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

Why isn't freefall based on the mass of the object falling?

  1. Oct 12, 2004 #1
    We all know that with newtonian physics that it doesnt matter in freefall, what the mass of an object is, the only thing that matters is the height from which it dropped (given no external force and all)

    But why is that true? Why doesn't it depend upon mass?

    i have read in a book or seen on television that general relaitivity answers this kind of question.
     
    Last edited: Oct 12, 2004
  2. jcsd
  3. Oct 12, 2004 #2

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Thought Experiment

    Suppose you do a Galileo and drop two identical weights from a tower. They fall side by side and being identical (and of course always neglecting air resistance!) they maintain that side by side till they hit the ground. You can even suppose they are touching each other as they fall. Now add the tiniest drop of glue to fasten them together so they are now ONE object, twice as heavy as either one of the twin weights. Are they going to fall differently? Is one little tad of glue going to make them fall twice as fast?

    Now do it in reverse, drop a single weight, then carefully bisect it and drop the two halves in touching position. You see that weight doesn't make the difference.
     
  4. Oct 12, 2004 #3
    i understand that

    but Why
     
  5. Oct 12, 2004 #4

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    General Relativity showed that the Inertial Mass and Gravitational Mass balance each other out.
     
  6. Oct 12, 2004 #5

    Plain old Newtonian physics answers this question.

    Here's why:

    Consider a mass we will call m1.

    Then we can write the equation for the gravitational attraction between m1 and the earth:

    F1 = Gm1Mearth/r^2

    By Newton's Second Law: F1 = m1a

    Substitute this in for F1 in the gravitational equation:

    m1a = Gm1Mearth/r^2

    Notice we can divide m1 out of both sides leaving:

    a = GMearth/r^2

    In other words the acceleration (or the rate at which a body falls) is independent of its mass. And assuming we drop objects from the same height; G, Mearth and r are all constants so all masses will fall at the same rate.
     
    Last edited: Oct 12, 2004
  7. Oct 13, 2004 #6
    We can find, at best, the laws of the universe.
    The "why the laws are like they are" it's something God should answer.
    But I don't think He will.

    blue
     
  8. Oct 14, 2004 #7

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    WHY is this question being in two or three places on PF? It's not hard to figure out! To further simplify what geometer illustrated already:


    Normally when a force acts on each of two objects, the more massive one will have a smaller acceleration by Newton's second law.

    However, what if the magnitude of the force acting on the two objects is proportional to the mass of the object?!? If the objects are falling bodies, the more massive one will also have a greater force acting on it! i.e. it will weigh more. It will weigh more by just the right amount to compensate for the fact that it is more massive and has more inertia...because the greater weight is due to the fact that it has that much more mass in the first place! The two effects (from Newton's second law and Newton's law of Gravitation) will cancel out:

    [tex] F = ma [/tex] Newton's second law

    [tex] F = mg [/tex] Magnitude of force acting on object in gravitational field

    [tex] a = \frac{F}{m} = \frac{mg}{m} = g [/tex]

    The assumption that both m's are equivalent hinges on what JasonRox said about the equivalence between inertial and gravitational mass, which I believe was verified experimentally even before GR, right?

    Now, for an object twice as massive:

    [tex] a = \frac{F}{2m} = \frac{2mg}{2m} = g [/tex]

    :rolleyes:
     
  9. Oct 16, 2004 #8
    That might be true if the gravitational field was a single value, g, throughout the universe but obviously this is not the case in the real world. If g is proportional to mass in the first place (the moon has a different g for example than the earth), two objects with different masses must accelerate at different rates toward the earth since each mass generates a different g field than the other with the Earth.

    The acceleration rates of two masses then should include the rate of acceleration of the earth towards the mass so that the total freefall
    rate is g + G(the acceleration of the Earth towards the mass). So
    although g is the same in any two mass system with the Earth as
    one mass, the G is different for different masses. The freefall
    rate of a baseball to the Earth then is g + b while the free fall rate
    of a bowling ball, g +B- they are different.

    Galileo jumped to conclusions then when he couldn't find a difference
    in freefall rates between a bowling ball and a baseball. The contributions
    of G to the Earth's g field between a bowling ball and a baseball is what?
    Pretty pretty small.

    And his critique of Aristotle's logic of gravity is also flawed for the same reason- he forgot about the gravity of the Earth. Trying to detect the free fall rate differences between a baseball and a bowling ball is like trying to measure the difference in displacement of the water in the Pacific Ocean caused by a flea and a frog. It's probably much worse actually. Even more absurd would be if one measured the dispacement of the Pacific Ocean when the flea and the frog are in the water at the same time and used that data to conclude there is no difference in weight between a flea and a frog, kind of reminds me of dropping two balls of different mass at the same time in Galileo's freefall experiment
    of gravity.
     
    Last edited: Oct 16, 2004
  10. Oct 16, 2004 #9

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm sorry Eyesaw, but I don't follow any of your above argument.

    What do you mean by "the" gravitational field? I'm well aware that every particle or body in the universe that has mass generates its own gravitational field. In this case, I was referring to Earth's gravitational field!

    What do you mean "generates a g field with the Earth"? The objects do not generate combined gravitational fields. Each of the three objects (Earth, the bowling ball, and the baseball) generates its own distinct gravitational field. However, since we are considering the accelarations of the two balls, only the forces acting on the balls are relevant. In this case there is one, and it is due to Earth's gravitational field. In detail:

    Let [itex] m_1 [/itex] be the mass of the bowling ball, [itex] m_2 [/itex] be the mass of the baseball, and [itex] M_E [/itex] be Earth's mass. Let [itex] m_1 [/itex] = [itex] 2m_2 [/itex] (the bowling ball has twice the mass of the baseball). Earth's gravitational field strength is:

    [tex] g = \frac{GM_E}{r^2} [/tex]

    This gravitational field strength is the force exerted per unit mass on any body under the influence of the field. So the force experienced by the baseball is:

    [tex] F_2 = G\frac{m_2 M_{E}}{r^2} [/tex]

    The force experienced by the bowling ball is:

    [tex] F_1 = G\frac{(2m_2) M_{E}}{r^2} = 2G\frac{m_2 M_{E}}{r^2} [/tex]

    The acceleration of each ball from Newton's 2nd Law is:

    [tex] a = \frac{F}{m} [/tex]

    For the baseball:

    [tex] a_2 = \frac{F_2}{m_2} = G\frac{m_2 M_{E} }{m_{2} r^2} = \frac{GM_E}{r^2} = g [/tex]

    For the bowling ball:

    [tex] a_1 = \frac{F_1}{m_1} = G\frac{(2m_2) M_{E} }{(2m_{2}) r^2} = \frac{GM_E}{r^2} = g [/tex]
     
  11. Oct 16, 2004 #10
    That's acceleration of a mass to the Earth caused by the G force of the Earth, you ignored the acceleration of the Earth towards the second mass caused by the g force of the second mass, which is different for two different masses. The total freefall rates then are different for different mass systems. g+l is not equal to g + m.
    In other words, a baseball does not fall at the same rate to the Earth as a bowling ball. By ignoring the acceleration of earth to claim UFF you contradict a precept in both Newton and GR- that mass is a cause of gravitational force (in Newton) or space-time distortion (in GR).
     
    Last edited: Oct 16, 2004
  12. Oct 16, 2004 #11
    That argument is flawed since the mass of the objects you are dropping is so small in comparison to the Earth. That's like arguing two drops of water in an ocean should cause the sea level to rise twice as much as one drop, then failing to observe such a change, to conclude that one drop of water weighs as much as two drops of water.
     
  13. Oct 16, 2004 #12
    Actually, I think we are talking about two different things here. All objects accelerate toward the earth at the same rate regardless of their mass. However, the total closing rate of the earth/object system varies with the mass of the other object in the system.

    By the way, with respect to the bowling ball and the baseball, the correction would be on the order of 10 exp (-24) meters per second squared - I would consider this negligible.
     
  14. Oct 16, 2004 #13
    Since gravitational field strength falls off with squared distance (or conversely increases with squared proximitiy), the larger acceleration on the Earth caused by the larger mass would bring the Earth closer to it at every point in time when compared with the situation with the smaller mass so at no time is the g caused by the Earth on the larger mass even equal to the g caused by the Earth on the smaller mass. So two different masses never fall at the same rate toward the Earth.

    E.g.:

    Baseball :

    a = mg/m = g.

    Soccer Ball:

    A = SG/S = G.

    G is not g because the r^2 is different in time for
    the soccer ball than the baseball. The g of the Earth on the baseball would
    grow larger as time passes than the G of the Earth on the soccerball due to
    the larger mass of the baseball causing the r^2 to decrease
    faster than in the case with the Earth and the soccer ball in this example.

    The Universality of Free Fall rates is a myth created by an unreasonable
    expectation on extant technologies from Galileo on to perform a task that is likened to an attempt to finding the differences in displacement of the Pacific Ocean caused by one drop of water and by two drops of water falling on it.

    Though the myth has helped generate much public interest in gravity with its mistique, it has done nothing but damage to the science of gravity due to its defiance of the logical methods.
     
    Last edited: Oct 16, 2004
  15. Oct 16, 2004 #14
    I just wasted fifteen minutes trying to refute this and failed. It then took me 30 seconds to figure why: The problem arises from the assumption that the eart is an inertial frame. If the earth is treated as an inertial frame, as is generally the case in solving newtonian problems, then the fall times will be equal. This is because an inertial frame is in uniform motion--not accelerating, so the earth could not be accelerating towards the two objects.
    Because the coordinate system is designated as fixed to the earth, there is no incorporation of the earth's acceleration--the origin should not be moving.

    So to make that shorter: you are right, because the earth is not an inertial frame, and technically it is a bad thing to fix a coordinate system. Anyone who tries to show that the two do will meet the earth in the same time with respect to a seperate inertial frame not fixed to either the earth or the balls, will faile. Further, if you attach the inertial frame to the balls, who have notably different masses (proportionally) you will get different accelerations, obviously.
     
  16. Oct 16, 2004 #15

    No, it is just erroneously based on the assumption of the earth as an inertial frame of reference. It is not some horrific conspiracy, just an over-simplification, that for all practical purpose makes little significant difference in terms of the calculated values because the earth is so large. Ignoring air resitance is far more significant than the mislabeling of the earth as an inertial frame.
     
  17. Oct 19, 2004 #16

    Yes, the moon landing demonstrated quite dramatically the error of attaching an inertial frame to a mass, aka the error of Galileo's two spheres. They demonstrated that the moon, with a mass 1/6th the Earth, "freefell" at a rate 1/6th that of the Earth towards the astronauts.
     
  18. Oct 19, 2004 #17
    Umm. Not sure what you meant by that. Given that, assuming an inertial frame attached to the alrger mass we get: g= GM/r^2 , then the acceleration is obviously going to be 1/6 if the mass of the larger object is 1/6. But that is based on the attachment of an inertial frame to the larger mass M. So your post is a little confusing.
     
  19. Oct 19, 2004 #18

    When discussing the Universality of Freefall what usually happens
    is we see a calculation like the one made in a post above :



    [tex] a_1 = \frac{F_1}{m_1} = G\frac{m_1 M_{E} }{m_{1} r^2} = \frac{GM_E}{r^2} = g [/tex]
    for a baseball,

    and
    [tex] a_2 = \frac{F_2}{m_2} = G\frac{m_2 M_{E} }{m_{2} r^2} = \frac{GM_E}{r^2} = g [/tex]

    for a bowling ball,

    demonstrating that two different masses fall at the same rate.

    I think we came to the agreement above that this result was achieved by ignoring the acceleration of the Earth towards the smaller masses. (The
    actual total accelerations should be g+baseball and g+bowling ball.

    My comment about the moon landing is that it demonstrated the folly, in
    real spectacular fashion, of neglecting the acceleration of 'the other mass' in free fall experiments.

    For if we but replaced the [tex] M_{E}[/tex] symbol in the above equations
    to mean the mass of say, an astronaut, and the [tex] m_1[/tex] and [tex]m_2[/tex] to mean the mass of the Earth and the mass of the Moon, respectively, we arrive at the same result of [tex] a_1[/tex] and [tex] a_2[/tex] both being equal to [tex] g [/tex] with 'g' having the meaning of the gravitational field generated by the astronaut. But we all saw on TV that it was not the case that the Moon "free-fell" at the same rate of [tex] g[/tex] towards the astronaut as when the Earth "free-fell" to him when he was on it. And surely this "experiment" is no different in principle than the usual consideration of two different masses falling
    toward the Earth?

    The landing on the moon then was truly a giant step for mankind for
    it really stood the Universality of Free Fall principle on its head.
     
    Last edited: Oct 19, 2004
  20. Oct 19, 2004 #19

    Phobos

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Explain how the "moon's free fall" was measured during the Apollo missions.

    As noted earlier, the difference in gravitational attraction between a planetary body and small masses like a baseball vs. a bowling ball is negligible for the purposes at hand.
     
  21. Oct 19, 2004 #20

    When you see the footage of the astronauts, it's obvious when they
    jumped up, it took longer for them to land than if they made the
    same jump on Earth. Their steps were more like leaps. This longer "landing"
    time on the moon equates to a large difference in the "free-fall" rates
    between the moon and the Earth toward the astronaut . That is,
    if we focused on the Astronaut instead of the planetary masses as the cause of the g (neglecting the G caused by the planetary masses on the Astronaut as we neglect the G nowadays of the smaller mass in "regular" free-fall experiments) we would obtain a "free-fall" rate that is dramatically different
    for different masses (Earth having a much greater "free-fall" rate toward
    the astronaut than the Moon).

    Maybe the above would be clearer if we spoke in terms of two different spheres being "dropped". In Galileo's experiment, one sphere, a baseball
    is dropped from a height on Earth, then another sphere, a bowling ball
    is dropped from the same height. The "identical" rate is used as experimental
    support for the claim of Universality of Free Fall. If we focused on the baseball instead and compared the "free-fall" rate of the Earth to the baseball
    with the "free-fall" rate of the Moon to the baseball (first dropping the baseball on the Earth from a height and then going to the moon and dropping
    it from the same height), we see that the Universality of Free Fall claim doesn't hold water.

    The former is like trying to find the differences in the dispacement of the
    water level of the Pacific Ocean between one and two drops of water falling on it, then finding none, concluding that all masses cause the same displacement on a body of water, while the latter is like comparing the differences in water level by filling the Grand Canyon with all the water in the Red Sea, draining it and then filling it again with all the water in the Yellow River and noting that since the larger mass of water filled a larger volume
    of the Grand Canyon, differences in mass must cause a difference in displacemnt on a body of water. Now, which experiment do you think is more instructive?
     
    Last edited: Oct 19, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?