Galileo was technically WRONG

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

  1. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    Start with variables

    r = radius of Earth = 6371000 m
    m_e = mass of Earth = 5.972E24 kg
    m_gb = mass of golf ball = .05 kg
    m_bb = mass of bowling ball = 6 kg
    m_bh = mass of black hole at center of Milky Way (est) = 8.2E36 kg
    m_o = mass of arbitrary object
    F_e.gb = force between Earth and golf ball = .491 N
    F_e.bb = force between Earth and bowling ball = 58.92 N
    F_e.bh = force between Earth and the black hole = 8.052E37 N
    F_e.o = force between Earth and arbitrary object
    a_e.gb = acceleration of the Earth towards the golf ball = .00008029 m/s^2
    a_gb.e = acceleration of the golf ball towards the Earth = 9.82 m/s^2
    a_e.bb = acceleration of the Earth towards the bowling ball = .00963481 m/s^2
    a_bb.e = acceleration of the bowling ball towards the Earth = 9.82 m/s^2
    a_e.bh = acceleration of the Earth towards the black hole = 1.348E13 m/s^2
    a_bh.e = acceleration of the black hole towards the Earth = 9.82 m/s^2 (corrected rounding error)
    a_o.e = acceleration of arbitrary object towards the Earth
    d = distance between objects = 10 m
    d_e = distance Earth falls before collision with object
    d_o = distance object falls before collision with Earth
    t_e = time Earth falls before collision with object
    t_o = time object falls before collision with Earth

    F = Gm1*m2/r^2, therefore the Force calculations above were used with the associated variables listed and can be verified here

    d = (a*t^2)/2
    t = sqrt(2d/a)

    For any object 10 m above Earth, we use subscript _o:
    t_e = t_o = sqrt(2d_e/a_e.o) = sqrt(2d_o/a_o.e)

    This gives us
    d_e*a_o.e = d_o*a_e.o
    d_o = d_e*a_o.e/a_e.o

    We also know that
    d = d_e + d_o = 10 m

    Substituting we get
    d_e + d_e*a_o.e/a_e.o = d
    d_e = d / (1+(a_o.e/a_e.o))

    This allows us to calculate t
    t = sqrt(2d_e/a_e.o) = sqrt(2(d/1+(a_o.e/a_e.o))/a_e.o) = sqrt(2d/(a_o.e+a_e.o))

    =========================================
    For the golf ball:
    t = sqrt(20/(9.82 + .00008029)) = 1.4271101 seconds

    For the bowling ball:
    t = sqrt(20/(9.82 + .00963481)) = 1.4264164 seconds

    For the black hole:
    t = sqrt(20/(9.82 + 1.348*10^13)) = .00000121806 seconds
    =========================================

    The reason most people get this wrong is because it is presumed that the Earth does not also move towards the falling object. It should be obvious that this is the case when the "falling object" is extremely massive (i.e. a black hole) - QED
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. rpenner Fully Wired Staff Member

    Messages:
    4,833
    Galileo was technically right if the two masses are dropped at the same time.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    Of course!

    Please Register or Log in to view the hidden image!

     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    From another thread:

    RPenner, as you can see, Dr_Toad and others are requesting your mathematical acumen in the evaluation of my proof. Thanks in advance!

    Please Register or Log in to view the hidden image!

     
    Last edited by a moderator: Oct 3, 2014
  8. Neddy Bate Valued Senior Member

    Messages:
    1,403
    RJBeery, are you saying that some folks looked at this equation...

    F = G (m_1 * m_2) / r^2

    ...and concluded that only one of the masses affects the attractive gravitational force between them?
     
  9. brucep Valued Senior Member

    Messages:
    4,098
    The rate an object falls (free fall) has nothing to do with when It was dropped or the mass of the object in free-fall.
    Since Galileo discovered this he's technically right infinite in extent. Unless somebody can show otherwise. Which RJBerry has't done with his juvenile analy
    You think there is some physics which will include force and acceleration in an inertial free fall frame. RJBerry analysis reminds me of a chinglu analysis where he either adds or subtracts, something that should be or shouldn't be included in the physics. Winding up with irrelevant nonsense.
     
  10. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    Yes, Neddy Bate, that's exactly the problem. They apparently believe that the force would only apply to one of the two masses.
     
    Last edited by a moderator: Oct 3, 2014
  11. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    If two objects are in free fall and one object falls a shorter distance before hitting the ground, I have to assume that you would agree it would do so in less time, correct? That's the crux here. The Earth is pulled up toward the bowling ball (or the moon, or the black hole) with a greater force than with the golf ball, therefore the bowling ball (et al) hit the Earth in less time because they are falling a shorter distance.
     
  12. OnlyMe Valued Senior Member

    Messages:
    3,914
    Except that while the earth is pulled up toward the bowling ball shortening the distance.., in the case of the golf ball, IT (the golf ball) falls toward the earth faster than the bowling ball does.., also shortening the distance...

    This is never an issue, if you always use the earth as the frame of reference for the fall in both cases.., because all of the closing velocity is then attributed to the smaller object. When you try to look at it from the rest frame before the drop you have to take into account that the golf ball falls faster toward the earth than the bowling ball does.., which is exactly balanced by the difference in how fast the earth falls toward the smaller object in each case.

    You are wrong! And as been pointed out Newton and Einstein are both right. Take the time to look over, The force of gravity is the same for atoms and baseballs. That should be a larger difference in mass than your bowling and base ball fantasy. And I am sure than even though I don't want to strain myself with the math, the researchers involved had no issue with the math... Without a reference I think I also remember a proposed test with neutrons, but I am unsure and a little skeptical given the short half life of a free neutron.
     
  13. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    MODERATOR NOTE:
    I've deleted one post and amended a couple of others.
     
  14. Aqueous Id flat Earth skeptic Valued Senior Member

    Messages:
    6,152
    So are you saying the Earth accelerates toward the bowling ball, even as the bowling ball accelerates toward the Earth, and, if so, what happened to that 9.8 m/s² g-field? Because as the Earth rushes toward the bowling ball, g, which is a function of r, now needs to be treated as g(r(t)). I don't see where you covered that in your mathematical analysis.

    What now?
     
  15. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    The 9.82 m/s^2 is unchanged in my analysis. The numbers I use in my analysis in the OP are DERIVED via F=Gm1*m2/r^2. They are NOT simply given by convention! The difference in g(r(t)) is indeed ignored, but that is inconsequential because the field gradation between 10 meters and the ground of the Earth would be the same for all objects, so it is considered to be a constant at the value derived at 10 meters. The result (or more specifically, the conclusion) would be the same either way. I'm holding out hope that you, Aqueous Id, are clever enough to grasp this proof because it appears the others who are (i.e. rpenner) don't have the intellectual fortitude to publicly admit it.
     
    Last edited: Oct 3, 2014
  16. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Let me get this straight. You are saying that if I drop a bowling ball from a height of 4.91 meters above the surface of the earth, that it will take 1 second for the ball to impact the surface of the earth?
     
  17. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    That's what Isaac Newtons is saying, yes.
     
  18. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    How much time do you think it takes?
     
  19. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    I reckon 4.91 meters would take 1 second, roughly; unless the mass was vastly larger than a golf ball. (presuming your post is correct! I did not verify)
     
  20. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    Roughly? No, I mean exactly?
     
  21. RJBeery Natural Philosopher Valued Senior Member

    Messages:
    4,136
    I've given you the tools to do the calculation yourself big guy.

    t = sqrt(2d/(a_o.e+a_e.o))
     
  22. paddoboy Valued Senior Member

    Messages:
    21,225
    The heavier item may have a greater attraction but it also has great resistance to change...Intertia:
    The lighter item has less attraction, and also less Inertia and less resistance to change/movement/falling.

    So we have one item with great gravitational attraction and more intertia, fighting that attraction...and the other have less gravitational attraction but also less Inertia fighting the gravitational pull.
    Guess what? The two effects balance each other out, so that they hit the ground at the same time.

    Galileo was right!
     
  23. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

    Messages:
    5,105
    I've given YOU the tools.

    Please Register or Log in to view the hidden image!

     

Share This Page