It is also electrically polarized, especially during high solar activity it seems. Ignoring for the moment all the processes going on at the surface and inside the sun. Jupiter is the same.

Who likes to think about the implications ?

 

[jagec]

That's not really true. At the (presumably hollow) center of a perfect sphere, gravitational attraction is equal in all directions. A small object placed in the center would therefore float in place, but if you were to slice the object in half without moving it, each half would drift off towards the closer side. Many other geometries have a similar cancellation point (think of the center of a donut, or the Lagrangian points between two orbiting objects).

Remember, each piece of mass is gravitationally attracted to every other piece of mass in the universe, and considering the mass of a perfect sphere to be located at the center is a mathematical simplification that only holds if you are OUTSIDE said sphere.

If you were to hollow out a black hole and place an object in the center, you would almost certainly exceed the Roche limit for that object, and it would get torn into pieces. Not to mention the difficulties of keeping the center hollow in the first place.

A charged sphere is radially symmetrical, but still "polarized". Charge repulsion ensures that more of the charge is carried on or near the skin than closer to the middle.

 

[TuxDave]

That makes me think that there is no point inside for example the earth or the sun where gravity is absolutely zero.

It's hard to understand the rest of your points when this initial point isn't right. The earth isn't a perfect sphere (because of it's shape, mass distribution) but there should still exist a point inside the earth with gravity balanced in all direction.

So if you remove that initial assumption of yours, how does the rest of your points follow?

 

[William Gaatjes]

It's hard to understand the rest of your points when this initial point isn't right. The earth isn't a perfect sphere (because of it's shape, mass distribution) but there should still exist a point inside the earth with gravity balanced in all direction.

So if you remove that initial assumption of yours, how does the rest of your points follow?

I do not find it difficult to grasp that there is a point where gravity balance is balanced in the sense that it is not zero. But can you provide proof that it is zero with a shape that is not a perfect sphere ?

 

[William Gaatjes]

That's not really true. At the (presumably hollow) center of a perfect sphere, gravitational attraction is equal in all directions. A small object placed in the center would therefore float in place, but if you were to slice the object in half without moving it, each half would drift off towards the closer side. Many other geometries have a similar cancellation point (think of the center of a donut, or the Lagrangian points between two orbiting objects).

Remember, each piece of mass is gravitationally attracted to every other piece of mass in the universe, and considering the mass of a perfect sphere to be located at the center is a mathematical simplification that only holds if you are OUTSIDE said sphere.

If you were to hollow out a black hole and place an object in the center, you would almost certainly exceed the Roche limit for that object, and it would get torn into pieces. Not to mention the difficulties of keeping the center hollow in the first place.

A charged sphere is radially symmetrical, but still "polarized". Charge repulsion ensures that more of the charge is carried on or near the skin than closer to the middle.

That must have been my mistake then... I assumed a solid perfect sphere.

When it comes to the polarization of an electrically charged hollow sphere, would you notice this outside the sphere ?

 

[TuxDave]

I do not find it difficult to grasp that there is a point where gravity balance is balanced in the sense that it is not zero. But can you provide proof that it is zero with a shape that is not a perfect sphere ?

You want a proof? Easy. Imagine the point in the center of a perfect cube. For every mass "m" with a vector distance "d" from the point, there will always exist another point of mass "m" with the vector distance "-d". Through proof by symmetry, the total sum of gravity due to all mass in the perfect cube will be zero.

 

[William Gaatjes]

You want a proof? Easy. Imagine the point in the center of a perfect cube. For every mass "m" with a vector distance "d" from the point, there will always exist another point of mass "m" with the vector distance "-d". Through proof by symmetry, the total sum of gravity due to all mass in the perfect cube will be zero.

I understand what you mean, and you are right in this example.
But this example is about point masses where every two opposite points are viewed independently yes ?

If i would take all point masses at the same time, i view and feel in my imagination an instability.
I have this strong feeling that the interaction of all point masses causes an instability that would make no point in the cube experience zero gravity. And there would always be a force in any direction besides the force to the center of the cube.

With a hollow sphere, i can see and feel when using my imagination that there would be automatically be a force to the center but at the center there would be a cancellation depending on the density, mass and thickness of the material and the radius of sphere.

With a solid sphere, i can see and feel when using my imagination that there would be automatically be a force to the center compressing the sphere to be smaller .

But you and jagec both made it clear, that i have overlooked something.

 

[TuxDave]

I understand what you mean, and you are right in this example.
But this example is about point masses where every two opposite points are viewed independently yes ?

If i would take all point masses at the same time, i view and feel in my imagination an instability.

It equally applies when you take it all into account at the same time. If I had to do a crappy word-like proof it would be something like.

The gravity felt at the center of a perfect cube is the sum of gravity due to all the mass around it.

Gravity = a1 + a2 + a3 + a4.....

And due to that symmetry mumbo-jumbo I mentioned, you can always pair up points of pass

Gravity = (a1 + b1) + (a2 + b2) .....

And since they're in opposite direction the they cancel each other out.

Gravity = 0 + 0 + 0..... = 0

Thus... zero.

 

[Biftheunderstudy]

The issue/question everyone seems to be dancing around here is called Newton's shell theorem, or in GR Birkhoff's theorem.

Spherically symmetric mass distributions have zero field inside....outside is another matter...

 

[JTsyo]

The issue/question everyone seems to be dancing around here is called Newton's shell theorem, or in GR Birkhoff's theorem.

Spherically symmetric mass distributions have zero field inside....outside is another matter...

this. Inside a hollow sphere there's no gravity due to the sphere, anywhere, not just the center. I'm sure that's freshman physics.

I don't follow what the rest of the OP is trying to get at. I don't think having a bigger mass will help you get more spherical since both gravity and the radial pull is proportional to mass.

 

[silverpig]

Bif already corrected the zero field in a sphere issue in this thread, but I'll add that there are in fact theoretical (and entirely possible for now) constructs of black holes that rotate so quickly they form a ring or disk.

I could easily see non-spherical neutron stars, even if they are pretty close to being spherical.

 

[William Gaatjes]

To be honest, it is just an interest and a hobby. When i had more time, i dug a lot deeper into the matter. Since a while i have chosen for a combination of hobby and work related subjects only, spending not much time about it.
I have only so much resources available. Inevitable choices. There will come a day again that i can use more resources.

I occasionally watch some of the Walter Lewin courses from MIT or the mechanical universe courses from Caltech to freshen up when i am tired.
That is how the idea in this thread spawned.