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Marion and thornton classical dynamics

  1. Apr 27, 2008 #1
    Hello,

    Does anyone have marion and thornton's classical dynamics book? I have a possible error that I wanted to point out. Actually, goldstein's classical mechanics book would work also since I found the same "error" in there as well.
     
    Last edited: Apr 27, 2008
  2. jcsd
  3. Apr 27, 2008 #2

    siddharth

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    I have the second edition of goldstein's book. Why don't you just ask your question here, and those who have the book/s will be able to help.
     
    Last edited: Apr 27, 2008
  4. Apr 27, 2008 #3

    George Jones

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    I have both books, but not in their latest editions. I might not be able to look at them until tomorrow.

    What error do these books have in common?
     
  5. Apr 27, 2008 #4
    OK. I am pretty sure this is not an not an error now that I found it in both books.

    equation 4.93 in goldstein:

    [tex]m\frac{d^2x}{dt^2}=-2m\omega v_z \sin \theta [/tex]

    They show that

    [tex]v_z = -gt[/tex]

    So the first equation becomes [tex]m\frac{d^2x}{dt^2}= 2m\omega gt \sin \theta [/tex]

    What I DO NOT understand is how they integrate that twice to obtain

    [tex]x = \omega g t^3 \sin \theta /3 [/tex]

    To do that, they need the initial condition [itex]\frac{dx}{dt}(t=0) = 0[/itex]. I do not understand how they get that initial condition!

    Also, in a footnote of Goldstein, they say "It is easy to show, using Eq (4.93), that a particle projected upward will fall back to the ground westward of its original launching spot." That is extremely confusing to me.
     
    Last edited: Apr 27, 2008
  6. Apr 27, 2008 #5

    Hootenanny

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    Could you be more specific? In what context is the above equation derived?
     
  7. Apr 27, 2008 #6
    x is the eastward coriolis deflection of a free-falling object in the Northern Hemisphere
     
  8. Apr 27, 2008 #7

    Hootenanny

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    What is the initial velocity of the free-falling body?
     
  9. Apr 27, 2008 #8
  10. Apr 28, 2010 #9

    Cleonis

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    Let me first discuss what is happening when an object is dropped from height.
    (In all of the following discussion I'm neglecting air friction.)

    The trajectory of an object that is released from altitude is a ballistic trajectory. It has an initial tangential velocity, from co-rotating with the Earth. You can think of the starting point of the trajectory as the apogee of an eccentric orbit.

    On its way down from apogee the angular velocity of the object in orbit increases. (Compare the extremely eccentric orbit of Halley's comet; falling towards the Sun it speeds up.)

    That is why an object released from altitude will not fall perpendicular to the ground, it will reach the surface eastward of its launch coordinates.


    Now the case of an object fired upward.
    That trajectory is like that of an orbiting object moving upward to its apogee.
    In climbing up the object loses angular velocity. So relative to the Earth the object will "fall behind". Later, during the descent, it will get that angular velocity back, but in the meantime the Earth has turned underneath all that.
    Therefore an object fired straight up will land westward of the coordinates that it has been launched from.

    There's a Java applet on my website designed for exploring precisely that kind of scenario: Ballistics and orbits

    Cleonis
    http://www.cleonis.nl
     
    Last edited: Apr 28, 2010
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