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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0according 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!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0same 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well it may be undefined at exactly 0 distance, but what about 0.0000001 distance?, applying the formula it would certainly raise a big number.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0at exactly zero gravity is also zero because of equal forces from all the sides , cancels out and resultant gravity is zero at the center

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now only one more thing, if you dont mind

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no no...it will oscillate about the center of earth (you can consider center of earth as mean position of the oscillation)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you've been very generous sharing your time to this conversation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i love discussing concepts because i too have learned something from this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lets imagine the earth now were hollow, meaning there is only empty space surrounded by a rocky shell, it a perfect spherical earth.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it would because it is equally attracted in all directions so the net force rests in 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok now what if you were not exactly at the center, lets say halfway between the surface and the center, what happens?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343891103421:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you mean it would accelerate towards the center and then make full stop at it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes...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..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343892463645:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0look 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will have too finish here for today, I have to go, but thankyou I hope we finish discussing it later.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea sure..but i would recommend you to read it once more..that will make it easier for you..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Every 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Short 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the equation g=g'(1D/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
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