Force and Space (Their Relation)

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SUMMARY

The discussion centers on the relationship between force and space, particularly in the context of gravitational forces exerted by different mass distributions, such as black holes and stars. Participants clarify that the gravitational force formula (F = GMm/r²) applies to point masses or spherically symmetric mass distributions, and that a black hole with the same mass as a star exerts the same gravitational force at a given distance. The confusion arises from the nature of black holes, which can exert overwhelming gravitational forces due to their compact size and the effects of their event horizon.

PREREQUISITES
  • Understanding of Newtonian physics and gravitational force calculations
  • Familiarity with the concepts of point masses and spherically symmetric mass distributions
  • Knowledge of black hole physics, including event horizons and Schwarzschild radius
  • Basic grasp of gravitational interactions between masses
NEXT STEPS
  • Study the implications of the Schwarzschild radius in black hole physics
  • Explore the differences between gravitational forces of black holes and ordinary stars
  • Learn about the concept of event horizons and their significance in gravitational theory
  • Investigate the mathematical derivation of gravitational force between spherical bodies
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Students of physics, astrophysicists, and anyone interested in understanding gravitational forces, particularly in relation to black holes and their unique properties.

Atran
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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.
 
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Atran said:
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.)
 
Doc Al said:
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.
 
Atran said:
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.
 
Doc Al said:
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...
 
Atran said:
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.
 
Doc Al said:
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.
 
Doc Al said:
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.
 
Atran said:
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.
 
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Doc Al said:
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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