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Period of falling through asteroid vs orbit

  1. Apr 20, 2014 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations

    See below

    3. The attempt at a solution
    My book derives the period of oscillation through the tunnel:
    T = 2pi/w = 2pi*sqrt(m/k) = 2pi*sqrt(3/(4pi*G*p)) = sqrt(3pi/(G*p))
    Where p is the density of the asteroid, and G is the Newton's gravitational constant.

    I know that the orbit velocity is found by equating the gravity force to the necessary centripetal force:
    mg = mv^2/r
    v = sqrt(r*g)
    So the period is 2*pi*r/v = 2pi*sqrt(r/g)

    I know these two periods are equal. Can anyone help me with putting them and similar terms and proving that they are?
     
  2. jcsd
  3. Apr 21, 2014 #2

    ehild

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    What is g on that spherical asteroid? Do you see any connection between "g" and the Universal Law of Gravitation?

    ehild
     
    Last edited: Apr 21, 2014
  4. Apr 21, 2014 #3
    If you click on "Go advanced" you will find that it is easy to use the letter π instead of the word "pi".
     
  5. Apr 21, 2014 #4
    Extra credit: what if the asteroid is differentiated, that is the density at its core is higher than at its mantle.
     
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