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Falling Through Earth's Center

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    With what speed would mail pass through the center of earth if falling in a tunnel through the center?

    2. Relevant equations

    W = -K

    K = 0.5mv^2

    (vf)^2 = (vi)^2 + 2aΔx

    3. The attempt at a solution

    My main question is why can't I use the last equation I listed (constant acceleration) to find vf where v initial is 0, a = 9.8ms^-2, and Δx = R = radius of Earth? This gives me a different answer than the book does.

    The book has vf = √(aR) whereas I have vf = √(2aR)
     
  2. jcsd
  3. Dec 14, 2013 #2

    mfb

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    a = 9.8ms-2 is valid on the surface of earth only, it is not true inside.
     
  4. Dec 14, 2013 #3
    How come the book uses the same value for their calculation then?
     
  5. Dec 14, 2013 #4
    no books will not use the value of 'g' as constant in general as the value of g changes with height with respect to the center of earth
     
  6. Dec 14, 2013 #5

    SteamKing

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    Books have been known to be wrong.
     
  7. Dec 14, 2013 #6
    Well that's great -__-. Also, if I use the conservation of energy, I still get v = √(2aR) but I don't understand how to solve it the same way my book wants me to.

    Using W = ΔK where K initial = 0 and so W = K final = 0.5m(vf)^2. If W = ∫ F dr, then what is F?

    Am I supposed to use F = GmM/(R^2) ?
     
  8. Dec 14, 2013 #7

    SteamKing

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  9. Dec 14, 2013 #8
    You may not find this helpful, but...

    As part of your course, have you written down a formula for the field strength at the surface of a planet in terms of the density and radius of the planet? If so, you can use that here.

    The trick is to realise that the strength of the gravitational field decreases as you approach the centre of the planet and and the centre is zero. If you think about what it would be like to be surrounded by a large uniform shell of matter - pulled equally in all directions - you will see that this is true.

    It turns out that when we go 'down' the tunnel into the planet we only have to worry about the mass that's enclosed by the sphere that is closer to the middle than we are. We've assumed that the planet is of uniform density so this is just like standing on the surface of a new, smaller planet. This is where your equation for g in terms of ρ and r comes in handy. You can write down an equation showing how g changes with r. Remember that g is the same as the acceleration? By thinking about the form of your equation you should recognise that this will cause a special kind of motion. Once you spot this, answering the question is easy.

    Alternatively, you could just look up what the potential energy would be in the middle of a planet and find the difference between that value and the PE on the surface.
     
  10. Dec 14, 2013 #9

    mfb

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    I am quite sure the book does not assume the acceleration to be constant everywhere inside earth.

    You start with wrong assumptions. The calculation itself is fine, but the approach is wrong.
    See the two posts above this for more details.
     
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