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Integration help please

  1. Oct 3, 2004 #1
    How do I intergrate this?

    [tex] t - to = \int_\Theta ^\Theta(t) \frac{m}{L} R(u)^2 du [/tex]

    u is [tex] \Theta(t) [/tex], [tex] \Theta [/tex] is the angle at time zero, and t is time. Can I solve for R in terms of t?

    Thanks.
     
    Last edited: Oct 3, 2004
  2. jcsd
  3. Oct 3, 2004 #2

    arildno

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    What you've written is completely incomprehensible.
    What are "theta" and R?
    What is u and t?
     
  4. Oct 3, 2004 #3
    I think he is doing solids of revolution. that is the only thing that I can think of as to why there would be an R in there...but from theta to theta makes no sense to me.
     
  5. Oct 3, 2004 #4
    Sorry for the confusion of the integral. I am new to Latex...and I don't know how to do the "integral from theta to theta(t)" (theta of t). I didn't mean theta to theta. =\
     
  6. Oct 3, 2004 #5

    arildno

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    A good guess; I think it is some sort of separable differential equation he started up with, but I'm not sure..
     
  7. Oct 3, 2004 #6

    arildno

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    Odyssey: Could you state the original problem?
     
  8. Oct 3, 2004 #7
    Thanks for the help. I appreciate it! =D

    OK, my problem is actually much more complicated than the integral.

    Suppose a point mass m experiences a net force F = -Amr^-3.
    How should I go about in describing its orbits for E>0, E <0, and E = 0 for the non-zero angular momentum cases? :confused:

    My prof gave me an example on the gravitational force law, which is an inverse square law. I am asked to find out how the particle would move (say Earth) if it is under the influence of an inverse CUBIC law. The integral I (tried to) give above is the angular momentum of the particle. And yes, it is a differential equation! :)
     
    Last edited: Oct 3, 2004
  9. Oct 3, 2004 #8

    arildno

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    Is this how it should be?
    [tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R^{2}(u)du[/tex]
     
  10. Oct 3, 2004 #9
    Precisely! Yes, that's how it should be! :rofl:


    Hummmm, I did some research on the Net, and I *think* it wouldn't be very pleasant if we live in a 1/r^3 world. The Earth would spiral into the Sun!?!?!? I am asked to find the equation of the orbit of such a world. Can you give me some pointers so I can start this problem please?

    Thx for the help again.
     
    Last edited: Oct 3, 2004
  11. Oct 3, 2004 #10

    arildno

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    Allright!
    So, am I correct in assuming that "R" is the unknown radius in some orbit expressed as a function of the angle?
     
  12. Oct 3, 2004 #11
    Yes, precisely. R is a function of the angle, which is not limited to just between 0 and 2pi.

    Let me get some more info on the problem. That might help! :) (will be posted here in minutes)
     
    Last edited: Oct 3, 2004
  13. Oct 3, 2004 #12

    arildno

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    Well, then you can't integrate any further.
    If you don't know what you integrate, you can't know what the integral would be.
    (That is, what you've got is a dead-end)
     
  14. Oct 3, 2004 #13
    Total energy is conserved for a conservative force.

    E = (1/2) m [r'(t)]^2 + (1/2) m [r(t)]^2 [theta'(t)]^2 + V (r(t))

    Angular momentum is conserved for a central force.

    L = m [r(t)]^2 [theta'(t)]^2

    we can view r may be a function of theta, assuming that L is non-zero.

    then E = (1/2) m[R'(theta(t))]^2 [theta'(t)]^2 + (1/2) m [R(theta(t))]^2 [theta'(t)]^2 + V(R(theta(t)))
    and
    L = m [R(theta(t))]^2 [theta'(t)]
     
  15. Oct 3, 2004 #14
    Hummm, under a 1/r^3 force, and using angular momentum/total energy conservation (writting the equations in polar coordinate form too), I separate the variables theta and t, and try to integrate to solve for R in terms of t (in trying to get an equation of the orbit). Am i heading into the right direction? That's how I got the integral at the very top in the first place.
     
  16. Oct 3, 2004 #15

    arildno

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    True enough; try to eliminate your time dependence, (using angular momentum equation),
    [tex]\frac{d\theta}{dt}=\frac{L}{mR^{2}}[/tex]

    You should then by able to find a differential equation for R expressed in the new independent variable [tex]\theta[/tex]
    (Basically, this is the same technique used to find the orbits for an inverse-square field)
    Good luck!
     
  17. Oct 3, 2004 #16

    arildno

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    Note:
    DO NOT KEEP t AS THE INDEPENDENT VARIABLE!!
    Switch to using the angle.
     
  18. Oct 3, 2004 #17
    OK, let me try to work this out........thanks for the help arildno!! =D

    I need the potential V(R) too. Just to verify the work. Is V(R) = Hmr^-2?
     
    Last edited: Oct 3, 2004
  19. Oct 3, 2004 #18

    arildno

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    If H is some constant related to your A, yes; then the potential goes as the inverse square of the radius.
     
  20. Oct 3, 2004 #19
    Actually I just found out the integral should be

    [tex]t-t_{0}=\int_{\theta_{0}}^{\theta(t)}\frac{m}{L}R(u)^{2}du[/tex]

    So do I still treat R as a constant? No?
     
  21. Oct 3, 2004 #20
    OOps! Sorry, I used two variables for the same constant! A = H! Silly me. :tongue:
     
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