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I have a query for any nuclear physicists out there.



 
 
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  #1  
Old June 24th 14, 01:16 PM
JAAKKO KURHI JAAKKO KURHI is offline
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Default I have a query for any nuclear physicists out there.

If the total mass of an atomic nucleus is made up of very small particles, and each particle would have a gravitational binding force, how many such particles would it take to make the total gravitational force equal to the strong force contained in the nucleus?
  #2  
Old June 24th 14, 08:11 PM posted to sci.astro
dlzc
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Default I have a query for any nuclear physicists out there.

Dear JAAKKO KURHI:

On Tuesday, June 24, 2014 5:16:52 AM UTC-7, JAAKKO KURHI wrote:

If the total mass of an atomic nucleus is made up
of very small particles, and each particle would
have a gravitational binding force,


Gravity is not a force.

how many such particles would it take to make the
total gravitational force equal to the strong force
contained in the nucleus?


The math you need for your answer is he
http://en.wikipedia.org/wiki/Chandrasekhar_limit

And I am not a nuclear physicist, but I was a Radiation Safety Officer (if that matters).

Gravity is more than 20 orders of magnitude smaller than the strong nuclear force. So "a lot".

David A. Smith
  #3  
Old June 27th 14, 04:50 PM
JAAKKO KURHI JAAKKO KURHI is offline
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First recorded activity by SpaceBanter: Apr 2013
Posts: 40
Default

Quote:
Originally Posted by dlzc View Post
Dear JAAKKO KURHI:

On Tuesday, June 24, 2014 5:16:52 AM UTC-7, JAAKKO KURHI wrote:

If the total mass of an atomic nucleus is made up
of very small particles, and each particle would
have a gravitational binding force,


Gravity is not a force.

how many such particles would it take to make the
total gravitational force equal to the strong force
contained in the nucleus?


The math you need for your answer is he
http://en.wikipedia.org/wiki/Chandrasekhar_limit

And I am not a nuclear physicist, but I was a Radiation Safety Officer (if that matters).

Gravity is more than 20 orders of magnitude smaller than the strong nuclear force. So "a lot".

David A. Smith
Dear David Smith

Evidently I didn’t lay out my equation very well, because your answer isn’t related to proposed equation.
Allow me to rephrase with detailed explanation: At the very early state when the components for the matter
are evolving, the nature has no other as primeval particle to work with as is the mass particle.
Even the smallest of all theorized particles the higgs boson, a fragment of proton is mostly made of mass,
also the nucleus is 99.7% of mass. So, it should be safety to say that mass particles are the major building
blocks in construction of the matter. Therefore, mass and it’s intrinsic property gravity is the only source for
all forces and all energy in the universe, including so called strong force.
Newton's gravitational constant, Big G was experimentally developed over 200 years ago, at the time the
science knew nothing about the solid particle assembly of the nucleus. An assembly where objects are bind
together with attractive force, having no distance in between the bundle of objects. So, the gravity within
the objects of nucleus is not comparable with gravity in between the objects of matter. Therefore, instead
of using the weight of the nucleons to calculate the gravity, the weight of individual mass particle and the
number of particles used in the nucleus system construction yield the correct force of gravity. A molecule
of matter is a assembly of atoms, similarly the nucleus can be the assembly of mass particles.
So, it’s possible using the available nuclear force as a target number and increasing the number of mass
particles till the total force of gravity is equal to required strong force. The question is, how to decide the
weight of a mass particle which size and density isn’t known.
Newton established the Big-G by experimentation, perhaps to experiment by numbers works for nucleus as well.

Jaakko Kurhi

Last edited by JAAKKO KURHI : June 27th 14 at 04:53 PM.
 




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