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Planetary Motion proof

  1. Oct 28, 2014 #1
    1. The problem statement, all variables and given/known data
    qvp4Q9t.png

    3. The attempt at a solution
    Honestly I'm completely lost, this is an assignment for my calc 3 class. I tried to do (a) but I think I'm completely off track so any helps appreciated.

    r x a=0
    |r x a|=0=|r||a|sin(theta)
    r cannot be 0 since it's in the denominator
    Assuming neither G nor M is 0, a cannot be 0
    so theta must be 0 or 180
    |r||a|sin(0)=0
    |r||a|sin(180)=0

    the above is my attempt at (a)

    I haven't tried any others yet :(.
     
  2. jcsd
  3. Oct 28, 2014 #2

    BvU

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    I don't see a proof in your attempt. You start with r x a = 0 which is what you are supposed to prove !?
    Since you have an expression for ##\vec a## the proof isn't all that dificult ...
     
  4. Oct 28, 2014 #3
    Uploading photos of more attempts
    x7j88aq.jpg HIGcRuC.jpg AKNzfd2.jpg
     
  5. Oct 28, 2014 #4

    BvU

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    Oh boy, a lot of reading to catch up. But I am delighted with a) and b): exemplary !
    I hope someone else steps in now to help you further; I really must get some sleep.
    What I ever knew abour planetary orbits has sunk so deep that I need some time to dredge it up; on the other hand: from what I've glimpsed it looks as if you're doing just fine. Any particularly big question marks left ?
     
  6. Oct 28, 2014 #5
    for (i) it says to compare |r|=|r x v|^2/(GM+|c| cos(theta) to the formula in the text book a(1-e^2)/(1+ecos(theta)), not sure what I'm suppose to do here.

    I'm also insecure about the correctness of my work because I can't remember all the cross product and dot product properties to see if I'm actually doing them correctly or I'm doing illegal operations.
     
    Last edited: Oct 28, 2014
  7. Oct 28, 2014 #6

    Ray Vickson

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    PF discourages the posting of pictures of your handwritten work. Some helpers (myself included) will not look at such responses. I know it is a hassle, but you really should type it out, and you can use LaTeX if you want to produce top-quality output.
     
  8. Oct 29, 2014 #7

    BvU

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    I fully agree with Ray. Perhaps you can pick up some TeX know-how on the way.
    See this link from the guidelines. What helps me is to move to a formula and use the right-mouse button | Show math as | TeX commands.

    However, I do want to help you, so let's plod on step by step

    In c) you have to prove something, namely what's on your first line. I find it hard to believe $${d\hat u\over dt} = {1\over |\vec r|}\; {d\vec r \over dt}$$ since that ignores a term $$ \left ({d\over dt}{1\over |\vec r|}\right )\; \vec r \; .$$ So your proof is therefore not correct, but repairable.

    In d) you start with the equality that is to be proven and fortunately continue with the rule to be used (we readers don't have page 624 at hand...):$$\vec u \times \left ( \vec v \times \vec w\right ) = \left (\vec u \cdot \vec w \right )\vec v - \left (\vec u \cdot \vec v \right ) \vec w$$It's vague and murky in the picture, but I think you want to fill in the expression for ##\vec a = -{GM\over |\vec r|^2}\hat u## and the result of c) to get $$\vec a \times \left [ \vec r \times \vec v \right] = -GM \left [ \hat u \times \hat u \times {d\hat u\over dt} \right ]$$With ##\hat u \cdot \hat u = 1## and ##\hat u \cdot {d\hat u\over dt}=0## you're in business. Nifty. I think you should clarify why it is that ##\hat u \cdot {d\hat u\over dt}=0## ?

    e) f) g) and h) are a piece of cake. I start to like this exercise; perhaps I will finally learn something about Kepler orbits by just letting you do all the work and sitting back comfortably :)

    In i) you are supposed to find expressions for a and e in terms of the variables used thus far: GM, ##\vec c##, ##\theta## (already there)

    By now, I am able (and hopefully you are too ) to read the Wikipedia article critically and find back all the stuff we've been through!
     
  9. Oct 30, 2014 #8
    Hey BvU, thanks for the help. I appreciate it. Anyway the project was due yesterday so I didn't have a lot of time retype everything. Mr.Vickson I will take the advice and be sure to type it in a better format in the future.
     
  10. Oct 30, 2014 #9

    BvU

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    My pleasure. Nice exercise; interesting subject, great Physics with a capital P
    :)
     
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