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Pulsars Don't Exist

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posted on Oct, 20 2010 @ 03:33 PM
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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.
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.



posted on Oct, 20 2010 @ 03:34 PM
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reply to post by mnemeth1
 





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.


It is indeed a bit counterintuitive, but you are wrong. Because the closer you are to either edge of a sphere, the more material is also behind you. Thus the net effect is indeed weightlesness.

en.wikipedia.org...



posted on Oct, 20 2010 @ 03:41 PM
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Originally posted by nataylor

Originally posted by mnemeth1
We are talking about spinning an object the size of an asteroid at near the speed of light.
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.


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.
Science doesn't work that way. Prove what you are saying with math. So far, what you are saying doesn't hold up.


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.



posted on Oct, 20 2010 @ 03:42 PM
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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.



posted on Oct, 20 2010 @ 03:51 PM
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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.




Do the math. You are wrong.

If you've got a few mile thick crust (say, 3 miles or 5,000 meters), and you occupy a half-square-meter area of the inner surface, you've got 1,250 cubic meters of material pulling you straight down. Assuming a sphere with outer radius of 15,000 meters, you've got 13,613,568,165,555 cubic meters pulling you tot he side or up.

Guess what, it all cancels out.



posted on Oct, 20 2010 @ 03:53 PM
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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)



posted on Oct, 20 2010 @ 03:54 PM
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Originally posted by mnemeth1

Originally posted by nataylor

Originally posted by mnemeth1
We are talking about spinning an object the size of an asteroid at near the speed of light.
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.


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.
Science doesn't work that way. Prove what you are saying with math. So far, what you are saying doesn't hold up.


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.



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.



posted on Oct, 20 2010 @ 03:55 PM
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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.



posted on Oct, 20 2010 @ 03:56 PM
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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.



posted on Oct, 20 2010 @ 03:57 PM
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reply to post by mnemeth1
 




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.


The product of these two forces can never give you positive vector pointing outward from the center. See Shell theorem , point 3. :


Inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero at the centre of mass.


Net gravitational force varies LINEARLY from maximum force at the surface of the sphere to zero force at the center, and always points to the center. PRESSURE increases from zero to on the surface to maximum at the center, and always points to the center, too.

The only force which points upwards in the centripetal force. And that force is zero at the center and increases to maximum on the surface.



posted on Oct, 20 2010 @ 04:00 PM
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reply to post by mnemeth1
 


Inverse square law force is actually the only force which SATISFIES The Shell theorem. See
edit on 20/10/10 by Maslo because: typo



posted on Oct, 20 2010 @ 04:03 PM
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reply to post by mnemeth1
 



I know, I'm talking about the center of the star, not the shell.

Re-read the questions.


As you use an analogy as being sand.....


You lost me at that point.



posted on Oct, 20 2010 @ 04:03 PM
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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?



posted on Oct, 20 2010 @ 04:03 PM
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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!



posted on Oct, 20 2010 @ 04:06 PM
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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?
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.





edit on 20-10-2010 by nataylor because: (no reason given)



posted on Oct, 20 2010 @ 04:07 PM
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reply to post by mnemeth1
 





So if one applies any modicum of centripetal force, you now have "gravity" holding you to the outer wall, right?


Yes, in a rotating hollow sphere, centrifugal force would push object onto the outer wall.
edit on 20/10/10 by Maslo because: (no reason given)



posted on Oct, 20 2010 @ 04:09 PM
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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!


It can't win out, because there is virtually no gravity.

At the center of the star, there is no gravity, only centripetal force.



edit on 20-10-2010 by mnemeth1 because: (no reason given)



posted on Oct, 20 2010 @ 04:10 PM
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unfortunatley for the rest of everyones egos here.... the OP is right. We are all gonna have to admit it.

The main stream physics community has been controlled for the last century, for good reason.



posted on Oct, 20 2010 @ 04:11 PM
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reply to post by mnemeth1
 


And there is virtually no centripetal force.



posted on Oct, 20 2010 @ 04:12 PM
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Originally posted by Maslo
reply to post by mnemeth1
 


And there is virtually no centripetal force.


So 700 hz rotation rates generates "virtually no centripetal force"

ok.



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