The Verification Venue / mass against air
Two Fingers
of Air
Everyone is taught the answer is no. In a vacuum, that answer is exact in the ordinary test-body model. In air, keep shape and size fixed and the heavier body really does arrive first. The air pushes equally hard on both equal silhouettes; the heavier one has more weight with which to ignore it.
The tower below is a stage, not a claim that Galileo performed the legendary Pisa drop. Press Pisa + air. Two 10 cm spheres leave together. Then remove the air and watch a nearly five-metre separation become zero.
model time: 3.393 s
- iron, 4.12 kg
- 3.393 s
- oak, 0.37 kg
- 3.570 s
- gap when winner lands
- 4.925 m
- selected world
- air
Same diameter, different mass: the iron sphere wins by 0.178 s.
The air does not know the mass
For the equal-sized spheres, area and drag coefficient are pinned. The bench solves the quadratic-drag equation live:
dv/dt = g - (rho Cd A / 2m)v²
vt = sqrt(2mg / rho Cd A)
With fixed Cd A, terminal velocity grows as the square root of mass. That is the regime your eyes usually see. It is not a general licence to compare arbitrary objects: a light dart can beat a heavy parachute. Small dust and droplets can also live in a different, Stokes-drag regime.
The second fall
In Two New Sciences, Galileo asks us to imagine a 100-pound and a 1-pound body falling through 100 braccia. He says the lighter trails by only two finger-breadths when the larger lands.
Press Galileo's claim + air. Treat both as iron spheres, convert 100 braccia to a declared 57 m, and take one finger-breadth as a declared 2 cm. The same model returns 0.942 m, or 47.1 finger-breadths. His claimed 4 cm is smaller by a factor of 23.5.
This is not a reconstructed experiment. Sphere sizes, the braccio conversion, constant Cd = 0.47, and still sea-level air are model choices. Adler and Coulter's 1978 analysis likewise found the famous small separation doubtful. The bench's exact 0.942 m belongs to the assumptions printed here, not to their paper.
The check
Pisa model preset
rho = 1.225 kg/m³
Cd = 0.47
diameter = 0.100 m
m = 4.12, 0.37 kg
height = 55.86 m
The analytic solution gives terminal speeds 133.679 and 40.060 m/s, landing times 3.39251 and 3.57014 s, and a 4.92519 m gap.
Remove the medium
t = sqrt(2h/g)
At 55.86 m, both read 3.375244 s and the gap is 0.000 m. At 1.6 m with lunar g = 1.62 m/s², both read 1.405457 s.
Free choices and uncertainties: constant density, still air, no buoyancy, spin, wind, or release error; spherical bodies; fixed drag coefficient; material masses rounded as supplied; 57 m per 100 braccia; 2 cm per finger-breadth. The 4.925 m result is a model output, not a Pisa measurement. NASA's stated 0.03 kg feather mass appears rounded and is not needed to obtain the vacuum equality.
The equality underneath
NASA reports that David Scott released a 1.32 kg hammer and its stated 0.03 kg falcon feather together from approximately 1.6 m on 2 August 1971. In the Moon's near-vacuum they arrived together within the accuracy of the release. The preset computes a model time; it does not pretend NASA measured 1.405457 seconds.
A pedantic, separate-drop footnote: in a two-body calculation the Earth moves toward the dropped object, so a 4.12 kg object's relative acceleration differs from a test particle's by about m/MEarth = 6.9 x 10-25. This is not composition-dependent falling. MICROSCOPE found no such equivalence-principle violation, reporting eta(Ti,Pt) = [-1.5 +/- 2.3(stat) +/- 1.5(syst)] x 10-15.