1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Drilling through the Earth problem

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Postal workers on planet Vashtup 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 containing mail into the tube from P.O. 1 and have others grab
    them at P.O. 2. Vashtup has g = 9.1 m/s2
    , and a radius of
    5200 km. When it is located within the shell of a planet, the
    weight of a particle of mass m is mgr/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 1088 s after
    it has been released, and b) its speed at that time. Note: r is
    positive 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.

    2. Relevant equations


    3. The attempt at a solution

    Note: My professor wants me to be using differential equations to solve this.

    I think I have a basic idea of how to do this problem, but am having trouble finding an equation that correctly models the motion of the object dropped into the tube. From what is given I have:



    a is the second derivative of position, so x''=gr/R

    Now, this is where I'm having trouble. I know I need to find r(t), or a function for position dependent on time but I'm not seeing how to make that connection.

    I tried using the above equation saying (R/g) r''(t) = r(t), y'(0)=0, y(0)=R

    but that definitely isn't right

    Any ideas?
  2. jcsd
  3. Apr 17, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Well, I don't know what those y's are doing there, might want to change those to r's, but if you rethink the signs of your equation I would not say it's 'definitely not right' because it is!

    Hint: You're sitting at r = R to begin with so think about F = ma with the correct sign (direction) in your coordinate system.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted