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santistebanc Group Title

if gravitational force increases with the inverse of the distance squared, does it mean the force is near infinite as distance approches 0?

  • one year ago
  • one year ago

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  1. mathavraj Group Title
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    according to te equation we get infinity but it doesnt mean that it is infinite..it is undefined at 0 distance...thats why we call infinity with other name undefined..if the force is infinite it is absurd the whole universe will collapse if you merge two rocks!!

    • one year ago
  2. ghazi Group Title
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    same is the case if you talk about center of charge, electric field is also inversely proportional to the square of distance ...so electric field should be infinite at center..but it ain't so...

    • one year ago
  3. ghazi Group Title
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    At the center of the earth, you would not feel any gravity. This is because the gravitational pull from every region of the earth is exactly counteracted by the gravitational pull from the corresponding region on the opposite side of you. This all adds up to a great bug zero.

    • one year ago
  4. santistebanc Group Title
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    well it may be undefined at exactly 0 distance, but what about 0.0000001 distance?, applying the formula it would certainly raise a big number.

    • one year ago
  5. ghazi Group Title
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    at exactly zero gravity is also zero because of equal forces from all the sides , cancels out and resultant gravity is zero at the center

    • one year ago
  6. santistebanc Group Title
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    the only explanation to this I can think of is that maybe it does approach infinity but it only happens for an infinitesimal amount of time which implies nothing extraordinary.

    • one year ago
  7. ghazi Group Title
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    there is no infinite gravity ...just follow a basic free body diagram at the center of earth ..and you're gonna get zero gravity..that is at the center of earth a body will be having damped oscillation and finally stable , with no gravity

    • one year ago
  8. santistebanc Group Title
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    ok well, gravity does the trick, but what about those of electrostatics? if you only had 2 opposite charged particles, they also use the inversely proportional to the square of the distance between them.

    • one year ago
  9. ghazi Group Title
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    see that's the different thing ...now you're talking about center of two charges it's a system...and distance between charges can't be zero ..because here, to keep the system stable charges will be at a certain minimum distance and if they will come closer forces between them will act in reverse direction (you can study potential energy curve)

    • one year ago
  10. santistebanc Group Title
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    you mean two oppositely charged particles (one + another -) will start repelling each other after they surpass a certain limiting distance?, like bounce off each other as if they were same charge? wouldn't they keep on bouncing off and back again indefinitely forever?

    • one year ago
  11. ghazi Group Title
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    yes exactly after surpassing a certain distance they will repel to get a position on minimum potential energy and they won't keep bouncing ...as soon as they reach that distance at which potential energy is minimum they will be stable...and bouncing is in case of gravity..that is if you will reach at the center of earth then bouncing will occur

    • one year ago
  12. santistebanc Group Title
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    ok I can live with that explanation. Thank you. And just out of curiosity, how is that bouncing with gravity? for example if you managed to build a tunnel or a hole that went straight to the center of the earth passed through it and to the other side of it, drop some object into it, wouldnt it just fall and exit through the other end, or would it bounce as you implied?

    • one year ago
  13. ghazi Group Title
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    well i manage to create a cavity or a hole and if i am dropping something there then the component of gravity at the other side will definitely try to pull it on its side but the opposite side will also apply force (gravitational force) to hold it back ..therefore object will keep swinging back and forth...|dw:1343890012680:dw| you can see that both the component will keep pulling object at its side...

    • one year ago
  14. santistebanc Group Title
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    oh ok, yes that is what I believed would happen, I thought that by "bouncing" you meant falling till reaching the center and then bounce off back to the same place where the object was dropped.

    • one year ago
  15. santistebanc Group Title
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    now only one more thing, if you dont mind

    • one year ago
  16. ghazi Group Title
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    no no...it will oscillate about the center of earth (you can consider center of earth as mean position of the oscillation)

    • one year ago
  17. santistebanc Group Title
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    you've been very generous sharing your time to this conversation

    • one year ago
  18. ghazi Group Title
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    i love discussing concepts because i too have learned something from this

    • one year ago
  19. santistebanc Group Title
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    lets imagine the earth now were hollow, meaning there is only empty space surrounded by a rocky shell, it a perfect spherical earth.

    • one year ago
  20. santistebanc Group Title
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    You have argued that if you were at the center of the earth there would be 0 gravity, which i think works exactly the same with the hollow earth

    • one year ago
  21. santistebanc Group Title
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    it would because it is equally attracted in all directions so the net force rests in 0

    • one year ago
  22. santistebanc Group Title
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    right?

    • one year ago
  23. ghazi Group Title
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    right

    • one year ago
  24. santistebanc Group Title
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    ok now what if you were not exactly at the center, lets say halfway between the surface and the center, what happens?

    • one year ago
  25. santistebanc Group Title
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    |dw:1343891103421:dw|

    • one year ago
  26. ghazi Group Title
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    then at that point you will be attracted towards the center...but as soon as you'll reach the center then gravity will work on you in all the directions and resultant force on you will be zero

    • one year ago
  27. santistebanc Group Title
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    you mean it would accelerate towards the center and then make full stop at it?

    • one year ago
  28. ghazi Group Title
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    yes...it will be accelerated at center but at center due to it's inertia it will swing back an forth...that's why i said it will oscillate..

    • one year ago
  29. santistebanc Group Title
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    ok ok, but why is it attracted towards the center if in the center there is nothing? wouldn't it be attracted to the surface walls which are the real source of attraction?

    • one year ago
  30. ghazi Group Title
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    see physically at center of earth there is a molten material..it's nothing like the center of earth is hollow....so it is obviously attracted to the surface...

    • one year ago
  31. santistebanc Group Title
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    that is one theory I had too, but I would like you to consider (and later discuss) the other theory I had which I am beginning to believe true: Maybe there is 0 gravity anywhere inside the hollow earth because the side that is closer pulls you stronger than the other farther side, but it is compensated by the larger surface that the other side has. So the forces all along might sum 0.

    • one year ago
  32. santistebanc Group Title
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    let me draw this

    • one year ago
  33. santistebanc Group Title
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    |dw:1343892463645:dw|

    • one year ago
  34. ghazi Group Title
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    now just see in your picture ..if that object reaches to the center of the earth all the forces will be equal and in all the directions therefore resultant will be zero

    • one year ago
  35. santistebanc Group Title
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    yes I know that, but now I want to know if anywhere is the same as in the center, meaning there is 0 gravity everywhere inside the sphere.

    • one year ago
  36. ghazi Group Title
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    i didn't get you?

    • one year ago
  37. santistebanc Group Title
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    look we know that the gravitational force increases as an object gets closer and closer to another and decreases as it goes away. My point is that as you can see in the drawing the object has more arrows pulling it away from the wall than pulling it closer to the wall, but those many arrows are weaker than the closer ones. I venture to suggest that if you sum the vectors of the forces you will end up with a 0.

    • one year ago
  38. ghazi Group Title
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    wait wait....gravity is concentrated at the center of earth how you are considering walls here...ideally center is a point and point is so small that it can't be divided further so if center of mass of any object coincides with the center of gravity it will experience zero gravity..that's it

    • one year ago
  39. santistebanc Group Title
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    I will have too finish here for today, I have to go, but thankyou I hope we finish discussing it later.

    • one year ago
  40. ghazi Group Title
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    yea sure..but i would recommend you to read it once more..that will make it easier for you..

    • one year ago
  41. fwizbang Group Title
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    For what it's worth, the Earth isn't a point charge, so the 1/r^2 force Law doesn't apply inside of it. The gravitational field/force on an object would be very small though(zero for a perfectly spherical earth.) The 1/r^2 doesn't go to infinity because, when you get to short distances, classical physics breaks down and you need to consider quantum effects that reduce the strength of the force.

    • one year ago
  42. ghazi Group Title
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    @fwizbang check this out ... acceleration due to gravity in terms of G, g=Gm/r^2 ..and who said earth is a point charge ...there was just an analogy in the the variation with respect to the radius between electric charge and earth...that's it

    • one year ago
  43. Dennylama Group Title
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    Every atom in the earth has a tiny gravitational field. gravitational force is strongest near the surface of the earth, and decreases to zero as you approach the center due to the gravitational pull of the atoms you've already passed. At any point inside the earth, gravitational force would be determined by the force vectors of every atom in the earth applied to your position.

    • one year ago
  44. fwizbang Group Title
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    @ghazi g = Gm/r^2 if r> Radius of the earth, which doesn't apply as r goes to zero. OK, I should have said point mass, but the basic idea is that, one way or another, the 1/r^2 force law(either electric or gravitational) is inapplicable at short distances.

    • one year ago
  45. ghazi Group Title
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    is it?

    • one year ago
  46. ghazi Group Title
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    if it is so then kindly define point and if you are defining for point mass it will surely be defined for R=zero and the basic thing is ..at R=0 resultant force from every side cancels out and net effect is zero

    • one year ago
  47. Dennylama Group Title
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    Short distances, as they say, are relative. At the center of a black hole exists, theoretically, a singularity where the force of gravity (as r approaches 0) does approach infinity, as predicted by the formula. An earth based application is more complicated.

    • one year ago
  48. fwizbang Group Title
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    That is what GR predicts, but the infinity is telling you that classical GR breaks down and that you need a new theory(quantum gravity or something) where everything works out to be finite.

    • one year ago
  49. Joseph91 Group Title
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    by the equation g=g'(1-D/R) ,where D is the depth and R is the radius of earth. By this we get g=0 at centre. because depth is measured from the surface. hence we get that R=D And we know that F=mgh since g=0, F=0

    • one year ago
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