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Thread: How long would it take for something to fall from one light year away into the sun

  1. #31
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    Also, I don't think it's possible to invert it explicitly, and get an expression for position as a function of time.

  2. #32
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    There, I think I like this one better.


  3. #33
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    Quote Originally Posted by 21st Century Schizoid Man View Post
    I saw the link with your derivation, but didn't realise there was a second page

    That would have saved me a lot of time and trouble . . .

    My method was rather different though.
    Always good to arrive at the same result by different routes, though. Gives you confidence in the Universe and your own maths.
    What did you do?

    Grant Hutchison

  4. #34
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    Quote Originally Posted by Hornblower View Post
    I found the Sky and Telescope report that included Brian Marsden's orbital determination. While still far from the solar system on the way in, it was in an extremely elongated ellipse with a period of about 5 million years. That is consistent with an aphelion of about a light year and e within a gnat's whisker of unity.
    That's interesting and, at 1 lyr., it's far enough to make me wonder if the comet was originally a "drifter (interloper) that came to town"?

    During the flyby the comet lost a slight amount of orbital energy to the planets and ended up in a much less elongated ellipse with a period of some 75,000 years. That would be an aphelion of about 3,600 AU, or 0.06 light year.
    Looks like it's here to stay. What a difference a few planets can make.
    We know time flies, we just can't see its wings.

  5. #35
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    Quote Originally Posted by George View Post
    That's interesting and, at 1 lyr., it's far enough to make me wonder if the comet was originally a "drifter (interloper) that came to town"?

    Looks like it's here to stay. What a difference a few planets can make.
    In theory an encounter with the Sun and other stars could have tweaked a slow interloper into that orbit but my hunch is that the odds are against it.

  6. #36
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    Quote Originally Posted by grant hutchison View Post
    Always good to arrive at the same result by different routes, though. Gives you confidence in the Universe and your own maths.
    What did you do?

    Grant Hutchison
    I am not sure I understand your very first step, with the equation of motion. Is it based on conservation of energy? (Also, what is “s”?)

    But in any event, I began with F=m a. F comes from inverse square law, a is second derivative with respect to time.

    Unable to solve this differential equation directly, considering time as the independent variable. However, rewriting it all with position as the independent variable, it works a lot better. Quickly got velocity, same as yours. From then on, derivation was similar, except that you used a change of variables and worked it that way, I asked Mathematica to solve it. I got a bizarre solution with complex numbers all over the place, but from the nature of the problem, we know there is a real solution. Focussed on the real part of the solution, ignoring the imaginary part and relearning how to take the logarithm of a complex number, simplified, and ultimately found something quite similar to my answer. However, my answer gave negative time as a function of distance, so it described an object moving away, not toward, the sun. Made the appropriate and very simple modification.

    Verified solution satisfies F=Ma.

    I think your method includes more physical constraints earlier than mine, and thus leads to the solution in a shorter path. So I think I recommend your method, unless one likes the scenic route.

  7. #37
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    Quote Originally Posted by 21st Century Schizoid Man View Post
    I am not sure I understand your very first step, with the equation of motion. Is it based on conservation of energy? (Also, what is “s”?)
    S is displacement. It's standard notation in school physics textbooks hereabouts. We were taught a crop of formulae for accelerated motion, relating velocity, acceleration, time and displacement, which I still know by heart:

    v = u + at
    s = (u+v)t
    s = ut + at
    v = u + 2as


    where u is initial velocity. Setting u to zero in the last formula gives v =√(2as), which is the equation I integrated with respect to distance, switching in r as my distance variable to reflect the fact that distance was being measured radially.

    Grant Hutchison
    Last edited by grant hutchison; 2019-Aug-31 at 08:26 PM. Reason: remembered the word "displacement"

  8. #38
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    Quote Originally Posted by 21st Century Schizoid Man View Post
    Also, I don't think it's possible to invert it explicitly, and get an expression for position as a function of time.
    Yes, I think we have an analogy to Kepler's equation for elliptical orbits, which requires iteration to derive a position from a time.

    Grant Hutchison

  9. #39
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    Quote Originally Posted by grant hutchison View Post
    Yes, I think we have an analogy to Kepler's equation for elliptical orbits, which requires iteration to derive a position from a time.
    Im guessing the intervals must be quite small to achieve accuracy when bear the Sun.

    iPhone on un-airborne plane
    We know time flies, we just can't see its wings.

  10. #40
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    Quote Originally Posted by George View Post
    Im guessing the intervals must be quite small to achieve accuracy when bear the Sun.
    You don't need time intervals - it's not that sort of iteration. It's just a matter of honing down on the value of d that corresponds to the desired t. The solution will be as precise as you care to make it, and is independent of earlier values of t and d.

    Grant Hutchison

  11. #41
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    Quote Originally Posted by grant hutchison View Post
    S is distance. It's standard notation in school physics textbooks hereabouts. We were taught a crop of formulae for accelerated motion, relating velocity, acceleration, time and distance, which I still know by heart:

    v = u + at
    s = (u+v)t
    s = ut + at
    v = u + 2as


    where u is initial velocity. Setting u to zero in the last formula gives v =√(2as), which is the equation I integrated with respect to distance, switching in r as my distance variable to reflect the fact that distance was being measured radially.

    Grant Hutchison
    Your formulae look to me like they are based on constant acceleration, which is not the case here (although you do reach the correct answer). So I am thinking of how to rewrite them, and get the following. (Does your use of bold suggest vectors? If so, I'm just going to do the one-dimensional version below.)



    Arg - think I'll skip the second one.





    Not exactly the same. However, differentiating with respect to ,



    Multiplying both sides by , and changing the independent variable to ,



    Then integrating with respect to ,



    or



    So, works for me. (I realise now I used both for initial velocity, and as a variable of integration. I'm too lazy to go back and change it.)

    But the reason I thought it was conservation of energy is, kinetic energy is



    But a force needed to move the object up is (negative because the acceleration is in the opposite direction from the gravity). Then potential energy is force acting through a distance, so



    Lower limit of integral left unspecified, because the choice of the "zero level" of potential energy is arbitrary. Total energy is conserved, so adding them up, the derivative with respect to should be zero,



    This can be rewritten as



    which is the same result as before. So it can be derived through conservation of energy.

  12. #42
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    Yes, potential energy per unit mass in an inverse-square field is -GM/r.
    So the kinetic energy per unit mass gained in falling from distance D to distance d is GM(1/d - 1/D).
    Multiply by two, take the square root, and you have the velocity at distance d, if velocity was zero at D.

    Grant Hutchison

  13. #43
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    Quote Originally Posted by 21st Century Schizoid Man View Post
    Your formulae look to me like they are based on constant acceleration, which is not the case here (although you do reach the correct answer).
    They are indeed based on constant acceleration.
    But the velocity formula I quoted provides the structure to go on and integrate acceleration with respect to displacement, which gives you a result that is half the difference in squared velocities (initial to final). So it's a way of calculating work from force times distance, and in this case all tied up with conservation of energy in an inverse-square gravitational field, as you point out, and I used it to get myself a final velocity from a stationary start.

    I could certainly have been more explicit, and the argument from conservation of energy would have worked fine too, but those two pages were just intended to be a personal aide-memoire for the solution of a problem I'd found tricky, in terms of finding a good substitution to solve the integral. (I have a whole file of pages like that!)

    Grant Hutchison

  14. #44
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    Quote Originally Posted by grant hutchison View Post
    Yes, I think we have an analogy to Kepler's equation for elliptical orbits, which requires iteration to derive a position from a time.

    Grant Hutchison
    Kepler himself encountered that complication when preparing his original planetary position tables. He worked from equally spaced position angles and calculated corresponding unequally spaced times. With iterative interpolation he could find the position for a given time to an arbitrarily high precision. This was covered by Sky and Telescope in 1971, the 400th anniversary of his birth.

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