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Scaling The Solar System By A Factor ##\alpha##

  1. Jan 13, 2017 #1
    I am given the solution to the first part of the problem, however not the second part - would appreciate for someone to double check my work! Cheers.

    1. The problem statement, all variables and given/known data

    If a scale model of the solar system is made using materials of the same respective average density as the sun and planets, but reducing all linear dimension by a factor ##\alpha##, how will the velocity and period scale with ##\alpha##? Assume a circular orbit in your calculations.

    2. Relevant equations

    $$F = \cfrac{GMm}{r^2}
    $$F = \cfrac{mv^2}{r}

    3. The attempt at a solution

    Starting off with the basic relationship required:
    $$\cfrac{GMm}{r^2} = \cfrac{mv^2}{r}$$

    Then: ##v^2 = \cfrac{GM}{r} \propto \cfrac{G \rho R^3}{r}##, where ##R## is the radius of the Sun.

    In general: $$v^2 \propto \cfrac{R^3}{r}$$

    Re-scaling the solar system by ##\alpha##:

    $$v^2 \propto \cfrac{\alpha^3 R^3}{\alpha r}$$

    i) Therefore ##v \propto \alpha##

    Now, ##v = \cfrac{2 \pi r}{T}##. Thus:
    $$\cfrac{2 \pi r}{T} \propto \alpha$$

    ii) Therefore ##T \propto \cfrac{1}{\alpha}##
     
  2. jcsd
  3. Jan 13, 2017 #2

    BvU

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    Well, apparently ##v\propto\alpha## is the right answer. If that is so, then I don't understand why you don't let ##r\propto\alpha## in part ii).

    Oh, and a belated :welcome: !
     
  4. Jan 13, 2017 #3
    Thank you! I'm not quite sure what you mean exactly. With some reasoning I could have also said that since ##v \propto \alpha##, then ##r \propto \alpha##. It thus follows that ##T \propto \cfrac{1}{\alpha}##.
     
  5. Jan 13, 2017 #4

    BvU

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    What I mean is

    In part 1 you use ##r' = \alpha r## to find ##v'= \alpha v##
    Why don't you do that in part ii) ? to find that ##T'= {2\pi r'\over v'} = ..##
     
  6. Jan 13, 2017 #5
    In which case ##T' = T## - thus ##T## is independent of ##\alpha##? What is wrong in my reasoning that results in me obtaining ##T \propto \cfrac{1}{\alpha}##?
     
  7. Jan 13, 2017 #6

    BvU

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    You left ##r## as is. In part i) you inserted ##\alpha r##:
    [edit] By the way: I had seen this part ii) result before, here
     
  8. Jan 13, 2017 #7
    Ah, how silly of me! Looked at the result you forwarded as well - got it now. Thank you so much and all the best for the new year!
     
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