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The question of letting two objects fall in vacuum and the effects on gravity is discussed on this site including questions like the gravitational pull of the objects towards each other, the Earth's spherical shape and so on. However I am interested into what forces are in play when you drop the object into atmosphere (assuming no wind). It seems to me that the naïve answer that they will fall at the same time is incorrect demonstrated by the extreme example of dropping a helium balloon and a bowling ball. It seems to me that Galileo's experiment is actually an approximation and both objects have masses large enough to dwarf other forces in play like air resistance and Archimedes' forces but in reality the heavier object will indeed fall tiny bit faster.

So if we want to calculate how much faster the heavier object will fall which forces do we take into account and how do they interact?

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  • $\begingroup$ By definition, if there's drag, it not a free fall. $\endgroup$ – Qmechanic Oct 15 at 14:20
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The main force to take into account is air resistance, which increases with the square of velocity.

When the balls are first dropped the air resistance increases. For the lighter ball there will come a point at which the increasing resistance due to its increasing speed exactly counteracts the force of gravity, so the ball will no longer accelerate but will continue to drop at a fixed speed.

For the heavier ball the force due to gravity is greater, so the ball must reach a higher speed before air resistance matches its weight. If the heavier ball was 100 times heavier, say, then its terminal speed would be ten times that of the lighter ball.

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  • $\begingroup$ Will both balls reach terminal velocity at the same time (just a different one)? What if they hit the ground before reaching it, seems like the heavier will still be first? $\endgroup$ – Stilgar Oct 15 at 11:37

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