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Net acceleration of the Earth

  1. Aug 22, 2012 #1
    I'm trying to explain to someone why I think the net acceleration of the earth (due to the sun) is non-zero.

    My reasoning is that the velocity of the earth is constantly changing. As the sun pulls the earth's orbit into an ellipse, the direction component of velocity is being changed. Therefore velocity is changing.

    It's easy to work out the centripetal acceleration from classical mechanics, but I'm trying to explain it intuitively.

    Their reasoning is as follows:

    -The direction component of all the infinitude of velocity vectors during the earth's orbit cancel out. (for every one pointing in direction X, there's another one, half a year later, pointing in direction X + pi, etc.)
    -Since the direction components all cancel out, the velocity is reduced to the constant speed.
    -Since the speed is constant, the acceleration is zero.

    I need a little help explaining why that reasoning is flawed. I'm pretty much at my limit of ability to explain things.
     
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  3. Aug 22, 2012 #2

    Doc Al

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    You are arguing--correctly--that the instantaneous acceleration of the Earth is non-zero. They are just making the point that the average acceleration (over a year, say) is zero. Two different issues.
     
  4. Aug 22, 2012 #3
    I see.

    What about the "net acceleration" though?

    This is the term they keep using.

    I say the net acceleration is non-zero since there is velocity changing work being done to keep the earth in orbit. (as opposed to having a constant velocity)
     
  5. Aug 22, 2012 #4

    A.T.

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    And they are making that point in a needlessly complicated way. If the Earth arrives at the same velocity it had a year ago, then the average acceleration since then was zero.
     
  6. Aug 22, 2012 #5

    jbriggs444

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    That is not a term that means anything to me. How do they feel that it differs from "acceleration"? What would "gross acceleration" mean?

    If one is playing tennis on a calm day, does that mean that the net acceleration is zero?
     
  7. Aug 22, 2012 #6

    phinds

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    Have them define "net acceleration" and you will likely have your answer.
     
  8. Aug 22, 2012 #7

    A.T.

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    No work is being done on an object in a circular orbit. In a elliptic orbit the net work done over a year is also zero.
     
  9. Aug 22, 2012 #8

    phinds

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    An excellent definition :smile:
     
  10. Aug 22, 2012 #9
    I don't think I understand that. Without the sun, the earth goes in a straight line. The sun comes along and gravitationally pulls the earth into an ellipse. Does it not take energy to continually yank the earth into orbit?
     
  11. Aug 22, 2012 #10
    When calculating work, you only include the component of force in the direction of velocity. In a circular orbit, the force is always perfectly perpendicular to velocity, and thus no work is done.

    If you want to think about it in terms of energy, think about the earth's gravitational potential with the sun. The further away the earth is from the sun, the more gravitational potential it has, so it would take an input of energy in order for the earth to move further away from the sun.
     
  12. Aug 22, 2012 #11
    Ahh gotcha. Yeah I was thinking of that completely wrong.
     
  13. Aug 22, 2012 #12
    Ok so they said when they say "net acceleration" they are referring to the sum of all the vectors in a rotational period.
     
  14. Aug 22, 2012 #13

    jbriggs444

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    Once you normalize that sum so that it is not infinite, it is the same thing as "average acceleration"

    If the average taken to be is over time then that is the same thing as change in velocity divided by elapsed time.

    By inspection, the velocity of the earth on January 1 is the same as it was on January 1 of the previous year [modulo various finicky caveats]. It follows that the change in velocity is zero and the average acceleration is zero.
     
  15. Aug 23, 2012 #14
    "net acceleration" usually refers to an instantaneous acceleration which results from multiple simultaneous acceleration vectors. They're referring to the average acceleration over a year; I wouldn't call that "net acceleration."
     
  16. Sep 30, 2012 #15
    "without the sun the earth goes in a straight line" is a false statement. Both sides of the argument are flawed. The answer may be hidden in a question, the why of an elliptical orbit. the star that is the central point to our solar system has a trajectory and also a velocity dictated by? at times this star is pulling us and at other times is on a collision course with us. Theoretically the farther away a planetary body, the more elliptical the orbit (considering mass). Is there a fixed point upon this star,as it travels in a circular path that defines one revolution?
     
    Last edited: Sep 30, 2012
  17. Sep 30, 2012 #16

    mfb

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    @slayerwulfe: Your post does not make sense at all.
     
  18. Sep 30, 2012 #17
    according to ? and this probably doesn't make sense either, Schrodinger's cat: Why the box, strap it to the cats neck. Is it now both dead and alive at the same time. Theory vs. reality where theory becomes what you want it to be. Thank you for your detailed analysis of no sense at all.
     
  19. Sep 30, 2012 #18

    mfb

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    Me, and probably the majority of all other physicists who might join that thread (again).

    Let's see:

    Which argument?

    How is the question "what happens to earth without sun" related to elliptical orbits? What about parabolas and hyperbolas, the other solutions of the Kepler problem?

    The sun is not the exact center of our solar system - it is quite close in terms of the center of mass, however.

    What?
    The trajectory of the sun is determined by the gravitational influence of other objects (and partially by its mass ejections, but that is not relevant here).

    The sun always attracts earth (and everything else), and we are not on a collision course.

    This is just plain wrong. Eccentricity is independent of the mean distance to the star. See Kepler problem for details.

    Fixed point where? Which circular path do you mean, and revolution of what?


    In order to get a superposition, you have to isolate the system from its environment. The box stands for that isolation, while a real box would not provide sufficient isolation anyway.
     
  20. Sep 30, 2012 #19

    phinds

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    I was going to say the same thing, but I'm trying to cut back on my use of the word "nonsense" on this forum. :smile:
     
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