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Planet's Orbits. |
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| Jan28-12, 08:39 AM | #1 |
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Planet's Orbits.
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."
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| Jan28-12, 08:45 AM | #2 |
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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. |
| Jan28-12, 09:19 AM | #3 |
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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. |
| Jan30-12, 12:48 PM | #4 |
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Planet's Orbits.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. |
| Jan30-12, 02:03 PM | #5 |
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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.
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| Jan30-12, 02:23 PM | #6 |
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| Jan30-12, 03:37 PM | #7 |
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| Jan30-12, 06:12 PM | #8 |
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This is the closest I've found: http://answers.yahoo.com/question/in...4063151AA4lBso but it does claim that the diff as ~20,000km. |
| Jan30-12, 06:25 PM | #9 |
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