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No. There is no net gravity inside a hollow sphere of uniform density and thickness. This is basic math, and I believe your inability to realize this doesn't speak well to your ability to disprove Einstein.
Originally posted by mnemeth1
Originally posted by Maslo
The sand in this hollow cavity would actually not be attracted anywhere, it would be weightless. Thats because the forces exactly cancel out, no matter the mass or radius or wall thickness of the sphere.
Nope. Since the sand on the inner side of the sphere is closer to the sand above it, it will have a greater attraction to the sand directly above it. It would indeed be under gravitational force - it would not be weightless.
A person would be able to walk around the inner sphere and arrive back where they started from.
Nope. Since the sand on the inner side of the sphere is closer to the sand above it, it will have a greater attraction to the sand directly above it. It would indeed be under gravitational force - it would not be weightless. A person would be able to walk around the inner sphere and arrive back where they started from.
Originally posted by nataylor
Let's take the centripetal force on a single neutron on the surface of neutron star with a 16-km radius, rotational frequency of 716 Hz, and a mass of 2 Solar masses. That would give us 5.4*10^-16 N of centripetal force. Now with what force would gravity be pulling that single neutron in? That would be 8.7*10^-16 N. As you can see, the pull of gravity is larger than the centripetal force.
Originally posted by mnemeth1
We are talking about spinning an object the size of an asteroid at near the speed of light.
Science doesn't work that way. Prove what you are saying with math. So far, what you are saying doesn't hold up.
Claiming gravity, which is one of the weakest forces in the universe, is capable of holding matter together as it rotates around at the speed of light is the most retarded theory I have ever heard in my life.
Originally posted by nataylor
No. There is no net gravity inside a hollow sphere of uniform density and thickness. This is basic math, and I believe your inability to realize this doesn't speak well to your ability to disprove Einstein.
Originally posted by mnemeth1
Originally posted by nataylor
No. There is no net gravity inside a hollow sphere of uniform density and thickness. This is basic math, and I believe your inability to realize this doesn't speak well to your ability to disprove Einstein.
Yes, there is.
If you have a few mile thick crust, that's certainly enough gravity to walk around on.
I calculated the centripetal force directly on the surface at the equator. It's 5.4*10^-16 N. There is no material "above" that. The gravitational attraction towards the center would be 8.7*10^-16 N. Gravity wins out over centripetal force there.
Originally posted by mnemeth1
Originally posted by nataylor
Let's take the centripetal force on a single neutron on the surface of neutron star with a 16-km radius, rotational frequency of 716 Hz, and a mass of 2 Solar masses. That would give us 5.4*10^-16 N of centripetal force. Now with what force would gravity be pulling that single neutron in? That would be 8.7*10^-16 N. As you can see, the pull of gravity is larger than the centripetal force.
Originally posted by mnemeth1
We are talking about spinning an object the size of an asteroid at near the speed of light.
Science doesn't work that way. Prove what you are saying with math. So far, what you are saying doesn't hold up.
Claiming gravity, which is one of the weakest forces in the universe, is capable of holding matter together as it rotates around at the speed of light is the most retarded theory I have ever heard in my life.
Well, lets see, what would be the centripetal force on that neutron just slightly off the spin axis at the center of the star?
Then add that to the gravitational attractive force of the matter directly above that point.
Then subtract the gravitational force of the matter below that point.
Originally posted by nataylor
I calculated the centripetal force directly on the surface at the equator. It's 5.4*10^-16 N. There is no material "above" that. The gravitational attraction towards the center would be 8.7*10^-16 N. Gravity wins out over centripetal force there.
Originally posted by mnemeth1
reply to post by nataylor
Not quite, because gravity obeys the inverse square law.
hyperphysics.phy-astr.gsu.edu...
When you throw that in, your math doesn't hold up.
Distance between sides plays a huge role.
edit on 20-10-2010 by mnemeth1 because: (no reason given)
Then add that to the gravitational attractive force of the matter directly above that point. Then subtract the gravitational force of the matter below that point.
Inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero at the centre of mass.
I know, I'm talking about the center of the star, not the shell.
Re-read the questions.
Originally posted by nataylor
Originally posted by mnemeth1
reply to post by nataylor
Not quite, because gravity obeys the inverse square law.
hyperphysics.phy-astr.gsu.edu...
When you throw that in, your math doesn't hold up.
Distance between sides plays a huge role.
edit on 20-10-2010 by mnemeth1 because: (no reason given)
It's exactly the inverse square law that makes it so that a mere 1,250 cubic meters of material can cancel out 13,613,568,165,555 cubic meters of material.
Someone already posted the link to the math: en.wikipedia.org...
Do the math, and learn.
Originally posted by mnemeth1
Originally posted by nataylor
I calculated the centripetal force directly on the surface at the equator. It's 5.4*10^-16 N. There is no material "above" that. The gravitational attraction towards the center would be 8.7*10^-16 N. Gravity wins out over centripetal force there.
I know, I'm talking about the center of the star, not the shell.
Re-read the questions.
Any centripetal force exerted against you would be completely independent of the mass or gravity of the sphere. It would solely be based on the radius of the sphere and rotation rate. Like the way "gravity" can be simulated inside a rotating space station.
Originally posted by mnemeth1
Originally posted by nataylor
Originally posted by mnemeth1
reply to post by nataylor
Not quite, because gravity obeys the inverse square law.
hyperphysics.phy-astr.gsu.edu...
When you throw that in, your math doesn't hold up.
Distance between sides plays a huge role.
edit on 20-10-2010 by mnemeth1 because: (no reason given)
It's exactly the inverse square law that makes it so that a mere 1,250 cubic meters of material can cancel out 13,613,568,165,555 cubic meters of material.
Someone already posted the link to the math: en.wikipedia.org...
Do the math, and learn.
So if one applies any modicum of centripetal force, you now have "gravity" holding you to the outer wall, right?
So if one applies any modicum of centripetal force, you now have "gravity" holding you to the outer wall, right?
Originally posted by nataylor
Originally posted by mnemeth1
Originally posted by nataylor
I calculated the centripetal force directly on the surface at the equator. It's 5.4*10^-16 N. There is no material "above" that. The gravitational attraction towards the center would be 8.7*10^-16 N. Gravity wins out over centripetal force there.
I know, I'm talking about the center of the star, not the shell.
Re-read the questions.
OK, here's the centripetal force for a neutron 1 meter from the center: 3.4*10^-20. And the gravitational force: 1.1*10^-19. Again, gravity wins out!
Originally posted by Maslo
reply to post by mnemeth1
And there is virtually no centripetal force.