Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Galileo's experiment and Equivalence Principle

  1. Nov 3, 2007 #1

    3-body problem (special case)

    According to the Newton's Mechanics:
    Let us consider a mass M of radius Ro. At a distance h from the center of this mass M, we place a spherical shell (e.g a spherical elevator) of mass m1 and radius R.
    Moreover, at the center of the spherical shell we place another point mass m2, (m1 ≠ m2).
    Now t= 0 (Phase I), we allow these three masses m1, m2, M to move freely under the influence of the force of universal attraction.
    Let us also assume that after a time dt (Phase ΙΙ), υ1, υ2 and V are respectively the velocities of masses m1, m2 and Μ, relative to an inertial observer Ο.
    Note:Masses m1, m2, and M are considered to be homogeneous and absolutely solid bodies.
    In addition, velocities υ1, υ2 and V are considered to be positive numbers (that is, only the meters of their magnitudes are taken into account), while mass m2 is considered to be found always within the spherical shell (spherical elevator) m1.

    The basic question that is being raised is the following:


    At what velocities do masses m1 (spherical elevator) and m2 (point mass) fall in the gravitational field of Mass M once simultaneously dropped in free fall from a height h, namely at the time dt>0 we have V1 = V2 or V1 ≠ V2 ?

    I await your anwer.


  2. jcsd
  3. Nov 3, 2007 #2


    This seems like the beginning of a discussion of General Relativity. I'd like to be in on it when it gets juicy, so I figured I'd get the ball rolling.

    For the example you gave, if I understand your meaning correctly, V1 = V2.

    You didn't ask for an explanation. Did you want one?
  4. Nov 3, 2007 #3

    My question is very simple:

    .....at the time dt>0 we have velocities , u1 = u2 or u1 ≠ u2 ?

  5. Nov 3, 2007 #4

    u1 = u2
  6. Nov 3, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    This doesn't seem to me to have much to do with General Relativity at all, except that it mentions the principle of equivalence.
  7. Nov 4, 2007 #6


    If, e.g m1 = 10^6 tn (mass a white draft) and m2 = 1gr (mass a feather), then, we have u1 =u2 ?


  8. Nov 4, 2007 #7


    Well, as pervect pointed out, this not a thread on SR or GR but Newtonian Mechanics. Perhaps it could be considered “leading up to” a discussion of SR or GR. Either way, I don’t mind answering as long as the moderators don’t get mad at me.

    Your last question is the very experiment that Galileo is described as doing. Remember the story about him dropping pairs of different size cannon balls off the Tower of Pisa? They always fell together and always hit the ground at the same time. He was forced to conclude that gravity makes u1 always equal to u2 no matter what m1 and m2 are.

    Why, you ask? I could present an algebraic proof. It would be the same as the ones that appear in freshman physics textbooks. But I don’t think that is what you are asking for. Philosophically speaking, well, I don’t know how to answer.
  9. Nov 5, 2007 #8

    As it is well-known, Einstein’s General Theory of Relativity postulates the following:
    “Gravitational fields have the remarkable property of imparting the same acceleration to all bodies independent of their material composition and amount of mass”. (See THE PRINCIPLE OF RELATIVITY, BY A. EINSTEIN – H.A LORENTZ – H. MINKOWSKI – H. WEYL, Page 114 DOVER PUBLICATIONS, INC). Einstein based this conclusion on the all-too-familiar Galileo’s experiment (the experiment conducted from the Leaning Tower of Pisa).

    Also, I have a question:
    If, m1 = 10^6 tn (mass a white draft) , m2 = 1gr (mass a feather), and the mass M of Earth is e.g M = 10^6 tn , namely m1 = M, then according to the Galileo's experiment, (experiment of the Tower of Pisa) we have again u1 = u2 ?

  10. Nov 5, 2007 #9


    Yes, u1 = u2 for your last example too.

    That applies to both Newton's and Einstein's version of gravity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook