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B Weight at different heights

  1. Aug 24, 2016 #1
    Let's say that a man standing on a scale on the surface of the earth weighs 1000 newtons. Now let's place the scale at the end of a very long pole extending far from the surface of earth, considering the earth to be static. Let's say that the man on the scale now weighs 100 newtons. If we were to now place the scale at the bottom of the pole at the surface of the earth with the man still on top of the pole, neglecting the weight of the pole itself, what would the scale now read?
     
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  3. Aug 24, 2016 #2

    phinds

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    What do you think and why?

    And by the way, this has nothing to do with relativity so your title is misleading.
     
  4. Aug 24, 2016 #3
    I wasn't sure what to title it, I was thinking general relativity. I'm thinking the reading will be the same as when the scale was placed at the top of the pole, but I have no idea how to show that so I'm just guessing.
     
  5. Aug 24, 2016 #4

    phinds

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    It's a good guess, and it's right, but why do you think this has anything to do with general relativity? This can be handled with classical Newtonian gravity.

    As to why it's right, it's because whichever place the scale is, it is measuring the same thing (neglecting, as you say, the weight of the pole) which is the weight of the man on top of the pole.
     
  6. Aug 24, 2016 #5
    It is weighing the same mass for the man in each case, right, but the acceleration of gravity for a mass placed upon the scale is different at each place. The mass of the man is the same at both places, so I want to be sure the reading upon the scale is due to the acceleration of gravity for where the man is placed rather than the acceleration of gravity for where the scale is placed. Are you sure the acceleration of gravity at the location of the scale itself has nothing to do with it?
     
  7. Aug 24, 2016 #6

    phinds

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    Why would it? It's not measuring anything where the scale is, it's measuring the thing at the top of the pole. If the pole were not there, the man would be in free-fall. What would his weight be then?
     
  8. Aug 24, 2016 #7
    Well for free-fall it would be zero I suppose. Okay great, so we are saying that in general relativity, there is no transition of the measure of force acting upon different radii, that weight is invariant when the location of the mass remains the same?
     
  9. Aug 24, 2016 #8

    Nugatory

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    The pole is not accelerating, it's just sitting there. Therefore the net force on the pole is zero.

    There's a downwards force of 100 N on one end of the pole, so that has to be balanced by an upwards force at the other end. That force comes from the spring in the scale, which is compressed enough to exert an upwards force of 100 N on the pole - so that's what the scale reads.
     
  10. Aug 24, 2016 #9
    Wonderful :) thanks guys. It seemed too simple an assertion for general relativity so I wanted to be sure.
     
  11. Aug 24, 2016 #10

    phinds

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    I keep asking you, WHY do you think general relativity has ANYTHING to do with this? It does not. This is a classical problem plain and simple.
     
  12. Aug 24, 2016 #11
    Well, I was working through some stuff with general relativity to begin with is all, and this topic popped up. I needed to know how a force measured locally would translate as measured at a different radii. Apparently it is invariant, but I didn't want to just assume it. Thanks :)
     
  13. Aug 24, 2016 #12

    phinds

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    In general relativity, gravity is not a force.
     
  14. Aug 24, 2016 #13

    Nugatory

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    That is not correct.

    Even in classical mechanics, there is no such thing as a force measured locally being translated to some other position - the concept has no meaning. In your problem, there is a force on the molecules at the top of the pole from the weight of the man; by Newton's third law these molecules exert an equal and opposite upwards force on the man and that's what holds him up. These molecules in turn exert a completely separate force on the molecules below them, and by Newton's third law the lower molecules exert an upwards force on the upper ones.... and so on all the way down to the bottom of the pole, where the bottommost molecules exert a force on the spring scale.

    Now, you have made the starting assumption that nothing is moving (the man is help up, the pole isn't bending or being crushed under the weight, ....). That can only happen if the net force on any part of the pole is zero, meaning that the downwards force on it from above is equal to the upwards force on it from below. Work that down every step of the way and you will see that the upwards force from the scale on the bottommost molecules of the pole has to be equal to the downwards force from the man on the topmost molecules of the pole. But that equality has nothing to do with the force being translated from one radius to another - they're completely different forces, and they just turn out to have the same strength because you set the problem up so that they would.
     
  15. Aug 24, 2016 #14
    Right. I was referring to weight.
     
  16. Aug 24, 2016 #15
    Okay cool :)
     
  17. Aug 24, 2016 #16

    phinds

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    And you think that weight has nothing to do with force? Look, I'm not trying to give you a hard time here, I'm trying to get you to see why the word relativity should never have entered this discussion, much less been repeated several times by you.
     
  18. Aug 25, 2016 #17
    I was thinking about this today and have run into a conceptual problem. The idea of the measure of weight being linear so measured the same anywhere within a gravity well sounded good, nice and simple for calculations, just integrate for the mass times the acceleration of gravity at each height for each point along a mass within the well and we would gain the total force acting upon the bottom of that mass. Let's say a boulder for example. We would just find the weight at the top where lesser acceleration of gravity acts by multiplying that mass times the acceleration of gravity there, then add to that the mass a little further down from the top with a slightly greater acceleration of gravity, and so on til we reach the bottom and that would give us the total weight acting upon the bottom of the boulder.

    Well, let's say we did that for the earth. We could take the mass of the top layer of earth (neglecting the weight of air and considering the surface of earth to be smooth), say one meter from the surface, multiply that times the surface acceleration of gravity, and that would give us the weight pushing down upon the spherical shell one meter from the surface. If we divide that by the surface area at r_earth - 1 m, which would be A = 4 pi (r - 1 m)^2, then we would gain the radial pressure acting upon that spherical shell. If we do that again for two meters down, multiplying the mass between the spherical shells at one and two meters times the slightly different acceleration there, then add that to the weight of the top layer, we now gain the force of weight pushing down upon the second spherical shell at r - 2 m. Dividing that by the area of the second shell, A = 4 pi (r - 2 m)^2, we gain the radial pressure at that spherical shell. It would be about twice that of the first shell since there is approximately twice the weight acting upon almost the same surface area. This is about right for pressure underwater and so forth, with the pressure increasing almost linearly with depth, so the idea of linear force of weight seems okay so far.

    But now let's consider all the way down almost at the center of the earth. At a spherical shell one meter out from the center, almost the entire integrated weight of the earth falls upon this shell, giving to incredible radial pressure at that shell. That still sounds okay because we know the pressure is large. But the problem I am having is if we move to the spherical shell two meters from the center, almost the same weight acts upon this shell as the first, but over four times the surface area, so that the pressure two meters from the center of earth should be four times less than that at one meter. And at the center itself, the pressure should work toward infinity if the weight is linear. But it would seem to me that the pressure within such a small volume at the center of the earth should be pretty much the same. So the force of the weight could not be linear there. What is the solution to this dilemma?
     
  19. Aug 25, 2016 #18

    jbriggs444

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    You are treating the sides of a wedge-shaped column of earth and stone as if those sides are vertical (which they each are) and as if they are parallel to the net force of gravity on the wedge (which they are not).
     
  20. Aug 25, 2016 #19
    Right okay, i thought about that. The only way to keep the forces linear is if some of it is transferred to tangential tidal force, right? So how do we find for that?
     
  21. Aug 25, 2016 #20

    Nugatory

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    It's not a tidal force, that's something completely different.

    At the pressures and temperatures you'll find near the earth's core, neither solid rock nor iron (which is what much of the core is made of) are rigid the way the pole in your previous thought experiment was rigid; instead, they behave like incompressible liquids. They will flow freely and change their shape until all the forces are balanced but they will always have the same volume. The pressure at any point in the liquid will be the same in all directions (google for "Pascal's Law", and we have a bunch of threads on that).

    Now imagine an upside-down pyramid that we've filled with water: What is the pressure at various depths, and what force does it exert on the sides of the pyramid? We can model the earth's core as a collection of these pyramids, all arranged so that their points are at the exact center of the earth. The force exerted against the sides of one pyramid from the pressure of the fluid inside is balanced by the pressure from the fluid in the neighboring pyramid.
     
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