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How much force a person must exert to escape the center of the earth?

  1. Jan 27, 2012 #1
    Hi I'm new to this forum. I've always wanted to know how much force it would take a person to escape the center of the earth. Let's say someone is buried deep into the earth's mantle nearing the outer core. How much weight of rock (be specific if you can) would be upon that person's body and how much force must he exert to escape back to the earth's surface. Yeah I know it's a weird question, but I would really appreciate it if someone answered this for me...
     
  2. jcsd
  3. Jan 27, 2012 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Your question is a bit confusing (at least to me). Are you asking what the weight of the column of rock is above a point at the center of the Earth?

    Or are you asking how much work a person would have to do to climb a ladder up a hole from the center of the Earth to the surface?
     
  4. Jan 27, 2012 #3
    These dumb questions need to be deleted.
     
  5. Jan 27, 2012 #4

    berkeman

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    Staff: Mentor

    That's a fair viewpoint, and sometimes we do delete questions like this. But I responded to see if the OP could be guided to some more reasonable version of their question. We'll see how the response goes. It may be that we can teach them some basic physics, which could be a good thing. :smile:
     
  6. Jan 27, 2012 #5
    Why don't you tell him how that question is dumb?
    It is fairly relevant if you consider the amount of force required by the person to climb a hole which leads to upper crust.
     
  7. Jan 27, 2012 #6
    Actually the answer is far from obvious. At the exact center of the earth, the gravitational attraction in any direction is cancelled out (assuming that the earth was purely spherical, rather than the oblate spheroid that it is) by the attraction from an equal attraction in the other direction. If the earth was hollow save for a thin shell, this would be true at any point within the earth, as it can be shown via calculus (I'll leave that exercise to you, but you can find it in any elementary physics book), that so long as you are within the sphere, there will be no force acting on you. However, anything outside of the shell see gravity as a point source at the center of that shell.

    However if the earth is filled (as it is), the moment that you move away from that center point, things get interesting. Consider the sphere r meters in radius centered on the earth's center of gravity. This sphere now has a mass of m[itex]_{s}[/itex] = (4/3)πr[itex]^{3}[/itex]ρ, where ρ is the density of the sphere, which means the gravitational attractions becomes f = Gm[itex]_{s}[/itex]m[itex]_{p}[/itex]/r[itex]^{2}[/itex] = (4/3)Gπr[itex]^{3}[/itex]m[itex]_{p}[/itex]ρ/r[itex]^{2}[/itex] = 4Gπrm[itex]_{p}[/itex]ρ/3.

    Now the mass in question (m[itex]_{p}[/itex]) is not the mass of the person, but the mass of the column of magma and rock. To a first approximation, for an area A, the mass will be m[itex]_{p}[/itex] = A(r[itex]_{g}[/itex] - r)ρ, where r[itex]_{g}[/itex] is the radius of the earth. Substitution, you get, f = 4GπrA(r[itex]_{g}[/itex] - r)ρ[itex]^{2}[/itex]/3, or the pressure P = f/A = 4Gπr(r[itex]_{g}[/itex] - r)ρ[itex]^{2}[/itex]/3.

    Now there are admittedly a number of simplifications here. ρ for instance, is technically a function ρ(r) which varies depending upon the properties of the material and the pressure acting on that property, but that's probably getting more complex than its worth. What it does imply is that the greatest pressure is not, in fact, at the center, but at the midpoint between the surface and the center of the earth.
     
  8. Jan 28, 2012 #7
    I'm asking about the weight of the column of rock. How much weight would be upon your body if you were buried in the core of the earth?
     
  9. Jan 28, 2012 #8
    Who decides what a dumb question is?
    Recently I have seen questions similar to this asking to find the motion of a particle dropped down an imaginary tunnel through the centre of the Earth.
    That was a perfectly legitimate question
     
  10. Jan 28, 2012 #9
    Here are the data you need to answer this question:

    http://geophysics.ou.edu/solid_earth/prem.html [Broken]

    Just multiply the pressure P with the base area of the column.
     
    Last edited by a moderator: May 5, 2017
  11. Jan 28, 2012 #10
    "How much force" is a difficult question to pose. If you wanted to know some detail about coming from the centre of the Earth to the surface then the quantity you need to consider is "Energy"
    If you assume that the density of the Earth is uniform then g will vary uniformely with depth from 9.81 (I will take it to be 10N/kg in what follows) to 0 at the centre of the Earth.
    This means that the ENERGY required to raise 1kg from the centre of the Earth to the surface will be 0.5gR where R = radius of Earth = 6.4x10^6m
    this comes out at 32x10^6 Joules to raise 1kg.
    If you need to take the variation in density with depth into consideration then the data recommended by DrStupid will help.
    You will then need to plot a graph of g against depth and find the area under the graph to calculate the energy to raise 1kg. I would 'count squares' but I am sure someone will tell you how to integrate the data.
    Hope this encourages you to remain interested in physics.
    I cannot add anything to you statement about 'pressure'....sorry
     
  12. Jan 28, 2012 #11
    So do I multiply 363.850 GPA by 3900 miles by 3 feet of rock to get the total weight?
     
    Last edited by a moderator: May 5, 2017
  13. Jan 28, 2012 #12
    I really suck at Physics you guys. I read Superman comic books, and in one issue he was trapped in the Earth's core. I just want to know how much weight was on his body and how much energy did he have to output to escape back to the Earth's surface.
     
  14. Jan 28, 2012 #13
    :rofl:
    :rofl: Certainly, you have interest for it!
     
  15. Jan 29, 2012 #14
    No, you multiply 636.85 GPa by height of Superman by width of Superman.

    That does not depend on the weight of the rock above him.
     
  16. Jan 29, 2012 #15
    Actually Superman is weightless at the center of the Earth, even though the pressure is so high.
     
  17. Jan 30, 2012 #16
    The column of rock upon his body is not and that's what kbclassof09 asked for.
     
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