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Planet's Orbits.

  1. Jan 28, 2012 #1
    There was one question i got in an interview, and I couldnt answer it at all:

    "There are 1 million stars, and 10 planets orbiting each star. Consider that the planets do not disturb each other, and the planet systems, including the stars, dont disturb each other. Then, how many of these planets will have perfectly circular orbits?"

    I said that its a case of coincidence, they replied:

    "Scientists dont stop at coincidence, they find its probability. Please tell how we can find the answer."

    :confused:
     
  2. jcsd
  3. Jan 28, 2012 #2

    DaveC426913

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    It isn't about the answer, it's about how you go about answering it.

    One question I would have is what is the tolerance for "perfectly circular". Because the orbital eccentricities will surely fall on a bell curve, so how many depends on how accurate we want.

    But you don't ask the question, what you do is pick a value and state your assumption.

    "OK, let's say 'perfectly circular' means an eccentricity of .01 or less, so..."

    They want to hear how you use your skills to crack an egg, even if the egg is hypothetical.
     
  4. Jan 28, 2012 #3

    D H

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    That is a nonsensical and nonscientific question, which is what my first answer would have been. If pressed, I would have responded with "Let's me answer that with the answer to your next question, 'What is your greatest fault?' I have a hard time dealing with nonsense. It makes me get snippy and short. I have learned to watch for that, but it is still trying."

    If pressed even further I would have answered that if the problem is looked at classically, with semi major axis and momentum drawn randomly from some distribution with a continuous CDF, the subset of orbits that are perfectly circular is a space of measure zero. Therefore, none of them. Looking at it quantum mechanically, semi-major axis and momentum are conjugate variables. There is no way to know whether an orbit is perfectly circular.


    If pressed even further with that nonsense, I would have gone back to option #1 and said something short and snippy such as "Scientists don't ask questions such as 'What do the laws of physics say will happen when the laws of physics are violated?'"
     
  5. Jan 30, 2012 #4

    No answer of any consequence may be formulated, as the associated principles pertaining to radial gravitational fields are violated by the very nature of the question. How then, would one formulate probabilities pertaining to that which violates long observed reality? Hence, probability concerning that which is, in essence, “make believe” is an exercise in futility consequently; the question is unanswerable to any degree of accuracy or expectation.
     
  6. Jan 30, 2012 #5

    tony873004

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    As others have mentioned, the answer is 0, unless you put a constraint on "perfectly circular" as Dave suggested. e=0.01 is circular enough that you couldn't visually discern that it wasn't perfectly circular. Earth's eccentricity is higher than this, yet Earth's orbit is out of round by less than 2 Earth diameters.
     
  7. Jan 30, 2012 #6

    Janus

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    Er, the difference between Earth's perihelion and aphelion is some 5,000,000 km or ~292 Earth diameters.
     
  8. Jan 30, 2012 #7

    tony873004

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    Perihelion and aphelion are measurements of Earth's distance to the Sun. The Sun doesn't sit at the center of the ellipse. It's on a focus. Earth's semi-major axis is about 21000 km greater than its semi-minor axis.
     
  9. Jan 30, 2012 #8

    DaveC426913

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    Source please? I'd have thought Earth's semi-minor axis would be easy to look up but it turns out to be quite elusive.

    This is the closest I've found:
    http://answers.yahoo.com/question/index?qid=20090524063151AA4lBso
    but it does claim that the diff as ~20,000km.
     
    Last edited: Jan 30, 2012
  10. Jan 30, 2012 #9

    D H

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    It's easily calculable. Semi minor axis is given by [itex]b=a\sqrt{1-e^2}[/itex]. The difference between semi major and semi minor axes is thus [itex]a-b=a(1-\sqrt{1-e^2})[/itex]. For the Earth, e is 0.01671123, a is 1.00000261 AU, and thus a-b = 20,890 km.
     
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