Galileo was technically WRONG

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

  1. RJBeery Natural Philosopher Valued Senior Member

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    Umm, but you know where it came from because you replied to it. I'm trying to bring Fednis over to this proper thread.
     
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  3. OnlyMe Valued Senior Member

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  5. exchemist Valued Senior Member

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    This is what I thought too, though I confess the more I thought about it the less confident I became! Beery's diagram makes the problem look clearer. I think we can settle this by means of looking at the maths for the two cases, i.e. one case in which the larger mass is taken as the frame of reference to describe the motions and the other in which the frame is that of the masses before release. Maybe in the morning my head will be clear enough - or maybe someone good at Newtonian dynamics can have a go in the meantime. It should be fairly straight forward.
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    Imagine one planet is chained to a magical wall which could not be moved. OnlyMe, what do you think the outcome of this picture would be? Do you think the fall time of the free planet would be less than, greater than, or equal to the measured fall time if both planets were free? Because the real answer is that this is what Galileo was describing. All objects from feathers to bowling balls to other planets TO BLACK HOLES would all be timed to descend to this chained planet at equal velocities from equal heights.

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    The fact that the Earth is free floating changes the math.
     
  8. RJBeery Natural Philosopher Valued Senior Member

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    This was posted in the other thread. For any intellectually honest person this discussion has been irrefutably settled and is over.
     
  9. Motor Daddy Valued Senior Member

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    Imagine a burning ball of force (read-"the sun") pushing objects known as planets away from it. The more time that elapses the further away the planets get from the sun, which is not to be confused with planets getting closer and crashing into the sun as time elapses. That doesn't happen! Planets don't crash into the sun, they get further away over time!

    The second law says I'm right!
     
    Last edited: Oct 3, 2014
  10. el es Registered Senior Member

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    When the mass of a test object is very small compared to the mass of a planet, Newton's Universal Law of Gravitation equation is changed to the gravitational acceleration equation. Notice that m is removed from both sides of the equation.

    http://www.physicsclassroom.com/class/circles/Lesson-3/The-Value-of-g

    The rings of Saturn are a great example of this. The radius is determined by the objects velocity, not it's mass.
     
  11. billvon Valued Senior Member

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    Now we are chaining planets to brick walls and Motordaddy has started with his crazy. It has become an authentic, homegrown Sciforums thread. All it needs now is Motordaddy to post "I have calculated it to be 4.9473409432423423430128520239953229110393746 seconds so I am right!" and Garbonzo to invoke God. Perhaps we could get Michael to hate Obama and Bells to call someone a misogynist just to round things out.
     
  12. Motor Daddy Valued Senior Member

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    If I understand you correctly, you're saying you agree with me.

    Edit: I don't hide my crazy like Miranda does.


    Love that video!

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    Last edited: Oct 3, 2014
  13. exchemist Valued Senior Member

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    Stop introducing more metaphors and let's just do the maths, for both frames of reference. Then we'll know.
     
  14. Neddy Bate Valued Senior Member

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    Okay, now I understand. They don't deny that the attractive gravitational force is a function of the product of both masses. But they assume the earth to be stationary when calculating the resulting acceleration of the other mass. I see.
     
  15. exchemist Valued Senior Member

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    The Like button has disappeared again, but….spot-on.
     
  16. Fednis48 Registered Senior Member

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    RJBeery, thanks for bringing my post over to this thread; I didn't mean to contribute to the derailing of a thread on black holes.

    tashja's quote of Prof. Wasserman in post #45 is an impeccable qualitative description, so I'm going to try and cover the math. One thing I hope everyone can agree on is the expression for the energy of the system, which is a combination of two kinetic energy terms and the Newtonian gravitational potential formula:

    E=kinetic+potential=P^2/2M+p^2/2m+GMm/(X-x)

    Here, positions and momenta are along the earth<->object axis, capital {M, X, P} refer to the Earth's mass, position, and momentum, respectively, lowercase {m, x, p} refer to the same properties of the falling object, and G is the universal gravitation constant. Without loss of generality, I have assumed that Earth's position coordinate is higher than the object's; if the reverse is true, the denominator in the potential energy would be (x-X) instead. The equation for a system's energy is its Hamiltonian, so we can use Hamiltonian mechanics (http://en.wikipedia.org/wiki/Hamiltonian_mechanics) to determine the dynamics. We want to compare accelerations, so we're looking for the rate of change of the relative velocity:

    d/dt(v_rel)=d/dt(V-v)==d/dt(P/M-p/m)=dP/dt/M-dp/dt/m

    Position and momentum are a canonical coordinate pair, so we can say dp/dt=-dH/dx. Note that on the Wikipedia page and in textbooks, position is usually denoted as q instead of x. Plugging this into the above expression:

    d/dt(v_rel)=-dH/dX/M+dH/dx/m=-Gm/(X-x)^2-GM/(X-x)^2=-G(m+M)/(x_rel)^2

    As RJBeery has been insisting, the result depends on both m and M. If we assume that M>>m, we can drop m from the equation to good approximation; this is the "Galilean approximation".

    edit: I anyone else experiencing an issue where anything written in parentheses compiles as TeX?
     
    Last edited: Oct 3, 2014
  17. Trippy ALEA IACTA EST Staff Member

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    How about addressing this:
    Instead of tilting at strawmen?
     
  18. brucep Valued Senior Member

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    read the post I wrote to you in the other thread. Tidal accelerations can change the free fall path through spacetime not the rate of free fall. That,
    Read my post answering yours in the other thread. Tidal accelerationslerations can change the path of the object through spacetime but it doesn't effect the rate of the falling object. The Newtonian analysis you're trying to do isn't something Newton would be doing. If it was then it would show up in the equation of motion derived from his theory.
     
  19. paddoboy Valued Senior Member

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    As well as the odd red herring or two.

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  20. RJBeery Natural Philosopher Valued Senior Member

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    Uhh, it DOES show up in the purely Newtonian analysis I did in the opening post of this thread. The equations you're posting about dr/dt are ignoring Newton's third law. And now you're refuting Fednis' math analysis as well as Professor Wasserman's qualitative description!

    And you want to be taken as an authority around here?

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  21. Trippy ALEA IACTA EST Staff Member

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    You're sure about that?

    From the perspective of a stationary observer taking measurements relative to themself, how can how fast I run towards you change how fast you run towards me? It can't. It's unphysical and nonsenseical.

    Two objects, A and B. They have a center of mass.

    as long as (m_a) >>> (m_b) then changing the value of (m_b) isn't going to change the rate at which B falls towards the center of mass of the system, it's only going to change the rate at which A falls towards the center of mass of the system.
     
    Last edited: Oct 4, 2014
  22. RJBeery Natural Philosopher Valued Senior Member

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    This is exactly right. Except you're not appreciating that if the Earth falls towards the bowling ball "more quickly" (than, say, towards the golf ball) then the bowling ball has a shorter distance to cover before it impacts the Earth. Not only did I show this in the opening post, Fednis gave a math-based analysis AND we have quotes from actual Physics professors supporting my claim.

    The case is closed. Further discussion is futile, but I'm sure that won't stop some folks here.
     
  23. Motor Daddy Valued Senior Member

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    The case is reopened.

    If you claim the bowling ball accelerates at the rate of 9.8 m/s^2, then in exactly 1 second the bowling ball traveled 4.9 meters, and has a velocity of 9.8 m/s. Agree?
     

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