A note on alternative approaches to solving a variable mass problem involving accretion

hailstones Variable mass problems are the archetypal problems of `rocket science’ in which aspects of the motion of a rocket need to be calculated as the rocket burns up its fuel. Variable mass problems also arise in situations in which a body gains mass by accretion, e.g., the way a hailstone (pictured) grows by condensation as it falls. The calculations are based on the impulse-momentum principle which states that the change in linear momentum of a system in the time interval dt is equal to the impulse of the external forces acting on the system in that interval. This principle derives from Newton’s second law for systems of particles, expressed in terms of momentum (intuitively, F = \frac{dp}{dt} gives us dp = Fdt).

In this note I want to explore a variable mass problem involving a falling hailstone gaining mass by accretion, focusing on the derivation and solution of the differential equation of motion. I find the problem interesting because it is illustrative of the fact that it is sometimes possible to solve for the required quantities by direct integration without going through the step of first finding a general solution of the differential equation. However, going through the longer process of first finding a general solution of the differential equation enables further insights to be obtained. In this note I will solve the problem in both ways, i.e., by direct integration and by finding a general solution of the differential equation. I will use the latter to delve a bit deeper into the situation.

The problem is as follows (the reader should attempt to solve it before reading on):

A spherical hailstone is falling under gravity in still air. At time t the hailstone has speed v. The radius r increases by condensation at the rate \frac{dr}{dt} = kr where k is a constant. Ignoring air resistance, derive the equation of motion of the hailstone and find the time taken for the speed of the hailstone to increase from \frac{g}{9k} to \frac{g}{6k}.

It helps to draw a sketch of the situation, such as the following:

Fig1

Using the impulse-momentum principle (dp = Fdt) and noting that the external force here is equal to mass times acceleration due to gravity, we can write

(m + dm)(v + dv) - (mv + 0dm) = (m + dm)gdt

which (treating products of differentials as zero) simplifies to

mdv + vdm = mgdt

Note that the sign of the acceleration due to gravity on the right hand side is positive because it is in the same direction as the velocity of the falling hailstone. Dividing through by dt gives

m\frac{dv}{dt} + v\frac{dm}{dt} = mg

At this point we can use the fact that mass is proportional to volume in this case, so

m = \lambda\frac{4}{3}\pi r^3

and therefore

\frac{dm}{dt} = \lambda 4 \pi r^2 \frac{dr}{dt} = \lambda 4 \pi r^3k

so

\lambda\frac{4}{3}\pi r^3 \frac{dv}{dt} + v \lambda 4 \pi r^3k = \lambda\frac{4}{3}\pi r^3g

which simplifies to

\frac{dv}{dt} + 3vk = g

This is the required equation of motion for the hailstone. As stated earlier, we can now proceed to find the time taken for the speed of the hailstone to increase from \frac{g}{9k} to \frac{g}{6k} either by direct integration, or by first solving the differential equation of motion to find an expression for v. The latter approach has the advantage that we can use the expression for v to obtain other insights. Using the direct integration approach, we would separate the variables and write

\int_0^T dt = \int_{\frac{g}{9k}}^{\frac{g}{6k}}\frac{dv}{g - 3vk}

and therefore the time taken for the speed of the hailstone to increase from \frac{g}{9k} to \frac{g}{6k} is

T = \big[-\frac{1}{3k}\ln(g-3vk)\big]_{\frac{g}{9k}}^{\frac{g}{6k}} = -\frac{1}{3k}\ln(\frac{3}{4})

To get the same answer by first solving the differential equation to obtain an expression for v, we would observe that the general solution of a first-order differential equation of the form

\frac{dy}{dx} + Py = Q

is

y = e^{-I}\int Qe^I dx + ce^{-I}

where

I = \int P dx

Here we have the differential equation

\frac{dv}{dt} + 3vk = g

so

I = \int 3k dt = 3kt

and therefore

v = e^{-3kt} \int g e^{3kt} dt + ce^{-3kt} = e^{-3kt} g \frac{1}{3k} e^{3kt} + ce^{-3kt} = \frac{g}{3k} + ce^{-3kt}

So the general solution is

v = \frac{g}{3k} + ce^{-3kt}

where c is an arbitrary constant.

Now, when v = \frac{g}{9k}, we have

\frac{g}{3k} + ce^{-3kt} = \frac{g}{9k}

and solving for t gives

t_1 = -\frac{1}{3k}\ln(-\frac{2g}{9ck})

Similarly, when v = \frac{g}{6k}, we have

\frac{g}{3k} + ce^{-3kt} = \frac{g}{6k}

and solving for t gives

t_2 = -\frac{1}{3k}\ln(-\frac{g}{6ck})

Then the time taken is

T = t_2 - t_1 = -\frac{1}{3k}\big[\ln(-\frac{g}{6ck}) - \ln(-\frac{2g}{9ck})\big] = -\frac{1}{3k}\ln(\frac{3}{4})

as before. Note that the arbitrary constant c disappears in the course of the calculation.

Although this latter approach seems more long-winded, it has the advantage that we can now use the general solution of the differential equation we have obtained to explore other features of the problem. For example, although it is not specifically required in the original problem, we could work out how far the hailstone travelled in the time it took for its speed to increase from \frac{g}{9k} to \frac{g}{6k}. To do this, we would rewrite the general solution of the differential equation obtained above as

\frac{dx}{dt} = \frac{g}{3k} + ce^{-3kt}

Separating the variables and integrating on both sides then gives the following expression for the position of the hailstone as a function of time:

x(t) = \frac{gt}{3k} - \frac{c}{3k}e^{-3kt} + d

where d is a second arbitrary constant. At the time t_1 calculated above we have

x(t_1) = -\frac{g}{(3k)^2}\ln(-\frac{2g}{9ck}) + \frac{c}{3k}\frac{2g}{9ck} + d

= -\frac{g}{(3k)^2}\ln(-\frac{2g}{9ck}) + \frac{2}{3}\frac{g}{(3k)^2} + d

and at the time t_2 calculated above we have

x(t_2) = -\frac{g}{(3k)^2}\ln(-\frac{g}{6ck}) + \frac{c}{3k}\frac{g}{6ck} + d

= -\frac{g}{(3k)^2}\ln(-\frac{g}{6ck}) + \frac{1}{2}\frac{g}{(3k)^2} + d

Therefore the distance travelled by the hailstone in the time it took for its speed to increase from \frac{g}{9k} to \frac{g}{6k} is

X = x(t_2) - x(t_1) = -\frac{g}{(3k)^2}\big(\frac{1}{6} + \ln(\frac{3}{4})\big)

Note that both the arbitrary constants c and d have disappeared in the course of the calculation.