The falling raindrop, revisited

A perennial homework exercise in differential-equation-based courses in Newtonian mechanics is the problem of a raindrop falling through mist, collecting mass. 1,2,3,4 If the rate of accretion is assumed to depend only on the raindrop's current mass (or radius) and not on its velocity, then the solution is fairly straightforward. When, by contrast, the rate of accretion is taken to depend on both the mass and the velocity, one is faced with a pair of coupled differential equations, and the trick for disentangling them can be surprisingly difficult to find -not only for the student, but also for the instructor who has forgotten the method after some years' absence and must rediscover it (as I can personally testify from recent experience). 5

Here I would like to show that a very general version of the raindrop problem [see ( 5)/( 6) below] can be solved by using a versatile technique that ought to have a place in all students’ (and instructors’) mathematical arsenals: namely, eliminating reference to the old independent variable (here the time t) and temporarily taking one of the old dependent variables as the new independent variable. Students may remember this trick from the analysis of one-dimensional motion with a force that depends on position (x) and velocity (v) but not explicitly on time:

By using the chain rule

we can temporarily eliminate t and instead take x as the new independent variable: this yields the first-order differential equation

for the unknown function v(x). In some cases this equation can be solved explicitly. 6 Once one has in hand the function v(x), one can reinstate time and solve (at least in principle)

the first-order separable differential equation

Let us now consider the raindrop problem, which involves a pair of coupled differential equations for two unknown functions: the raindrop’s mass m(t) and its velocity v(t). The first equation is the Newtonian equation of motion for the raindrop,

which is obtained by the standard procedure of looking at the same collection of water particles (the “system”) at two nearby times, t and t + ∆t, and writing that the rate of change of the system’s total momentum equals the total external force on the system. The second equation states the hypothesized law of accretion for the raindrop: here I shall consider the very general form

where λ > 0 is a constant and α and β are (almost) arbitrary exponents. This form includes the two most commonly studied cases -namely the easy case (α, β) = ( 2 3 , 0) [accretion proportional to the surface area of a spherical raindrop, with resulting acceleration g/4] and the hard case (α, β) = ( 23 , 1) [accretion proportional to the volume swept out, with resulting acceleration g/7] -but is much more general. 7 I will show that all these problems can be solved by a unified technique.

First, a few preliminary remarks. Since (5)/( 6) is a pair of first-order differential equations for two unknown functions, the general solution will contain two constants of integration.

Since this system is time-translation-invariant, one of the constants of integration simply sets the origin of time. The other constant of integration fixes the relation between the initial mass and the initial velocity: that is, it fixes the mass at the moment when the velocity has a specified value (or vice versa). The simplest solution arises by demanding that m = 0 when v = 0; but we will make some partial progress toward finding the general solution as well.

As mentioned above, all the cases with β = 0 are easy: the accretion equation ( 6) can be solved immediately for m(t) [it is separable], and the Newtonian equation ( 5) can then be solved for v(t) [it is linear first-order with nonconstant coefficients]. The trouble arises when β = 0, as now the equations ( 5) and ( 6) are coupled.

To decouple them, we employ the technique mentioned earlier: use the chain rule

forget temporarily about time, and instead consider the velocity v to be a function of the mass m (i.e. we temporarily use m as the independent variable). Inserting (7) into the Newtonian equation ( 5) and using the accretion equation ( 6) to eliminate dm/dt (which now multiplies both terms on the left-hand side), we obtain

Making the change of variables w = v 1+β , we find 8

which is a first-order linear differential equation with nonconstant coefficients for the function w(m). The integrating factor is m 1+β , and after standard manipulations we obtain the general solution

It is most convenient to express the constant of integration C in terms of the raindrop’s mass m 0 at the moment its velocity is zero: we have

The simplest case is m 0 = 0. Then we have 11

where K ′ is a constant that we need not write out. We now insert (12) into the Newtonian equation ( 5). Since each of the terms in this equation is linear in m, the constant K ′ drops out, and we get

So we are done: we have proved that the raindrop falls with constant acceleration 1-α 2+β-α g. [When (α, β) =

This content is AI-processed based on open access ArXiv data.