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Particle mechanics (Calculus)

  1. May 3, 2009 #1

    C.E

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    1. Hi, I was hoping someone could offer some guidance with the following, I don't even know how to start it.

    The motion of a planet of mass m around the sun of mass M is governed by the following equation (note r is a vector):

    d^2r = -k r
    dt^2 ||r||^3

    where K=Gm (G is the gravitational constant) and r(t) is the position of the planet relative to the sun.

    1. Show the following quantities are conserved.

    a. Total energy (the sum of the potential and kinetic energies of the planet).

    b. The angular momentum of the planet J.

    c.The lenz- Runge vector
    dr x J-mk r = L
    dt ||r||
    Hint: (a x b) x c = (a.c)b- (b.c)a (note a.c is scalar product of a and c similarly b.c is scalar product of b and c).

    2. (a) Interpret the constancy of J geometrically.
    (b). Assume the planet moves in an ellipse with one focus at the sun, show by considering the point when the planet is furthest from the sun that L points in the direction of the major axis of the ellipse.
     
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  3. May 3, 2009 #2

    tiny-tim

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    Hi C.E! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Hint: dot-product both sides with dr/dt, and integrate :wink:
     
  4. May 3, 2009 #3

    C.E

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    When I dot product both sides with dr/dt I get the following:

    dr[tex]^{2}[/tex]/dt[tex]^{2}[/tex] . dr/dt = -kr/||r||[tex]^{3}[/tex] . dr/dt.

    Can I simplify this? What am I integrating it with respect to?
     
  5. May 4, 2009 #4

    tiny-tim

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    No, it's simple enough as it is. :smile:

    And you're integrating with respect to t.

    Hint: "squared" :wink:
     
  6. May 4, 2009 #5

    HallsofIvy

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    dr[tex]^{2}[/tex]/dt[tex]^{2}[/tex] . dr/dt /dt.
    is of the form u du/dt (with dr/dt as u). That's easy to integrate.
     
  7. May 4, 2009 #6

    C.E

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    Ok so on the left hand side I get u[tex]^{2}[/tex]/2 is this right?
    I'm sorry but I don't know how to integrate the right hand side the ||r|| is really confusing me. Any more hints?
     
  8. May 4, 2009 #7

    HallsofIvy

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    Yes, which in this case is (dr/dt)2.

    You don't need to integrate the right side. Just recognize it as being the negative of the potential energy!
     
  9. May 4, 2009 #8

    C.E

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    I thought dr/dt was the velocity, why is its square the negative of the potential energy? How does this show energy conservation?
     
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