Let us calculate the equation for the change of the Reynold's number (R = v \, L/\nu) as a function of some dimensionless time t = \lambda \, \tau, \ [\tau] = 1:
<br />
m \, \frac{\nu}{L \, \lambda} \frac{dR}{d\tau} = m \, g - \rho_{0} \, V \, g - \rho_{0} \, \left(\frac{\nu \, R}{L}\right)^{2} \, L^{2} f(R)<br />
where:
<br />
m = \rho \, V = C \, \rho \, L^{3}<br />
and C is a constant that depends on the shape of the body. We simplify the equation to be:
<br />
\frac{dR}{d\tau} = \lambda \left(1 - \frac{\rho_{0}}{\rho}\right) \frac{g L}{\nu} - \lambda \, \frac{\rho_{0} \nu}{\rho \, L^{2}} \, g(R), \ g(R) \equiv \frac{R^{2} \, f(R)}{C}<br />
We choose \lambda so that the coefficient in front of g(R) is equal to 1:
<br />
\lambda = \frac{\rho \, L^{2}}{\rho_{0} \, \nu}<br />
We see that, other things equal, this unit of time is proportional to the square of the linear dimension of the body. The equation for the Reynold's number becomes:
<br />
\frac{dR}{d\tau} = K(L) - g(R), \ K(L) \equiv \left(\frac{\rho}{\rho_{0}} - 1\right) \, \frac{g \, L^{3}}{\nu^{2}}<br />
with the initial condition:
<br />
R(\tau = 0) = 0<br />
The solution of this dimensionless first order ODE is given in analytic form as:
<br />
\tau(R; L) = \int_{0}^{R} {\frac{dR'}{K(L)-g(R')}}<br />
The question is now, given a fixed value of the true velocity v, how does the true time t depend on L. Formally, we try to find the derivative:
<br />
\left(\frac{\partial t}{\partial L}\right)_{v} = \left[\frac{\partial}{\partial L}\left(\lambda(L) \, \tau(R; L)\right)\ \right]_{v} = \lambda'(L) \, \tau(R; L) + \lambda(L) \, \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{v}<br />
where we used the product rule. We will evaluate each of the derivatives. First of all, because \lambda(L) is proportional to the square of of L, we have:
<br />
\lambda'(L) = \frac{2 \lambda(L)}{L}<br />
Next, we note that \tau(R; L) and R(v, L). Therefore, when we keep v constant (as indicated in the subscript of the partial derivative) and change L, we actually change R as well. Using the chain rule for partial derivatives, we may write:
<br />
\left(\frac{\partial \tau(R; L)}{\partial L}\right)_{v} = \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{R} + \left(\frac{\partial \tau(R; L)}{\partial R}\right)_{L} \, \left( \frac{\partial R}{\partial L}\right)_{v}<br />
\tau depends on R through the upper bound of the integral and on L as a parameter. Using the rules of calculus for parametric integrals, the partial derivatives are:
<br />
left(\frac{\partial \tau(R; L)}{\partial R}\right)_{L} = \frac{1}{K(L) - g(R)}<br />
<br />
\left(\frac{\partial \tau(R; L)}{\partial L}\right)_{R} = -\int_{0}^{R}{\frac{K'(L)}{(K(L) - g(R'))^{2}} \, dR'} = - \frac{3}{L} \, \int_{0}^{R}{\frac{K(L)}{(K(L) - g(R'))^{2}} \, dR'}<br />
where, in the last step we used that K(L) is proportional to L^{3} and, therefore:
<br />
K'(L) = \frac{3 K(L)}{L}<br />
Similarly:
<br />
\left( \frac{\partial R}{\partial L}\right)_{v} = \frac{R}{L}<br />
Substituting all these expressions, combining all the remaining integrals under a common integral and simplifying the result, we finally get:
<br />
\left( \frac{\partial t}{\partial L}\right)_{v} = \frac{\lambda}{L^{2}} \, \left\{ \frac{R}{K(L) - g(R)} - \int_{0}^{R}{\frac{K(L)+2 g(R')}{(K(L) - g(R'))^{2}} \, dR'}\right\}<br />
What does this derivative tell us? It tells us how does the time t(v) necessary to reach a particular velocity v change if we change the linear dimension L of the body, other things equal (densities, viscosities, gravitational acceleration and shape). If it is positive, it takes longer to reach the same velocity if we increase the length. This means larger bodies will fall slower. If it is negative, it is quite the opposite: larger bodies fall faster. If it is zero, the time of fall does not depend on the linear dimensions of the body. By equating this with zero for all values of R, I found out that this is possible only if:
<br />
g_{0}(R) \propto R^{-3} \Leftrightarrow f_{0}(R) \propto R^{-5} \Leftrightarrow F_{\textup{drag}} \propto v^{-3}<br />
i.e. the drag force is inversly proportional to the cube of the relative speed of the body through the viscous medium. This is a monotonically decreasing function and unphysical. Let us compare the result when there is no drag (g = 0). By considering inequalities instead of equalities, we can find that if g(R) changes faster than O(R^{-3}) then the derivative is negative and larger bodies would fall faster.
