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Angular momentum and central force

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves under the influence of a central force
    F(r)=-mk[(3/r^2)-2a/r^3]rhat

    Show that if the particle is moving in a circular orbit of radius a, then its angular momentum is L=mh=m√(ka)
    2. Relevant equations
    L=mvr for circular orbit

    3. The attempt at a solution
    From the equation, we extract the second derivative of the position vector r''=-k[(3/r^2)-2a/r^3]rhat. We integrate this in order to find the velocity to use in L=mvr.
    (At this point, can I dot product both sides by rhat to get rid of the vector?)
    r''=-k[(3/r^2)-2a/r^3]
    Integrating:
    r'=-k[-3/r+a/r^2]=-k[(-3r+a)/r^2]
    Using the angular momentum equation L=mvr=m*r'*r:
    L=m*-k[(-3r+a)/r^2]*r.
    We have r=a, so
    L=m*(-k[(-3a+a)/a^2]*a)=m2k.

    So obviously, the integration is wrong because I haven't included the integrating constant and also the answer we get is wrong. And intuitively by substituting in r=a, we are saying that r doesn't change.
    How could I correctly approach this problem?

    Thank you!
     
  2. jcsd
  3. Nov 24, 2015 #2

    BvU

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    That would be strange: ##\vec L \equiv \vec r\times \vec p## so dot product would be zero !

    What is essential for a circular orbit ?
     
  4. Nov 24, 2015 #3
    dL/dt=0 because there is no external force acting on the particle.
     
  5. Nov 24, 2015 #4

    BvU

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    That's right, but: think back a lot further, when you first learned about circular motion. This is really a very easy exercise.
    Note that the radius of the orbit is given!
     
  6. Nov 24, 2015 #5
    I've found the answer by using the equation of orbital motion -r=(r^4F(r))/(h^2m) where h=|L|/m. Rearranging the equation and setting r=a, I got L=msqrt(ka)

    Thank you :)

    If you don't mind me asking another question (using the same central force):

    The particle is projected from point A, at distance a from the centre of force O, in a direction perpendicular to OA, its projection velocity being half of the velocity required for a circular orbit of radius a.
    Compute the projection velocity and use it to determine the angular momentum.

    The equation I have from the notes is velocity=sqrt(-aF(a)/m) assuming that r=a. Since the projection velocity is half that, v(projection)=(1/2)(sqrt(-aF(a)/m)). Then using the fact that v=h/a, we solve for h, and then use h=L/m to compute for L.
    Are the steps correct?
     
  7. Nov 24, 2015 #6

    haruspex

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    As BvU indicated, it's a bit simpler than that. Just write that the force equals the necessary centripetal force.
    The second part seems even simpler. You have already found the angular momentum for a circular orbit at radius a. You are told the initial velocity is now half the velocity that corresponds to that angular momentum. So how large is the angular momentum?
     
  8. Nov 24, 2015 #7
    Would the angular momentum be just 1/2 of that of the angular momentum we found before? So (m/2)(sqrt(ka))
     
  9. Nov 24, 2015 #8

    haruspex

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    Looks right to me. The only thing that bothers me is that it seems too easy.
     
  10. Nov 24, 2015 #9
    Great! Thank you :) It gets a bit trickier. I was able to derive the radial equation of motion as r''-(9/4)ka/r^3=-3k/r^2 using the equation r''-h^2/r^3=Fr(r)/m and rearranging a bit.

    Then the next question asks to define u(θ)=1/r(t) and show that r''-(9/4)ka/r^3=-3k/r^2 could be expressed as u''(θ)+9u(θ)=12/a.
    Do I just let r(t)=1/u(θ), differentiate it twice to find r''(t) in terms of u(θ) to substitute into the equation of motion? Even though u and r have different values, namely θ and t, am I allowed to just use chain rule? For example, would r'(t)=-u'(θ)/u(θ)^2? I get an unusual coefficient for r'' if I do it that way.

    The question after that asks to solve the equation obtained (u''(θ)+9u(θ)=12/a) and show that the particle follows the orbit:
    r=3a/(r-cos(3θ)).
    I am not quite sure how to start this one.
    Any help will be appreciated as always!
     
  11. Nov 24, 2015 #10

    haruspex

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    Yes, but you have to be consistent about the independent variable with respect to which you are differentiating. If wrt t then terms like dθ/dt will appear: (d/dt)u = (du/dθ)(dθ/dt). So you need to find a relationship between r, θ and t which will provide a substitution for dθ/dt.
    You could try multiplying through by u'. Another approach is to drop the inhomogeneous term on the right of the equals and see if the equation reminds you of anything.
     
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