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Galileo stated that objects dropped from the same height will hit the ground at the same time, that the rate of gravity is constant for all objects no matter the mass. But I think there is a flaw in this statement.

All objects with mass have gravity. If a object has a greater amount of mass, it has a greater gravitational pull. Therefore, a hammer has more gravitational pull than a feather.

Given that the hammer and feather have gravitational pull, we can prove that the hammer falls faster than the feather, because it has more mass.

Because the hammer has a gravitational field, it is tugging the Earth toward itself a tiny bit, and the same with the feather. But since the hammer has more mass than the feather, it will tug the Earth toward it faster, meaning it would "fall" to the Earth faster if you dropped it.

Doesn't this mean that objects with greater mass fall faster, disproving Galileo?

Qmechanic
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    Maybe this might answer your question: http://physics.stackexchange.com/q/3534/ I will vote to close, as I feel your question already has an answer –  Oct 07 '15 at 22:12
  • Galileo's observation (and therefore his theory) are valid for object of mass small compared to that of the Earth and in locations near the surface of the planet and separated by a horizontal distance much smaller than the radius of the Earth. – dmckee --- ex-moderator kitten Oct 07 '15 at 22:17
  • Yes, your reasoning is correct, though, it might be interesting to calculate the gravitational acceleration of a hammer and compare it to a feather. I'm also pretty sure Newton figured this out too and he put numbers to it. – userLTK Oct 07 '15 at 22:17
  • @dmckee It's applicable for small masses like a hammer and a feather, but it's not true for any mass. –  Oct 07 '15 at 22:19
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    @PythonGuy: it depends strongly on how you set up the problem. Usually one would claim that the hammer and feather falling side-by-side will hit at the exact same time; this is not really affected and still quite true. However indeed $g$ should strictly be approximated as $G(M + m)/R^2$ when considering the mass of the object itself; you see this $M + m$ all the time in orbital motion calculations -- and might be greater by something like $10^{-17}\text{ m}^2/\text{s}^2$, leading to a falling-time difference of the order of one part in a quintillion or so. – CR Drost Oct 07 '15 at 22:49
  • @ChrisDrost Yes, a very small difference. I agree. –  Oct 07 '15 at 22:51

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