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SHM+Gravitation Problem!

  1. Feb 11, 2013 #1
    Problem: Assume that a tunnel is dug across the earth(radius=R) passing through its centre. Find the time a particle takes to cover the length of tunnel if it is projected into the tunnel with a speed of [itex]\sqrt{gR}[/itex]



    2. Relevant equations:
    Basic SHM equations:
    1.F=-kx
    2.T=(2*pi)/(omega)
    Gravitation:
    [itex]F=GMm/r^2[/itex]




    Attempt:
    V(potential due to earth)=GMx/R(R+x)
    1/2*m*g*R=GMx/R(R+x)
    x=R
    So particle will first go inside the tunnel accelerating till centre then decelerating till surface and further till it reaches distance R from surface.
    I am confused now As this is not a SHM(Outside surface the field is non linear) So how do I find the time?
     
  2. jcsd
  3. Feb 11, 2013 #2
    g`=GMx/R3
    Thats the acceleration at a distance x from the center.
    So the acceleration of the particle is directly proportional to the (-ve) of the distance from the center.
    So the particle will execute a Simple Harmonic Motion with its circular frequency, ω2=GM/R3
    Now could you work it out?
     
  4. Feb 11, 2013 #3
    But outside the surface the field is non linear, so it wont be a shm.
     
  5. Feb 11, 2013 #4
    The particle won't be able to go outside! The force of gravity will keep on increasing just enought to make its velocity 0 at the surface! Think about it
     
  6. Feb 11, 2013 #5
    But there is no change in potential energy from one end to other end as both are on surface. So shouldn't K.E. remain same?
     
  7. Feb 11, 2013 #6

    ehild

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    Read the problem carefully: it asks how long the particle is inside the tunnel.

    In the tunnel, the particle performs SHM. Why? What force acts on the particle at distance r<R from the centre of Earth? Assume that the density of the Earth is constant. You have to know that the force exerted by a homogeneous sphere is the same as if all mass enclosed in the sphere of radius r concentrated in the centre.

    ehild
     
  8. Feb 11, 2013 #7

    haruspex

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    No. At radius r from the centre of a solid sphere radius R>r (assuming each concentric shell is in itself uniform), the gravitational pull from the portion of the sphere at radius > r exactly cancels itself. So only consider the pull from the part of the earth at radius < r.
     
  9. Feb 12, 2013 #8
    yeah that's what i did to get the equation!
     
  10. Feb 12, 2013 #9

    haruspex

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    Sorry - misread it. Too hasty.
     
  11. Feb 13, 2013 #10
    Understood. But the little doubt that remains in my mind is that particle was projected with a speed in the tunnel, can the equations of SHM be used?
     
  12. Feb 13, 2013 #11

    ehild

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    The initial velocity and initial position determines the amplitude and phase constant of the SHM, but the equation ma=-kx does not change.

    ehild
     
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