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    mcc91's Avatar
    mcc91 Posts: 1, Reputation: 1
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    #1

    Dec 4, 2007, 09:46 PM
    Circular Motion & Law of Gravity
    An asteroid in orbit about the sun has a linear speed of 4 km/s when at a distance of closest approach d from the sun. What is its linear speed when it is at its greatest distance from the sun, a distance 2d?

    Answer = 2 km/s

    How do you get the answer?

    Must know by tonight!
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
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    #2

    Dec 5, 2007, 01:44 AM
    Is this a case of CIRCULAR motion?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #3

    Dec 5, 2007, 02:04 AM
    Use the fact that angular momentum, r x p, is conserved.
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
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    #4

    Dec 5, 2007, 03:33 AM
    If it is circular motion, with Sun as centre, then how can distance from Sun change?

    Capchin's method will not give the required answer, unless we assume that in the 2 cases velocity is perpendicular to the line joining the Sun and the asteroid.
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #5

    Dec 5, 2007, 05:28 AM
    Jiten, velocity will be perpendicular to the line joining the Sun and the asteroid at both the perihelion and aphelion (the 2 cases we are looking at), I wouldn't say this was an assumption, I would say it is plain from the question.
    It does give the required answer.

    Also jiten, I'm assuming this isn't circular motion, the OP probably just assumed that it was because they didn't know how to tackle the question.
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
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    #6

    Dec 5, 2007, 02:34 PM
    I do not see how it is clear from the question that the 2 vectors of angular momentum
    Are perpendicular to each other.

    There must be some assumption regarding the nature of the orbit, e.g. perhaps an ellipse.

    Please tell me where is it clear from the question?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #7

    Dec 5, 2007, 02:55 PM
    It's a characteristic of every orbit in nature. Can you draw an orbit of a single object around another stationary object which does not have perpendicular r and p vectors at the aphelion and perihelion?
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
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    #8

    Dec 5, 2007, 03:05 PM
    Are you claiming that the student who asked the question, and all physics students, know about this "WELL-KNOWN LAW OF NATURE?"

    Nature is full of examples where what seems obvious is NOT correct!

    For example, Relativity, Black Holes, etc.

    Many Greek scholars presumed the planets' path was circular!
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #9

    Dec 5, 2007, 03:30 PM
    I thought it obvious that at both aphelion and perihelion, p and r would be perpendicular, not so much a law of nature, but more a logical reasoning.
    Was this not obvious to you?

    How would you solve it without applying conservation of angular momentum?
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
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    #10

    Dec 5, 2007, 03:43 PM
    NO, it was NOT obvious to me, simply because it was not Mathematically obvious.

    My solution would be :

    Use Conservation of Angular Momentum, and assume that the oribit is such that vectors
    Involved are perpendicular to each other.

    OR

    The given answer is possible only if the vectors involved are perpendicular to each other.
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