Was Aristotle Actually Correct About Gravitation? Apparently, Aristotle reasoned (without experimentation) that heavier objects must fall to Earth faster than lighter objects.  For example, a 1,000 kg ball of iron would fall faster than a 1 kg ball of iron.
Galileo is reputed to have disproved this experimentally by dropping two differently heavy objects from the Tower of Pisa, and they were observed to hit the ground at the same time.  Case closed, right?
Well, it suddenly occurred to me this morning that Aristotle was actually correct, and Galileo's experiment was incapable of discerning any difference in arrival times between the two weights he dropped.  And I wonder if our measuring devices today could discern the difference in arrival times in this case?  
The formula for force due to gravitational acceleration is:
$$F = \frac{Gm_1m_2}{d^2}$$
For both the 1000kg and 1kg masses, the product of them both with the Earth's mass is clearly the Earth's mass, so $F$ will be virtually the same for them both.  But if we take the 1000kg mass out to the Moon's orbit, and we stop the moon in its orbit so it stands still with respect to the Earth, the force $F$ will clearly be much larger between the Moon and the Earth than between the 1000kg mass and the Earth.
The result of this is that the Moon and the Earth will meet much faster than the 1000kg mass will meet the Earth.
Am I right, and is Aristotle vindicated, or is there something I'm missing here?
 A: Consider three bodies $Earth$, $B$ and $C$. Now, we take two cases. In the first case, we will observe gravitational attraction between body Earth and the heavier body ($B$); and in the second case, we will observe the gravitational attraction between body Earth and the lighter body ($C$).
Case 1
Consider body $Earth$ and $B$, separated by a distance of $r$. Therefore,
$$F_G=\frac{Gm_{Earth}m_B}{r^2}$$
Acceleration of $B$ is :
$$a_B=\frac{Gm_{Earth}}{r^2}$$
And acceleration of $Earth$ is :
$$a_{Earth}=\frac{Gm_B}{r^2}$$
Case 2
Consider  $Earth$ and body $C$, separated by a distance of $r$. Therefore,
$$F_G=\frac{Gm_{Earth}m_C}{r^2}$$
Acceleration of $C$ is :
$$a_C=\frac{Gm_{Earth}}{r^2}$$
And acceleration of $Earth$ is :
$$a_{Earth}=\frac{Gm_C}{r^2}$$
We see that $a_C=a_B$. But $a_{Earth}$ is changing when the mass is changing.
$$a_{Earth}=\frac{Gm_B}{r^2} ~~~\text{and}~~~ a_{Earth}=\frac{Gm_C}{r^2}$$
Therefore, $a_{Earth}\propto m_{body}$ and $a_{Earth}= \frac{Gm_Cm_{body}}{r^2}$
The acceleration of Earth is directly proportional to the mass of the body. So as the mass of the object increases, the acceleration of Earth increases, decreasing the time of impact and hence, decreasing the time taken by the body to fall.

Note - Although Aristotle was partially correct, he was far from the real understanding of Gravity. Galilean Gravity was closer to the truth. Aristotle deduced that heavy objects fall faster because his observation was affected by air resistance, which is not the real cause of heavier objects falling faster. Galileo had already refuted his explanation of gravity, explaining the affect of air resistance on falling bodies. Galilean Gravity  was correcter.
Please refer to this answer for a detailed explanation.
A: Clearly missed the point in this statement.

For both the 1000kg and 1kg masses, the product of them both with the earth's mass is clearly the earth's mass, so F will be virtually the same for them both.

The product is definitely not the same and the force on 1000kg ball is exactly 1000 times greater that the force on 1kg ball and much much more on the moon. But the acceleration being ${\frac{F}{m}}$ is going to be same for all these objects.
A: If you assume that the earth is immobile, the fall time if the same.
If you consider that the earth moves because of the mass of the item, yes, the heavy item fall time is shorter (infinitesimal difference). If both items are dropped at the same time, they will hit the floor at the same time though.
(This question has already be answered : Don't heavier objects actually fall faster because they exert their own gravity?)
