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Central Force II

  1. Dec 16, 2005 #1
    The orbit of a particle in a central field is known to obey the following relationship:

    r = A/(1+sin(theta))

    a) determine the form of the central force F(r) that is responsible for this motion.
    b) What is the distance of closest approach between the particle and the point that acts as the origin of the force? What is the furthest distance that the particle can be found form the origin of the force?

    a) After applying the equation of motion, you get f(r) = -(A^3*l^2*m)/r^2

    But for part b, how do I find the r-min?

    Also A is not mentioned as a constant so do I assume it is?
     
  2. jcsd
  3. Dec 16, 2005 #2

    siddharth

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    For part b,

    You have
    [tex] r(\theta)=\frac{A}{1+\sin \theta} [/tex]
    So for what value of [itex] \theta [/itex] is r minimum and what is the minimum value?
    HINT: You know [itex] \sin \theta [/itex] can only take values between -1 and 1.
     
    Last edited: Dec 16, 2005
  4. Dec 17, 2005 #3
    What is the distance of closest approach between the particle and the point that acts as the origin of the force?

    R is min at theta = -pie/2, therefore R = 0

    What is the furthest distance that the particle can be found form the origin of the force?

    R is max at theta = pie/2, therefore R = A/2

    Is this correct?
     
  5. Dec 17, 2005 #4

    siddharth

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    No, it is not correct.
    When [tex] \theta = \frac{-\pi}{2} [/tex], [itex] \sin \theta [/itex] is minimum (-1).

    Now,
    [tex] r = \frac{A}{1+\sin \theta} [/tex]. So 'r' will not be 0 when [itex] \theta = \frac{-\pi}{2} [/itex].
     
  6. Dec 17, 2005 #5
    Okay so R is minimum when theta is equal to pie/2, thus A/2


    But R is max when theta is equal to pie, thus R max = A ?
     
    Last edited: Dec 17, 2005
  7. Dec 17, 2005 #6

    siddharth

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    Nusc,

    What is the value of [itex] r [/itex] when [tex] \theta = -\frac{\pi}{6} [/tex] ?

    Compare this value to your "r max".

    Can you now figure out what happens to [itex] r [/itex] as [tex] \sin \theta [/tex] approaches [itex] -1 [/itex]?
     
    Last edited: Dec 17, 2005
  8. Dec 18, 2005 #7
    My bad, so when theta = -pie/6 Rmax = 2A
    When theta = pie/2 R min=A/2


    Can you now figure out what happens to r as sin(theta) approaches -1 ?

    sin(theta) = -1 if theta is equal to -pie/2

    R = A/(1 + sin(-pie/2)) = A/(1-1) = A/0

    How can r not be zero?
     
    Last edited: Dec 18, 2005
  9. Dec 18, 2005 #8

    siddharth

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    No, that's not right at all.

    I did not say A/0 (Which is not 0). I meant as the denominator approaches 0 (ie, really close to 0 but not 0).

    What is the value of the fraction as the denominator approaches 0? Try using a calculator to find this value for smaller and smaller values of the denominator. Can you see any relation?
     
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