Why do two bodies of different masses fall at the same rate (in the absence of air resistance)? I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.
According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.
BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.
I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".
Can someone explain this to me? I feel pretty dumb right now :)
 A: Because force 'pushing' object closer to earth is proportionally bigger for 'heavier' object. 
But heavier object is also have higher gravitation force.
So these two factors perfectly compensate each other: Yes you need more force for a set acceleration, but more force is here due to heavier mass.
A: Newton's gravitational force is proportional to the mass of a body, $F=\frac{GM}{R^2}\times m$, where in the case you're thinking about $M$ is the mass of the earth, $R$ is the radius of the earth, and $G$ is Newton's gravitational constant.
Consequently, the acceleration is $a=\frac{F}{m}=\frac{GM}{R^2}$, which is independent of the mass of the object. Hence any two objects that are subject only to the force of gravity will fall with the same acceleration and hence they will hit the ground at the same time.
What I think you were missing is that the force $F$ on the two bodies is not the same, but the accelerations are the same.
A: it is because the Force at work here (gravity) is also dependent on the mass
gravity acts on a body with mass m with 
$$F = mg$$
you will plug this in to $$F=ma$$ and you get
$$ma = mg$$
$$a = g$$
and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.
This is a peculiar aspect of gravity and underlying this is the equality of inertial mass and gravitational mass (here only the ratio must be the same for this to be true but Einstein later showed that they're really the same, i.e. the ratio is 1)
A: There are two ways that mass could effect the time of impact:
(1) An object which is very massive has a stronger attraction to the earth. Logically, this might make the object fall faster and so reach the ground sooner.
(2) An object which is very massive is difficult to get moving. (I.e. it has very high inertia.) Thus one might logically expect the very massive object to be more difficult to get moving and so to lose the race.
The miracle is that in the world we live in, these two effects exactly balance and so the heavier mass reaches the ground at the same time.

Now let me give a simple explanation for why it's natural that this comes about. Suppose we have two very heavy masses. If we drop them separately they take some time T to fall. On the other hand, if we attach them together, will they take the same length of time? Think about a sphere split into two halves:

Two two halves of the sphere would fall at the same speed as each other. So if you dropped them next to each other, they'd fall together. And dropping them next to each other isn't going to be any different from screwing them together and dropping them together. That is, there won't be any force on the screws. So the combined (or screwed together) sphere has to fall at the same rate as the split sphere.
A: Lets say two separate mass $M_1$ and $m_2$ where $M_1$ >> $m_2$, both fall, from the same instant in a gravitational field 
Force on $M_1$ is $F1$ = G $M_{\text{earth}}$    $M_1$/ $R^2$
Force on $m_2$ is $F2$ = G $M_{\text{earth}}$  $m_2$/ $R^2$
Therefore the forces are  $F1$ >> $F2$
So, most people think $M_1$ should accelerate much faster than $m_2$ 
But as you wrote above a = F/m and substituting $F1$, $F2$, $M_1$ and $m_2$ into that formula we find:
$F1$/$M_1$ = $F2$/$M_2$ = G $M_{\text{earth}}$/ $R^2$
Therefore the acceleration is independent of the masses we drop, and is a constant. 
EDIT
Bother, by the time I had written this down, it was answered, obviously the other authors had a different acceleration in their typing.
A: I am answering this in my own way critics are welcomed.
So, Generally consider two bodies of same shape,surface area (same) but MASSES DIFFERENT.
Now if you release both the bodies from a similar height say h, with u =0m/s for both (initially rest or aka free fall) they travel h height in same time .
So,what is happening here is we know Newton's law of gravitation like F = (GMm)/R^2
Where, G is Grav.const. and M and m are masses along which gravitation (attraction only always) acts and R is the distance between the center of masses or simply centers of body (We just take point objects usually dw bout it).
Now F=[(GMē) m]/R^2 Mē= earth mass and R dist.between earth center and object
You can simplify (GMē/R^2) as g' usually when distance of object from earth surface is very less than earth radius you can take g'=g =9.8m/s^2
So,finally F=gm=mg
For two different masses This F so called gravitational force of attraction is different but look g is same for both masses that is acceleration of two bodies is same even though there is different mag.of force on them .
Mostly ppl are confused here they generally interpret this as greater the force greater the acc.n which is not right to be honest
Greater force doesn't mean greater acc.n greater force maybe because of their masses being different but with same acc.n so it is possible for two bodies of different masses experiencing different forces having same acc.n
It is more right if you put this explanation of so called g being same for two bodies rather than a1=a2=g YOU BETTER USE
a1=F1/m1 and a2=F2/m2 and then you can see that g=F1/m1=F2/m2 You will be getting a lil clarity if you do it this way .
Therefore since same acc.n it says both bodies have same change in their velocities over equal time interval so these bodies attain same final velocities over a particular time since they started with 0m/s at same time that's why they cover equal disp.during their journey down hence they reach ground in same time.
A: Let us think about this by contradiction.
Suppose the two masses fall at different rates (say, heavier mass falls faster), then if you tie the two masses together, what will happen?
Solution #1. if you tie the masses together, they form a even larger mass, thus they fall faster
Solution #2. if you tie the masses together, the lighter mass will give the heavier mass a drag force, thus they fall slower.
The two solutions contradict each other; so they must fall at the same rate.
