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One of out Ideas of Gravity Falls

  1. Jan 9, 2009 #1
    By Chuck St. Louis January 5,2009
    Guatemala City, Guatemala

    Gravity what we have always believed may have just taken a fall !
    Can there be anything new as far as gravity goes. Have not all the great minds of physics pondered and experimented in great detail the effects of this elastic phenomena? The idea I am about to propose is so simple and obvious once you understand the question I almost hesitate to submit it here. After hearing many physics lectures pertaining to gravity I felt compelled to express and relay information that may be very disconcerting for many but really just confirms that F= M*A.
    How many times have you heard that all objects in a vacuum fall to the earth at the same rate? The classic example taught to me in my grade 11 physics class was; "If you drop a lead ball 1 inch in diameter and a cotton ball of the same size together in a vacuum, they will drop at the same rate." That is perfectly true and not negotiable. Now drop a cotton ball first and then the lead ball second and I propose with all certainty that the cotton ball will fall faster. The time that both objects take to hit the earth will be the same. The difference is that the cotton ball with its small mass has very little inertia and the earth with its great mass has great inertia. So when the cotton ball is dropped, the earth moves an infinitesimal amount and the cotton ball experiences all the motion. When a really massive lead ball is dropped for example, the lead ball does most of the moving but now the earth is accelerated and moves a bit, all-be-it very little, so now the lead ball has just traveled a shorter distance in the same time compared to the cotton ball. Conclusion: The cotton ball falls with a greater velocity. Distance divided by time equals velocity.
    Reference frame
    If you are standing on the earth and looking up and watching different objects fall in a vacuum, you will notice that all objects fall to you at the same rate from your point of reference. But if you are suspended and fixed in space at a non movable point above the surface of the earth, and see an object fall towards earth, from your reference frame all objects with a different mass fall at a different velocity. Likewise you would notice that the earth would move a little towards each object as it falls.
    Fig.1 (see attachment) is an example of a planet sitting in a vacuum in space that is not orbiting a sun and has two objects released towards it. The question is which of the two objects will have a greater velocity?
    Conclusion;
    All objects with different masses pulled by gravity and falling individually toward any planet or star will accelerate at different velocities.
     

    Attached Files:

    Last edited by a moderator: Jan 9, 2009
  2. jcsd
  3. Jan 9, 2009 #2
    Actually I dont know whats new in that...
    You are just considering a two body problem instead of a one body problem...
    But think about it, the mass of earth is about 10^24 kg, so the two-body effects can be very-very well neglected....

    "confirms that F= M*A"

    F=M*A is not right in the general sense either. If we are talking about such precision as not neglecting the two-body effects, in the earth- falling body system, then we should be precise throughout the post.
    Instead it should be:

    [tex]\vec F = \frac{d \vec p}{dt}[/tex]

    Where p is the momentum vector.


    "Distance divided by time equals velocity"

    Thats not at all true in general. If our motion is even pace motion then its good. But throughout your post you are talking about "acceleration due to gravity". The definition of velocity is the first derivative of the "space-time" function, and in this case thats what we should use.
     
  4. Jan 9, 2009 #3

    Doc Al

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    Staff: Mentor

    Thought experiments wherein one makes the ludicrously small correction to the relative acceleration of falling object and Earth due to the Earth's acceleration, have been discussed ad naseum here on PF. Nothing new there.
    Actually, if you include the Earth's acceleration, it's the lead ball that will reach the Earth faster.

    Let's simplify the model. Treat the Earth as a non-rotating, uniform sphere with mass M. Let the object of mass m be at a distance R from the Earth's center and released from rest. Forget about air resistance, et cetera.

    Viewed from an inertial frame in which the center of mass of the Earth-object system is at rest:
    - the acceleration of the object is: a_o = GM/R^2
    - the acceleration of the Earth is: a_e = Gm/R^2

    Thus the relative acceleration of the object with respect to the Earth is:
    a_o + a_e = GM/R^2 + Gm/R^2 = G/R^2(M + m)

    Thus in this simplified thought experiment, the heavier object will have the greater acceleration with respect to the Earth and thus will hit the ground first. But not by much! :wink:

    (Note that here I assume that you drop the different objects separately, in two different thought experiments. If you put them side by side and release them, then they will fall together and hit the Earth at the same time.)
    Nah.
    I'd say you have that backwards.
    That statement is true.
    See above.
    It depends on your reference frame.
     
    Last edited: Jan 9, 2009
  5. Jan 11, 2009 #4
    Thanks for reply.

    "confirms that F= M*A"

    F=M*A is not right in the general sense either. If we are talking about such precision as not neglecting the two-body effects, in the earth- falling body system, then we should be precise throughout the post.
    Instead it should be

    "Distance divided by time equals velocity"

    Thats not at all true in general. If our motion is even pace motion then its good. But throughout your post you are talking about "acceleration due to gravity". The definition of velocity is the first derivative of the "space-time" function, and in this case thats what we should use.[/QUOTE]


    The reason I put forward F=M*A is that the force that is acting between the objects during a fall at a given moment of time is the same. They are always pulling at each other with the same force.

    When I posted the thread the attachment was not available for viewing until much later in the day. Did you manage to see the Diagram?

    Have a great day.
    Chuck
     
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