Force and Space (Their Relation)

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Hi,

Let's say a 10 kg particle (X) whose volume is 1 m3, and another 10 kg particle (Y) whose volume is 1*10-24 m3.

Using the general force formula (F = GMm/r2), the same force is applied if I stand 1 meter before X or Y, but Y has smaller volume so it has to exert more force, right?

If so, how can I calculate that?
Thanks for help.
 

Answers and Replies

  • #2
Doc Al
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Using the general force formula (F = GMm/r2), the same force is applied if I stand 1 meter before X or Y, but Y has smaller volume so it has to exert more force, right?
No, why do you think that? (Note that that formula only applies strictly to point masses or for spherically symmetric mass distributions.)
 
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No, why do you think that? (Note that that formula only applies strictly to point masses or for spherically symmetric mass distributions.)
Because, from what I know, black holes exert much stronger force than the time they were stars. Stars becomes compressed to singularities to form black holes.
 
  • #4
Doc Al
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Because, from what I know, black holes exert much stronger force than the time they were stars.
They exert a stronger force at their surface, since their radius is so much smaller. But replace an ordinary star with a black hole of the same mass and it will exert the same gravitational force on you as long as you stay the same distance from it.
 
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They exert a stronger force at their surface, since their radius is so much smaller. But replace an ordinary star with a black hole of the same mass and it will exert the same gravitational force on you as long as you stay the same distance from it.
I'm little confused, now if a star comes over the horizon of black hole's sigularity then the star is sucked because of strong force exerted by the surface? How comes a surface of particle playing role in exerting force?

In newtonian physics using the force formula. Is (r) the distance between two centered points of particles? If so, what is the point of a mass? Since (r) never equals zero...
 
  • #6
Doc Al
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I'm little confused, now if a star comes over the horizon of black hole's sigularity then the star is sucked because of strong force exerted by the surface? How comes a surface of particle playing role in exerting force?
I'm not sure I understand what you're asking. A black hole is tiny--thus you can get way too close, where the gravity is overwhelming. A black hole with a mass equal to our sun would have a Schwarzschild radius of about 3,000 m.

In newtonian physics using the force formula. Is (r) the distance between two centered points of particles? If so, what is the point of a mass? Since (r) never equals zero...
Again, I'm not sure what you're asking. Think of the 'r' being the distance between two spherical bodies. Shrinking one of those bodies into a point won't change the distance between them.
 
  • #7
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I'm not sure I understand what you're asking. A black hole is tiny--thus you can get way too close, where the gravity is overwhelming. A black hole with a mass equal to our sun would have a Schwarzschild radius of about 3,000 m.
I mean, how is the gravitational force very strong if you are within black hole's event horizon? Even if its mass is about our sun's mass.
 
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Again, I'm not sure what you're asking. Think of the 'r' being the distance between two spherical bodies. Shrinking one of those bodies into a point won't change the distance between them.
I mean, for example: Two charged spheres are attached/attracted to each other, is the distance between them 0 or the sum of their radius (sphere(1) radius + sphere(2) radius)?

I think that the distance never gets to zero, so the force approaches infinity as the distance approaches zero.
 
  • #9
Doc Al
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I mean, for example: Two charged spheres are attached/attracted to each other, is the distance between them 0 or the sum of their radius (sphere(1) radius + sphere(2) radius)?
I don't know why you've introduced charge, but for two spherical masses the gravitational force will be calculated using the distance between their centers.
 
  • #10
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I don't know why you've introduced charge, but for two spherical masses the gravitational force will be calculated using the distance between their centers.
I did that just for make it less arguable, because If it were gravitational force then someone might respond that the force between the two masses is too small, unless the masses were have been said to be huge enough.

My central problem is: If a black hole exerts the same force as a same-mass star, then how the black hole absorbs everything, even light?
 

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