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Physics Diff'eq Word Problem, Help Setting Up

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Question in full detail:

    Postal workers on planet Gortak want to drill a straight
    tube through the planet, starting at Post Office 1, passing
    through the center of the planet, and ending on the other side
    at Post Office 2. They plan to release small packages contain-
    ing mail into the tube from P.O. 1 and have others grab them
    at P.O. 2. Gortak has g = 9.6 m/s^2, and a radius of 6400 km.
    When it is located within the shell of a planet, the weight of a
    particle of mass m is m*g*r/R, where r is its distance from the
    center of the planet. Assume that there is no air resistance.
    Compute a) the position r of the package 1883 s after it has
    been released, and cool.gif its speed at that time. Note: r is posi-
    tive if the package is on the same side of planet as P.O. 1, and
    negative if it is on the same side as P.O. 2.


    Attempts:

    I don't want anyone to really solve it, but what I want is just help setting up the differential equation or at least a push in the right direction.

    I believe its going to be a second order diff'Eq, and related to using F=ma.
    so ma=mg since there is no air resistance.
    m dV/dt-mg=0, no external forces other than gravity so it should be a homogenous equation. I get stuck here because I feel im missing something in the equation, and not sure how to take into account the object going back and forth.

    Oh wait maybe its a Simple harmonic oscillator problem?
    Any help would be nice, its due in about 6 days. Again no solutions to the Diff'eQ please. I want help in setting up the equation or helpful advice in the right direction. My professor said to use Mathematica for part of it so I don't know if that means the solutions are only going to be obtained numerically or graphically.

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 15, 2013 #2

    jedishrfu

    Staff: Mentor

    One thing you need to consider is that as the mail falls toward the center then the effective mass pulling it down decreases until it passes thru the center of the planet then as it climbs out toward the other post office the effective mass increases in other words its a function of the distance from the core center.

    F = GM(r)m/r^2

    with M being the mass of Gortak

    M(r) = density * 4/3 * pi * r^3
     
  4. Apr 15, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Is "the weight of a particle of mass m is m*g*r/R, where r is its distance from the center of the planet" actually in the problem? That is not true- the weight is proportional to the mass of the planet between it and the center of the earth. And that is proportional to the volume and so to (r/R)3, not r/R.
     
  5. Apr 15, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The mass is proportional to r^3, but the distance of r^2 to the center is the distance to this mass. Combined, the force (~m/r^2) is proportional to r, assuming a constant density of Gortak.

    A force which is proportional to the distance from some point is a very common problem, and has nice solutions...
     
  6. Apr 15, 2013 #5
    Yeah, that's how it is stated in the problem.
     
  7. Apr 15, 2013 #6
    So I'm thinking I will have to do two DEs? as the mail gets closer to the center, the mass will be changing? So dM/dr= density * 4/3 * pi * r^3 ------>dM= density * 4/3 * pi * r^3*dr.

    Then plug in above, use m*dv/dt for force and solve it again.
     
  8. Apr 15, 2013 #7
    actually nvm, F= mgr/R. Says right in problem. So I can do the DE of that with dv/dt=m*g*r/R, then do another DE with the previous answer to get its position.

    I guess what I'm not understanding is how to account for the oscillating motion of the mail from one end of the planet to the other.
     
  9. Apr 16, 2013 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The solution r(t) will include that oscillation. You just have to plug in the given time to get the position.
     
  10. Apr 16, 2013 #9
    Okay thank you! I got it now. It came to a cosine wave.
     
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