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TomLikesPhysics Group Title

I am supposed to integrate Kepler´s second Law to find T (period) as a function of R (radius of a circle). However, this is a special case where we talk about a circle and not about an ellipse. See the attachment for details.

  • one year ago
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  1. TomLikesPhysics Group Title
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    • one year ago
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  2. TomLikesPhysics Group Title
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    Here is what I did so far.

    • one year ago
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  3. TomLikesPhysics Group Title
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    But I have to no good reason to drop the sin when I integrate. However I think the end is correct. But I can not justify my steps to get there.

    • one year ago
  4. ivanmlerner Group Title
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    What did you get when you integrated?

    • one year ago
  5. TomLikesPhysics Group Title
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    The thing you see on the sol.jepg picture. On the left side it will be A(T) which is a full circle because we are talking about a circular orbit so when the Period T passes the planet is where it was before and has therefore made a full circle area which is pi*r^2. On the right side I am kind of clueless how to integrate that stuff properly.

    • one year ago
  6. Xishem Group Title
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    Alright, if we use the equation you're using: \[dA=\frac{1}{2}r v \sin(\alpha)dt\]Which of those quantities changes in time?

    • one year ago
  7. TomLikesPhysics Group Title
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    A and the angle alpha. r is constant because we talk about a circular orbit, and therefore the velocity is also constant (or the speed).

    • one year ago
  8. Xishem Group Title
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    Now, I want you to be cautious. In this example, you know that the angle changes linearly in time, but you know exactly how it changes?

    • one year ago
  9. Xishem Group Title
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    I'm curious how you derived your differential equation. Is that the equation that is given to you for Kepler's second law?

    • one year ago
  10. TomLikesPhysics Group Title
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    Yes, if you look at Kepler´s law and say that dt is very small than you can say that the area made by the radius is a triangle.

    • one year ago
  11. TomLikesPhysics Group Title
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    here is a picture of it

    • one year ago
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  12. TomLikesPhysics Group Title
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    However, I am still clueless how the angle depends on time. Is it the integral of omega (the angular velocity)?

    • one year ago
  13. Xishem Group Title
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    I'm not sure. Alpha is going to be a function of r(t) and r(t+dt).

    • one year ago
  14. TomLikesPhysics Group Title
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    alpha=omega*t

    • one year ago
  15. TomLikesPhysics Group Title
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    That one should be helpfull...

    • one year ago
  16. TomLikesPhysics Group Title
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    than I could rewrite 0.5*v*r*sin(alpha) as 0.5*omega*sin(omega*t) but the integral of that after dt is only -0.5*cos(omega*t)

    • one year ago
  17. Xishem Group Title
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    Alpha, as an angle, should be the radial angle if it's omega*t. |dw:1352577895185:dw|

    • one year ago
  18. Xishem Group Title
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    Because the angular velocity is defined as: \[\omega =\frac{d \theta}{dt} \rightarrow \omega\ dt = d \theta\]

    • one year ago
  19. TomLikesPhysics Group Title
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    but that gives me:

    • one year ago
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  20. Xishem Group Title
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    I would do it this way. Since this is a circle, you can define the area of any arc as: \[A=\frac{1}{2}\theta r^2\]If we look at a small piece of A, then it becomes:\[dA=\frac{1}{2}r^2d\theta\]\[dA=\frac{1}{2}r^2\omega dt\]

    • one year ago
  21. TomLikesPhysics Group Title
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    but I want to find T as a function of r. How does this help?

    • one year ago
  22. Xishem Group Title
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    \[\int\limits_{A(0)}^{A(T)}dA=\int\limits_{0}^{T}\frac{1}{2}r\omega\ dt\]\[A(T)=\frac{1}{2}r\omega T\]\[\pi r^2=\frac{1}{2}r\omega T\]

    • one year ago
  23. Xishem Group Title
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    Sorry, sorry. I missed an r term. One second.

    • one year ago
  24. TomLikesPhysics Group Title
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    like that?

    • one year ago
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  25. Xishem Group Title
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    Yep. Looks okay to me.

    • one year ago
  26. TomLikesPhysics Group Title
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    It seems that we only used mathematics but not Kepler´s second law.

    • one year ago
  27. Xishem Group Title
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    Well, Kepler's second law just states that the area swept out by the radius vector sweeps out equal areas in equal time intervals.

    • one year ago
  28. TomLikesPhysics Group Title
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    ... ah so we used that to claim that omega is constant!

    • one year ago
  29. Xishem Group Title
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    Usually, the more general form of Kepler's law that I use most is: \[dA=\frac{1}{2}|\vec r \times d\vec r|\]No, not exactly. We could only claim that because it's a circle.

    • one year ago
  30. TomLikesPhysics Group Title
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    But than we did not use any stuff of kepler. The formula for A follows from the unit circle and the rest was mathematics.

    • one year ago
  31. Xishem Group Title
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    It's not that. It's that we made a simplification based on our understanding of the problem.

    • one year ago
  32. TomLikesPhysics Group Title
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    Where did we simplify?

    • one year ago
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