
Originally Posted by
Janus
Obviously, in air, a rock and piece of lint don't hit the ground at the same time air drag will effect them differently. The rock is much denser and thus has a smaller surface area relative to its mass than the lint does* However, do the same experiment in a vacuum, and they will.
The fall at the same speed assumes that only gravity is at play (or at least that other effects are minimal enough to ignore)
It is based on two equation F=ma And F= GMm/r^2
The first equation tells us that A given Force of strength F will accelerate a Mass of m at an rate of a.
The second gives us the gravitational force acting between two masses, M and m with a distance of r between their centers.
This force will cause M and M to accelerate towards each other. If M is the mass of the Earth, and m the mass of the object we are dropping, r is just about the radius of the Earth then the acceleration of m can be found by equating the two equations above or ma=GMm/r^2.
The ms cancel out and you get a=GM/r^2.
Basically what this means is that if you increase the value of M by some factor, you both increase the force trying to accelerate it and the amount of force needed to accelerate it by the same factor, and you end up with the same acceleration.**
*Even when the two objects are made of the same substance, this can have an effect. If you were to chip off a small bit of dust from that rock, it would also tend to fall slower through air than the whole rock. This is because as things get smaller, their volume (and thus their mass) decreases faster than their surface area And the speck of dust will "float: float on air more readily than the rock does.
** in this example we assume that our dropped object is much, much less massive than the Earth. This way, we don't have to worry about how much the Earth accelerates towards the object since it will be an incredibly small amount and can safely be ignored. In situations where our object where a significant fraction of the mass of the Earth then you would have to account for this and an additional complication is added to the problem.