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After looking at this question:

Don't heavier objects actually fall faster because they exert their own gravity?

A thought occurred to me that due to the increased gravitational pull of the heavier object, will that cause more air particles to be attracted, thus increasing friction?

If so, will it then -in any circumstance- affect the falling rate of the heavier object to the point that a lighter object may fall faster than it?

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2 Answers 2

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If you consider that gravity is weak compared to the electromagnetic force because

$G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2} $

and

$k_e \approx 8,987 \times10^9 N m^2 C^{-2} $

it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, repelling the air molecules (bouncing them off). Even if some of the mass is accreted , it would be several orders of magnitude smaller than the object's mass, besides, acceleration at the surface of the earth is independent of the objects mass because

$ \frac{GMm}{r^2} = ma$

so its acceleration does not depend on $m$.

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  • $\begingroup$ Why is it not independent of the earths mass $\frac{GMm}{r^2} = Ma$? $\endgroup$
    – t0xic
    Jul 31, 2015 at 18:58
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    $\begingroup$ That would be Earth's acceleration. You have to realize that mass is what "regulates" how much force affects an object, but since the force deppend on the objects mass the acceleration is dependent only on the field in which the object is placed. $\endgroup$
    – Andre Maizel
    Aug 1, 2015 at 16:39
  • $\begingroup$ Yes, and the earth is not accelerated because according to Newton there is no force on the earth... $\endgroup$
    – t0xic
    Aug 1, 2015 at 17:48
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    $\begingroup$ I'm not sure if you're being serious or not, Newton said the exact opposite. Action-reaction, Earth is subject to the exact same force as the falling object, only with different acceleration. $\endgroup$
    – Andre Maizel
    Aug 2, 2015 at 19:14
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If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.

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  • $\begingroup$ The counterargument (which I don't personally believe) is that air can be a LOT closer to your falling object than the earth generally is (at least while it's still falling). Halving the distance quadruples the force. It would be great if your answer could discuss why this doesn't matter. $\endgroup$
    – Daniel Griscom
    Jul 31, 2015 at 0:29
  • $\begingroup$ A counter-counter argument: in reality, all earth's mass acts like it is coming from the center of the earth, which is why there is an r^2 in the denominator of Mr. Maizel's equation. Since air is distributed as a layer completely around the earth, the same argument would hold ... all of the mass of the air that is below the falling object could also be treated as if it was located at the center of the earth. $\endgroup$
    – David White
    Jul 31, 2015 at 0:41

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