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Speed through the Center of the Earth

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  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose we had a straight tunnel, through Earth's center, to a point on the opposite side of the planet, and used it to deliver mail to the other side. With what speed would our packages pass through Earth's center.

    2. Relevant equations
    ag = (GM)/R2
    Mins = (4/3)πR3

    3. The attempt at a solution
    So, I'm assuming (but not completely sure) that the question is asking for the man's speed at the center (but again, I could be interpreting the question completely wrong. Either way, I'm still stuck on how to solve it.)

    I figured that I could take the integral of ag with respect to R (since the distance from the center of the earth is continuously changing as you pass through the center.)

    So: ag = ∫[G*ρ*(4/3)π *R3]/R2
    (from 0 to R)

    My problem here is that ρ is just M/V which is 3M/(4πR3), which basically cancels out a whole bunch of R's and leaves you with:

    ag = GM∫[1/R2
    (from 0 to R...I'll worry about signs later)

    AND:
    ag = GM/R, which doesn't actually work out since you end up having m2/s2, which is not acceleration.

    I know I probably messed up the integral, but I'm not sure how else you would solve this problem.
     
  2. jcsd
  3. Apr 4, 2015 #2

    collinsmark

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    There are two radii to consider (this keeps much of the cancellation from happening):
    • There's the average density* calculation which you have already done, [itex] \rho = \frac{M}{V} [/itex], where [itex] V = \frac{4}{3} \pi R^3 [/itex]. Here, this [itex] R [/itex] is constant, and doesn't change as the position of the mail package progresses. You have already calculated this in your original post. (And of course, the mass of the Earth is constant.)
    • Then there's the mass of Earth "underneath" the mail package. And the volume of earth "underneath" the package is [itex] \frac{4}{3} \pi r^3 [/itex]. This [itex] r [/itex] changes with the mail package's position, however. So this mass underneath is this volume times the previously calculated [itex] \rho [/itex]. We're making the assumption that this [itex] \rho [/itex] density is constant and does not change as a function of the mail package's position.*
    What I haven't discussed is how the "shell" of earth above the mail package fits into the picture, if at all. Can you comment on what the gravitational forces felt by an object inside a massive hollow, spherical shell?

    *In this problem we are assuming that the Earth has a uniform density. This assumption isn't true, but it's not too horrible an approximation. Since we're already ignoring the Earth's rotation, friction and such anyway, it should be okay.
     
    Last edited: Apr 4, 2015
  4. Apr 4, 2015 #3
    Ok, that makes sense. I'll try to work that out now.

    I've only briefly discussed shells. But in the case of the earth, once you get into the innermost shell (of the Earth, in this case) the net force acting on the object is 0. So in a hollow, spherical shell, once you get past the outer layer, the net force is 0?
     
  5. Apr 4, 2015 #4
    I tried it again, it still doesn't work. No matter what I try, too many R's end up cancelling out:

    Even if I set it so that Mins = ρ * (4/3)π*r3
    w/ ρ = 3M/(4πR3)

    ag = (4/3) π Gρ ∫ r3dr/r2
    (from 0 to R)

    ag = (4/3) π Gρ ∫ rdr
    ag = (4/3) π Gρ * R2/2
    ag = (4/3) π * G * 3M/(4πR3) * R2/2

    ...Do you see what I mean?
     
  6. Apr 4, 2015 #5

    gneill

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    Staff: Mentor

    In order to determine velocity by integrating acceleration you must integrate w.r.t. time. Integrating acceleration as a function of position is not correct.

    Rather than deal with the chain rule and mucking about with resulting equations, why not consider a conservation of energy approach? What's the work done in moving an object from the surface to the center? You still need to deal with the mass of the planet below the current position in order to determine the force as a function of position.
     
  7. Apr 5, 2015 #6
    I'm just really confused about how specifically to deal with the mass of the planet below the current position.
     
  8. Apr 5, 2015 #7

    gneill

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    How so? The mass below the current position is the mass contained within the spherical volume with radius equal to that of the current position.
     
  9. Apr 5, 2015 #8
    I mean, how would I integrate with time when I don't even know the time it will take to get down to the center of the earth?
     
  10. Apr 5, 2015 #9

    gneill

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    That is why I suggested using conservation of energy (work-energy theorem) instead. Then time is not involved.
     
  11. Apr 5, 2015 #10
    ...So you don't use an integral, then? But how do you deal with the changing mass below you?
     
  12. Apr 5, 2015 #11

    gneill

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    No, you must use an integral. You want to compute the work done by gravity while moving a test mass from the surface to the center.
     
  13. Apr 5, 2015 #12

    mfb

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    It is closely related to the approach via energy conservation - and I agree that it is the easiest approach here.
     
  14. Apr 7, 2015 #13
    I still am completely lost. I don't understand how to make this integral, nor how to integrate it.
     
  15. Apr 7, 2015 #14

    gneill

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    What is it you need to integrate to find work?
     
  16. Apr 7, 2015 #15
    The potential energy of an object on the earth's surface down until the center of the earth?
     
  17. Apr 7, 2015 #16

    gneill

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    Not quite.

    Work = Force x Distance, right? The work done in moving a test mass from one position to another without allowing it to accelerate will be what you want. (Think of the test mass as being moved at a constant velocity, applying a force that exactly balances the gravitational force so no acceleration takes place)

    So the work done by gravity is equal to the integral of ##F_G(r) \cdot dr##. You need to find an expression for ##F_G(r)## to integrate. This is where the mass of the planet below the current position comes into play so you can use Newton's Law of Gravitation to find that force at any radial position.
     
  18. Apr 7, 2015 #17
    So:
    W = ∫(FG(r)*dr
    (from 0 to R...I'll figure out the signs later)

    W = ∫(GMm)dr/r2
    W = ∫(GρVm)dr/r2
    W = Gρm ∫(4πr3)dr/(3r2)
    W = Gρm * (4π)/3 * ∫rdr
    W = Gρm * (4π)/3 * R2/2
    W = Gm * (ρ) * (2π)/3 * R2
    W = Gm * (3M)/(4πR3) * (2π)/3 * R2
    W = Gm * M/(2R)

    ...And that's the work?
     
  19. Apr 7, 2015 #18

    gneill

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    Looks okay to me. How does that amount of work translate to a velocity for an object that undergoes the same change in location via free-fall?
     
  20. Apr 7, 2015 #19
    Can't you use the kinetic energy theorem (0.5mv2)?
     
  21. Apr 7, 2015 #20

    gneill

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    Staff: Mentor

    Try it and find out. What answer do you get?
     
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