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Gravity: a simply inquiry

  1. Oct 10, 2004 #1

    etc

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    i'm sure for you veterans of the physical sciences you'll have no troubles with my answer .. i digress.

    i don't fully understand the concept that gravity (on earth as it may be) is independent fully of mass. would not the mass of the object (say a hammer VS. a house), if gravity is the force of attraction between some two objects (the earth and the hammer/house), have an effect? should not the house and hammer be pulled to the earth at different accelerations?

    if not (so says my physics text), is it because both their masses are just negligible as compared to the earth's?

    i hope i posed my question in a clear manner. if not, i'll eagerly rephrase.

    thanks guys.
     
  2. jcsd
  3. Oct 10, 2004 #2
    Can't say I know the answer with any certainty, but I think you are on the right track. The general formula is:
    [tex]F_{g}=G\frac{m_{1}m_{2}}{r^{2}}[/tex]
    As you can see, the masses of both objects are significant. I believe you are correct in saying that the mass of an object on earth is so insignificant compared to the mass of the earth that motion on earth's surface is governed by [itex]F_{g}=mg[/itex].

    Good question. I may be incorrect so, fellow forum goers, please verify this.
     
    Last edited: Oct 10, 2004
  4. Oct 10, 2004 #3

    Tide

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    etc,

    The house and the hammer are each accelerated at the same rate for precisely the reason the force of gravity on the house is so much greater - it has much greater mass! It is harder to accelerate a more massive object. The astonishing aspect is that inertial mass and gravitational mass appear to be one and the same!
     
  5. Oct 11, 2004 #4

    rcgldr

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    The simplified theory is that gravitational acceleration is the same for both objects regardless of size. There's more force, but also more mass to resist the acceleration of the larger force. [tex]F = m a[/tex] and this can be restated as [tex]a = F / m[/tex]. In the case of gravity, [tex]F[/tex] is proportional to [tex]m[/tex], so [tex]a[/tex] remains constant.

    However, this doesn't take into account that the earth is also being accelerated toward the objects. In the case where the earth has a huge mass compared to the objects being accelerated this fact can be ignored, as the difference is probably unmeasurable (but could be calculated).

    Say you have something with 1/4 mass of the earth on one side of the earth, and something with the mass of a pool ball on the other side, both a few thousand miles away from the earth. Both the moon like object and the pool ball object are accelerated towards the earth at say [tex]X m/s^2[/tex]. The earth is being accelerated towards the moon like object at [tex]1/4 X m/s^2[/tex], and the pool ball is being accelerated towards the moon like object at [tex]1/16 X m/s^2[/tex]. The pool ball's mass is so small that it's pull on the earth and moon like object can be ignored. The moon like object and earth accelerate towards each other at [tex](1+1/4) X m/s^2[/tex], while the earth and pool ball acclerate towards each other at [tex] (1+1/16) X m/s^2[/tex] Bottom line is that the moon like object and earth collide before the pool ball and earth collide.

    So to be technically correct, the more massive object collides first, because it accelerates the earth towards itself more than the less massive object. It's also possible if the two objects are large or close enough that they collide with each other first, before colliding with the earth.
     
    Last edited: Oct 11, 2004
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