Seeing as this conversation has drifted a bit to the esoteric side of physics....let me mention that two objects of differing mass dropped simultaneously in a vacuum do NOT reach the ground at the same time. The one with more mass will drop faster...if only by a tiny amount. The word 'same' is only as valid as your ability to measure is. Two rods measured with a tape will look to be the 'same' if within 1/64" or so.....but easily seen to be different with a more precise measuring device. So it goes with the gravity drop test.....measure close enough and the heavier object will win the race to the ground by a tiny fraction.
I don't believe that is correct, in a vacuum both the light and heavy objects will accelerate at exactly the same rate and thus fall at the same velocity assuming they are released at the same time from the same altitude. That math is relatively easy to show.
Gravitational Force F = G M m / r^2
G = Gravitational Constant (fix value across the universe)
M = Mass of the earth (or any planet or object we want to be attracted to)
m = mass of object we are going to release
r = Distance between the centers of mass of the M and m
Fundamental equation of motion (ie Newton's second law of motion). This equation governs the motion of all objects and is accurate for nearly all situations (moving at a large fraction of the speed of light breaks it, so does close proximity to black holes, and a few other situation but otherwise its good to go.)
F = m a
F = Force
m = Mass (in this case the mass of the object we will drop)
a = Acceleration
Set these two equation equal to each other. Which is basically asking what will the m object's motion be like under the force of gravity.
m a = G M m / r^2
Algebra tells us we can cancel the m's on both side of the equation. This then leaves us with the acceleration due to gravity
a = G M / r^2
Thus the mass of the m object does not factor into the acceleration the m object experiences toward the M object. ie a heavier m object produces a proportionally larger force thus maintaining the same acceleration. All m objects at the same radius from M will experience the exactly the same acceleration independent of their own mass. Barring other forces at work both object should experience the exact same acceleration and fall exactly together.
G = 6.67408 × 10^-11 m3/kg-s^2
M = 5.972 × 10^24 kg (for the earth
r = 6.371 x 10^6 m
a = 9.820 m/sec^2 (this is an average value since we used an average radius. The earth is not a perfect sphere. This value is slight higher than the usual measured value due to not accounting for the centrifugal acceleration most of the surface experiences due to the planet's rotation, the poles zero, max as the equator. Science often uses the average measured value 9.80655 m/s^2
Rambling...