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#37
Sep1104, 01:12 PM

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I have another good question, (at least I find it puzzling). I agree that the pull of the earth on either side of you will be equal and opposite so you wont go anywhere. But to a rigid body, it WILL feel too forces trying to pull it appart internally. But at the same time, the body will also feel the force pushing inwards due to the pressure of all the material above it. Do you think these two forces would be exactly equal in magnitue and opposite in direction, and so you really would feel weightless, despite all the hydrostatic pressures? You wouldent be crushed due to the weight maybe.



#38
Sep1104, 03:06 PM

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#39
Sep1104, 03:17 PM

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Well, there will be two forces due to gravity wont there. You will have the gravitational force of all the mass above you pulling you upwards, and you will also have a gravitational force of all the mass below you pulling you downwards. The sum of these two forces add to zero. But for a body at the center, it will not be accelerated in any direction since the net force is zero, but that does not mean that it wont feel a tug on each side where the force is present. I am refering to someones post where two people tug on your arms. The net force is zero, but both of your arms sure do feel a force in either direction. By analogy, the force tugging on your arms could be the force of gravity above and below you. You dont go anywhere, because the net force is zero, but wont you feel it on both sides.



#40
Sep1104, 03:27 PM

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What post are you referring to? 


#41
Sep1104, 03:49 PM

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someone earlier said something about that in tab 1 of this post. You should see it there.



#42
Sep1104, 06:17 PM

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Perhaps you're thinking of Hurkyl's analogy in post #10? If so, he's saying something quite different.



#43
Sep1104, 09:09 PM

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I imagine the specific gravity of liquid nickel is very high (8.902?). If you were to make it to the center of the earth, you would likely be forced towards the upper layers of the liquid mantle.
Edit: Solid inner core? ...my bad sorry. 


#44
Sep1104, 09:32 PM

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hey doc al, maybe im just interperting something incorrectly then. I thought that newtons law states that [tex] F = \frac{GmM} {R^2} [/tex]. Lets do a simple case. Lets say im at the center of the earth, and there is just a column of earth above my head and below my feet that all has the same uniform density. The weight that is trying to push down on my head will equal the weigh trying to squash me from me feet, I think we both agree on that point. And that force would just equal the sum of the mass at its distance, times the gravity at that distance, added together over the distance. That would be the force pushing me down from my head, likewise there would be that force equal in magniude present trying to push me from my feet as well.
Now for the gravity. I understand that the NET gravitational force I will feel will be any mass that is below me and is unacounted for above me. What i mean is that if i dont stand directly in the middle of that column of earth, there will be more earth one one side of me ( either above or below). As a result, there will not be a proper balance of attraction between myself and all the particles of earth, so there will be a net force in the direction where there is the larger amount of earth, and I will feel like I want to fall in that direction. If I were in the center, there would be equal mass and distance both above and below me, so I would not feel any net gravitational force. You said it does not have to do with net gravitational force, but I do not see how you can avoid it. I thought it is TOTALY dependent on having that net force cancel out. For every particle trying to pull me towards it above my head, there is the same amount of force trying to pull me towards the column of earth below my feet, and for the reason, the NET force on me is zero, and I go nowhere. But that does not necesairly mean that i would not feel myself trying to be pulled both up and down does it? Hmm, now after all this typing I kind of see what you mean .haahhaaha. I will post it anyways. (SIGHHHHHHHHHHH) . If every particle on my body is feeling as thought it is being pulled both up and down, then as a whole, each particle in my body relative to each other, dont want to go anywhere. I think what I was thinking about would be a gradient across my body. Because I take up space, I could not really be a point masss at the center, therefore, Techincally speaking, I should feel some small component of force from my head down to my toes. The thing was that I was considering the entire space I occupied as the center. In reality the center is a point, so a rigid body like me cant be the center, by definition, some parts of my body will be off center, by 3 feet or so on each side, im a tall guy :). So the parts of my body that are off center should feel some force due to the unbalance of force above and below it, but were talking about maybe 3 feet difference at most, I would think this to be a very very very very small force difference, and in affect, so small I would still think that my entire body is weightless. Thanks Doc Al you were right. I did not think about the situation hard enough. As a side note, Shouldent that single particle feel a force on each side trying to pull it appart like the anlogy I made earlier? Would this force be strong enough to rip electrons off of that point mass, and be a reason why the center of the earth would have a liquid plasma? Or am I still looking at the situation incorrectly. Edit: I just made the same exact mistake to this situation. Each electron relative to the atom will not experience a great force tugging at it, thus this cannot be the reason for the plasma. The pressure has to be the only result of the plasma. great pressure = great temperature = energy to rip electrons from their orbits and turn stuff into a plasma state. 


#45
Sep1104, 10:48 PM

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Perhaps the problem here is that all points within your body cannot be in the exact center of thwe Earth; some will be above it , and some below. Is that the heart of your question?



#46
Sep1104, 11:32 PM

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I think Cyrus is trying to describe tidal forces! :)
They will be extremely weak however as the tidal force varies as the gradient of the gravitational force. 


#47
Sep1104, 11:57 PM

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what is a tidal force? Its 12:55 am, I am studying general chemistry. I just went to 711 and bought a sobe energy drink and now im wired and good to go for another two hours. Tommorow morning is going to suck! ;P Anyhoo, I would think that if i were off center from the center of the earth by, say one atom, then the only force due to gravity would be the equivelant of the force between me and that atom and our distance squared, since all the other mass' gravitaional affects can be ignored, since they have a counterpart on the other side of me. So it really would be a weak force even if i am off center by just a tiny amount no? I made this into a hot thread, 600 views yes!



#48
Sep1204, 12:09 AM

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If gravity is pulling more strongly on one part of an object than on another part of the object then the object is effectively being pulled apart due to the differential. That is the tidal force.



#50
Sep1204, 07:54 AM

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#51
Sep1204, 02:21 PM

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HI Doc AL, I have a question for you. You said that the only gravity pulling on a body would be within that sphere due to the mass within the shere itself. Could this also be equal to the mass outside the sphere that does not have a counter part, if we take the origion to be the place of displacement of the particle. I attached a pic to show you what i mean. We could do like you said, and account for each point mass inside the blue sphere and its distance from the center of mass. But is it not equally correct to say that we could consider the point that is off center of the sphere to be defined as the new center of the sphere. And this new center point could have a max radius that would be equal to radius of the big sphere, minus the displacement from the center of the big sphere. So that it would now be possible for me to draw a new sphere of smaller radius inside the bigger sphere, centered at the new point. So now all the mass inside my sphere of radius 22mi, has a counterpart on the opposite side to cancel out the force gravitiationaly. The only parts that dont have something to cancel them out would be the highlighted green area. If I did the point mass times the distance squared for each of these green point masses and added them up, shouldent I get the same value of gravitational force as if we did it your way, by using all the point mass inside the blue sphere?



#52
Sep1204, 02:44 PM

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I think I'm missing your point with your offcenter sphere, so I can't comment. Is it a hollow sphere? An imaginary sphere? Beats me. Please note that all of these arguments about zero gravity inside a hollow cavity within the earth depend on a spherically symmetric distribution of mass.
I'll repeat: Assume a spherically symmetric distribution of mass, of radius R. Now imagine an imaginary sphere of radius r (r < R) cocentered with the big sphere of radius R. The mass outside of the small sphere will exert no gravitational field within the small sphere. So any gravity experienced at any point within the small sphere is entirely due to the mass within that small sphere. (We are of course ignoring any other bodies in the universe.) 


#53
Sep1204, 03:05 PM

P: 4,780

Hey Doc,
Did you click on the picture i attached that displays what I mean. For simplicitys sake lets just think of them as circles in a common plane. At that every point inside the big circle has a point mass that is equal. According to what you said, my red dot, is a point mass that is not located in the center of the big circle. The black dot IS the center of the big circle. According to you, the gravitational force will equal to the sum of all the point masses inside the circle that I can draw around the red dot, with a center of the black dot. (this corresponds to the blue circle in the picture.) I hope this part is clear. Now, what I am saying is that, what if i IGNORE the black dot, and consider my RED dot as the center of some arbitrary circle. Then the maxiumim circle I can draw with the center at the RED dot without going outside of my big circle will equal the radius of the big circle, minus the distance from the red dot to the black dot. Now If I draw this circle, (based on my picture), it has a radius of 22miles. And what I am saying is that at the red dot inside this circle of 22 miles, there should be no gravitaional forces present because for every point inside the circle of 22 miles, there is the same point opposite it to cancel out any net force due to gravity. So clearly, it can be seen that the only things that cant cancel out, will be the points that lie within the green area. And what I am wondering is that if I sum the distances squared times the GMm for each particle of the green area, and sum it for all of the points in the green space, should this not produce a gravitational force that is equal to the way we do it your way, by including the effects of all the points in the blue circle and adding them up. It says 0 views on my picture so I think you did not see it in my previous post. Please click on it as it will make things alot clearer than my confusing text ( sorry about that ) 


#54
Sep1204, 06:13 PM

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I get you now.
Yes, I'd say you are correct, assuming that the large sphere has uniform density. If you want to find the gravitational field at point x (red dot) then it is equal to the field from a sphere of mass of radius 3 miles (in your example). But you can certainly draw a sphere around the red dot and say that all mass inside the sphere will contribute no field at the red dot. So the sum of the field from the remaining mass in the big sphere must add up to the same. Very clever! 


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