One of two coefficients of fit of relationship density to fall time.

I worked the problem the hard way. Using the densities of pure elements ranging from lithium to osmium, I computed the fall times using numerical solution of the differential equation $,\left\{y''(t)=\frac{\text{na}}{(2 y(t))^2},y'(0)=0,y(0)=5.\right\}$ . The value of $\text{na}$ is the acceleration due to gravity at the time of contact. Then I did the log-log fit and looked at the answer. Duh! Since gravity is an inverse square phenomenon, the value of $a$ has to be the reciprocal of $-2$ , i.e., $-\frac12$ . That is what the fit showed. The value of $b$ based on the units and the experimental setup is $1\,425\,242.3116$ seconds.

The list of elements used: Actinium, Aluminum, Americium, Antimony, Arsenic, Barium, Berkelium, Beryllium, Bismuth, Boron, Bromine, Cadmium, Calcium, Californium, Carbon, Cerium, Cesium, Chromium, Cobalt, Copper, Curium, Dysprosium, Erbium, Europium, Gadolinium, Gallium, Germanium, Gold, Hafnium, Holmium, Indium, Iodine, Iridium, Iron, Lanthanum, Lead, Lithium, Lutetium, Magnesium, Manganese, Mercury, Molybdenum, Neodymium, Neptunium, Nickel, Niobium, Osmium, Palladium, Phosphorus, Platinum, Plutonium, Polonium, Potassium, Praseodymium, Promethium, Protactinium, Radium, Rhenium, Rhodium, Rubidium, Ruthenium, Samarium, Scandium, Selenium, Silicon, Silver, Sodium, Strontium, Sulfur, Tantalum, Technetium, Tellurium, Terbium, Thallium, Thorium, Thulium, Tin, Titanium, Tungsten, Uranium, Vanadium, Ytterbium, Yttrium, Zinc and Zirconium. I only used element known to solids or liquids at standard temperature and pressure and which had known densities. I could have used arbitrary densities just as well.

$g=\frac{6.674 \grave{ } * {}^{\wedge} -11 \text{Quantity}[\text{Meters}]^3}{\text{Quantity}[\text{Kilograms}] \text{Quantity}[\text{Seconds}]^2}$

I actually solved the differential equation twice once by a single Riemann sum to get a guess of how long it takes for the balls to meet and using Mathematica's internal differential equation solver to get a more accurate answer. The reason for this is that the differential equation goes singular shortly after contact when the centers distance goes to 0.

$\text{table}=\text{Table}\left[ \\ \text{density}=\text{ElementData}[\text{element},\text{Density}]; \\ \text{mass}=\text{density} \text{Volume}[\text{Ball}[\{-5,0,0\},\text{Quantity}[1,\text{Meter}]]]; \\ a=d\to \text{Evaluate}\left[-\frac{g \text{mass}}{d^2}\right]; \\ \text{step}=300\text{s}; \\ t=0\text{s}; \\ s=5.\text{m}; \\ v=0.\text{m}/\text{s}; \\ \text{While}\left[s>\text{Quantity}[1.,\text{Meter}], \\ \text{acceleration}=a(2 s); \\ s\text{+=}\frac{\text{acceleration} \text{step}^2}{2}+\text{step} v; \\ v\text{+=}\text{acceleration} \text{step}; \\ t\text{+=}\text{step}; \\ \right]; \\ \text{guess}=\text{QuantityMagnitude}[t]; \\ \text{ClearAll}[y]; \\ \text{ClearAll}[t]; \\ \text{na}=\text{QuantityMagnitude}[a(1)]; \\ y=\left(y\text{/.}\, \text{NDSolve}\left[\left\{y''(t)=\frac{\text{na}}{(2 y(t))^2},y'(0)=0,y(0)=5.\right\},y,\{t,0,0.9991 \text{guess}\},\text{Method}\to \text{ImplicitRungeKutta}\right]\right)[[1]]; \\ \text{ClearAll}[\text{time}]; \\ \text{time}=\text{time}\text{/.}\, \text{FindRoot}[y(\text{time})=1,\{\text{time},0.98 \text{guess}\}]; \\ \{\text{element},\text{density},\text{time}\text{s},\text{UnitConvert}[\text{time}\text{s},\text{MixedUnit}[\{\text{Days},\text{Hours},\text{Minutes},\text{Seconds}\}]]\},\{\text{element},\text{elements}\} \\ \right]$

The output is too large for easy cutting-and-pasting. Here is the log-log plot:

The fit is about 10 digits accurate.