Galileo was technically WRONG

Discussion in 'Physics & Math' started by RJBeery, Oct 1, 2014.

  1. OnlyMe Valued Senior Member

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    True, but he was applying the results of the large object case to conclusions drawn from a small object case, Galileo's. The results of the small object hypothetical would not differ in any amount, no matter which FoR you choose.
     
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  3. Neddy Bate Valued Senior Member

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    True, for small objects. Thus the only question is whether RJBeery wanted to limit the scope to small objects or not. It is evident from the OP that he did not want to limit the scope that way.
     
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  5. Neddy Bate Valued Senior Member

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    If it is your position that the effect on small objects is too small to measure, then that is fine. I only object to people saying that it is somehow invalid to calculate the two different accelerations and add them together, the way I showed. If one of the accelerations is not measurable, it will be zero, and that can be added to the other acceleration just fine.

    I should also note that I am considering an arbitrarily small drop distance, so that the accelerations can be considered to be constant during the drop.
     
    Last edited: Oct 7, 2014
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  7. Trippy ALEA IACTA EST Staff Member

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    This is only true if your reference frame is that of the center of mass. In both the reference frame of the earth, and the reference frame of the golf ball, the center of mass accelerates towards the center of the earths mass.
     
    Last edited: Oct 7, 2014
  8. Aqueous Id flat Earth skeptic Valued Senior Member

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    Correct. When Fednis asked this, I sarcastically sniped "gravity", recognizing that the c.m. translates across the field of the Earth, and that this had to constitute acceleration. As soon as you posted this I realized my thinking error: that the field itself moved, since the ball is an element of that field; that the thing which I was using as my origin got away from me, while I was busy railing on imaginary presumed trolls, and it became the enemy's origin. exchem was right -- I was reacting as a paranoid.

    Sorry everybody for getting shrill. Sorry RJ, Fednis, Neddy, tashja and the rest of you fine folks. And since I was borderline nasty I should get sanctioned. Trippy, I'd accept a card or a ban. I think it would help the bad folks out there to see me willingly take my medicine.

    Now I only need to convince myself that Galileo's frame of reference, even if it started out inertial, must have become non-inertial. And for that I'll apply the rule I had in mind -- that the field at his origin has to be homogenous and isotropic. And now I see what my altered mind was ignoring: that the ball itself disturbed the field at Galileo's origin, invalidating it by my own criteria.

    Thanks rpenner and exchem for correcting me.

    Yeah, sorry, Neddy. I was willing to accept it up to the point I got wrapped around the axle believing you were translating across the field. What you're saying was Fednis' main point which I attacked fiercely. He must of been taken aback having just posted his innocuous suggestion of another way to look at the problem, as a simplification.

    Again, sorry folks, for being a -- well, something worse than a blowhard.
     
  9. Fednis48 Registered Senior Member

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    Wow, this thread got civil quickly! Apology accepted, Aqueous Id; I'm still a little skeptical that there's a swarm of creationist trolls out there trying to sow discord in the science-forum community, but if you thought I was such a troll, I can see why you'd treat me with hostility.

    The resolution of the inertial frame issue has left me with a question, though. It's a bit off topic, so I'll start a new thread. Don't worry: I'll frame it as a pure thought experiment from the OP, so we won't have to waste any time arguing over whether Galileo would have given a $#&* about the conclusion!

    Please Register or Log in to view the hidden image!

     
  10. OnlyMe Valued Senior Member

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    True, and it was one of the reasons I raised the FoR issue early on... Without much success.

    Acceleration is a tricky subject because it can be thought of as absolute where it can be measured with an accelerometer, which is of no use in any part of this problem.., everything is in free fall... Or it can be relative when observed or measured as a change in distance between two objects or coordinates.

    But as rpenner pointed out.., in my own words, when looking at the two body problem as a whole, the center of mass remains stationary while the earth and dropped objects fall toward it. Whatever it looks like from any FoR that much remains true.

    And I still don't believe that anything here invalidates Galileo's conclusion, in fact or technically, because in his case you could not distinguish between the earth's center of mass and the two body center of mass, before or after the drop.
     
  11. Trippy ALEA IACTA EST Staff Member

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    You haven't met Chinglu or Wellwisher yet, have you?
     
  12. Trippy ALEA IACTA EST Staff Member

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    Agreed.
     
  13. paddoboy Valued Senior Member

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    And that's overlooking the ones that still remain partially closeted, although still rather obvious.
     
  14. brucep Valued Senior Member

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    My claim is based on GR. Gravity is a local theory of gravity. Where the gravitational mass of M and m interact is local. They both contribute to the local spacetime curvature [gravity]. The change in the local spacetime curvature determines the path of the falling object not the rate which it falls. It's the same for both bodies being modeled. The earth orbital path is determined by the local spacetime curvature. Deviation in the path is associated with a change in the local spacetime curvature. For example when another body contributes to the local spacetime of the earth the earths path is 'perturbed'. It's not perturbed because the bodies are accelerating towards each other as Newtons model predicts. The perturbation is a result of the the local spacetime curvature telling the objects how to modify the inertial path. I don't think you're a crank. You have more knowledge about physics than I do just not gravitational physics. So which analysis makes more sense for evaluating the physics of local inertial frames. GR or Newton. But I get it. In this forum when I introduce GR everybody clams up. There's an example of how the GR analysis corrects the Newton analysis in the thread I started which is being ignored. Like I said I get it. LOL. Sorry for calling you a crank.
     
    Last edited: Oct 8, 2014
  15. OnlyMe Valued Senior Member

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    They need a cancel button for the reply topic. This new engine seems to remember what you have been trying to forget.
     
  16. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    (shakes head)
     
  17. rpenner Fully Wired Staff Member

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    My bad. I neglected a square root when transcribing and got suspicious answers that turned out to be wrong.

    Correction: \(t(m_2, r_1, r_2, h) = \sqrt{ \frac{(r_1 + r_2 + h)^3}{2 G m_2}} \cos^{\tiny -1} \sqrt{\frac{r_1 + r_2}{r_1 + r_2 + h}} + \sqrt{ \frac{(r_1 + r_2 + h)(r_1 +r_2) h}{2 G m_2}}\)

    \(t = 0.948234 \textrm{s}\) is the corrected result.

    Should read \(t(m_1, m_2, r_1, r_2, h) = \sqrt{ \frac{(r_1 + r_2 + h)^3}{2 G ( m_1 + m_2)}} \cos^{\tiny -1} \sqrt{\frac{r_1 + r_2}{r_1 + r_2 + h}} + \sqrt{ \frac{(r_1 + r_2 + h)(r_1 +r_2) h}{2 G ( m_1 + m_2)}}\)
     
  18. rpenner Fully Wired Staff Member

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    I discovered my error while substituting \(G = g r_2^2 / m_2, h = g t_0^2/2, r_1 = 0, r_2 = R, m_2 = M \) into \( t = \sqrt{ \frac{(r_1 + r_2 + h)^3}{2 G m_2}} \cos^{\tiny -1} \frac{r_1 + r_2}{r_1 + r_2 + h} + \sqrt{ \frac{(r_1 + r_2 + h)(r_1 +r_2) h}{2 G m_2}}\) getting a bad limit as \(t_0 \to 0\). The corrected relationship is:

    when \(t_0, R, g > 0\) we have:
    \(t = \sqrt{ \frac{(R + g t_0^2/2)^3}{2 g R^2}} \cos^{\tiny -1} \sqrt{ \frac{R}{R + g t_0^2/2}} + \sqrt{ \frac{(R + g t_0^2/2)(R)(g t_0^2/2)}{2 g R^2}} = t_0 \left[ \sqrt{ \frac{(1 + \frac{1}{2} \frac{g t_0^2}{R})^3}{2 \frac{g t_0^2}{R}}} \tan^{\tiny -1} \sqrt{ \frac{g t_0^2}{2 R}} + \sqrt{ \frac{1 + \frac{1}{2} \frac{g t_0^2}{R}}{4}} \right] \approx t_0 \left[ 1 + \frac{5}{12} \frac{g t_0^2}{R} - \frac{1}{160} \left( \frac{g t_0^2}{R} \right)^2 + \dots \right] \)
    without the missing square root the answer is badly off in the limit of small \(t_0\).
     

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