The raindrop that falls at a seventh of g

Collisions are easier than forces found that momentum conservation lets a collision be solved without knowing anything about what happens inside it. The push that needs nothing to push against applied the same accounting to a rocket, a body that throws mass away and is pushed forward by what it throws. The pile that lands heavier than it weighs ran it the other way, for a chain falling onto a scale: each link arriving at the pile is stopped in an instant, the scale has to supply the impulse, and the reading is three times the weight of what has landed.

This essay turns the falling chain inside out. Instead of a body losing mass to a pile, it follows a body gaining mass as it falls, by gathering what lies in its path at rest. The classic case is a raindrop falling through a still mist of smaller droplets and absorbing every one it touches. With no air resistance, the question is how fast it falls. The answer is a seventh of g — not approximately, and not for one particular drop or mist, but as the acceleration every such drop settles into.

Momentum that has to be shared #

A drop of mass $m$ moving at speed $v$ has momentum $mv$, and gravity adds momentum at the rate $mg$. If the drop’s mass were fixed, all of that would go into speed, and it would fall at $g$. But the drop is collecting droplets that were at rest. Each one, absorbed, must be brought up to the drop’s speed, and the momentum it acquires comes from the drop. The equation of motion is

$$\frac{d(mv)}{dt} = mg,$$

with the mass growing as the drop sweeps up mist: in a time $dt$ it passes through a cylinder of cross-section $\pi r^2$ and length $v\,dt$, and gathers the water in it. If the mist holds $\rho_m$ kilograms of water per cubic metre, the mass grows at $\rho_m \pi r^2 v$.

Those two lines contain the whole problem, and its first surprise comes before any solving. The mass of a drop of density $\rho$ is $\frac{4}{3}\pi\rho r^3$, so its growth rate is $4\pi\rho r^2\,dr/dt$. Setting that equal to the rate of sweeping, the $\pi r^2$ cancels and

$$\frac{dr}{dt} = \frac{\rho_m}{4\rho}\,v, \qquad \text{so} \qquad \frac{dr}{dx} = \frac{\rho_m}{4\rho}.$$

The radius grows by a fixed amount for every metre fallen. It does not matter how fast the drop is going: a fast drop sweeps a given length of path in less time, and gathers the same water.

Growing by the metre #

For a real cloud the numbers are sobering. A typical cloud holds about half a gram of liquid water in each cubic metre, spread among droplets a few micrometres across. A drop collecting all of it grows by $0.5 \times 10^{-3}/(4 \times 1000)$ metres per metre — 125 nanometres for every metre it falls, an eighth of a millimetre per kilometre. To grow from a cloud droplet ten micrometres across to a drizzle drop of a millimetre’s radius it must fall about eight kilometres through the cloud; to reach a two-millimetre raindrop, sixteen. Clouds are rarely that deep, and real drops collect only a fraction of the droplets in their path, because many are swept aside in the air flowing round the drop.

That is why rain takes time to form and why a cloud must be deep or turbulent to make it. A drop that grows by collision needs a long path, and a cloud supplies one by carrying its growing drops up in its updraughts and letting them fall back, over and over. The linear growth with distance, a consequence of nothing more than geometry, sets the scale of the whole process.

A seventh of g #

With the growth law in hand, the motion can be found. Write the momentum equation in terms of the radius rather than the time, using $dt = dr/(kv)$ with $k = \rho_m/4\rho$. The equation becomes one for $v^2$ as a function of $r$, and its solution for a drop that starts from nothing is

$$v^2 = \frac{2g}{7k}\,r.$$

The speed grows as the square root of the radius, and the radius grows in proportion to the distance fallen, so the speed squared is proportional to the distance fallen — which is the signature of uniform acceleration. Differentiating gives the acceleration: exactly $g/7$. The density of the mist has cancelled. So has the density of water, and the size of the drop.

A drop that starts with a finite size does not begin at a seventh of g. At first it has gathered almost nothing compared with its own mass and falls freely; as it grows, the gathered water comes to dominate, and its acceleration falls towards $g/7$, which is approached from above for any starting size. The figure follows three drops in a dense idealised mist, chosen so the approach takes seconds: the largest takes longest, the smallest almost none, and after eight seconds all three are falling at the same 0.143 g. A thinner mist makes the approach slower and leaves the destination where it is.

Why the answer had to be a pure number #

The cancellation that leaves $g/7$ can be seen before any equation is solved, and seeing it explains why the result is so universal. The only place the mist’s density enters the problem is in the growth law, $dr = k\,dx$, which ties the drop’s radius to the distance fallen through the single constant $k = \rho_m/4\rho$. Measure distances in units of $1/k$ — for a real cloud, eight thousand kilometres — and $k$ disappears from the equations. What is left contains $g$ and nothing else that has units, apart from the drop’s starting size.

So once the starting size has been forgotten, the acceleration can only be $g$ times a pure number. Nothing in the problem could make it depend on the mist, the water or the drop. The number itself has to come from the geometry: the 3 in $m \propto r^3$ and the 2 in the area $\pi r^2$, which together make the mass grow as the cube of the distance. That the number comes out as one-seventh is arithmetic; that it is a number at all is dimensional analysis.

The same argument predicts how the drop ages. Falling at $g/7$ from rest, it covers a distance $gt^2/14$ and so its radius grows as the square of the time and its mass as the sixth power. In a real cloud, holding half a gram of water per cubic metre, a drop in vacuum would need a minute to fall two and a half kilometres and grow by a third of a millimetre. The idealised drop is not a fast one either; it is merely free of the air, which would hold it to a few metres a second and make the same growth take many minutes.

Where three-sevenths of the work goes #

Gravity does work $mg\,dx$ on the drop as it falls a distance $dx$. A drop of fixed mass would turn all of it into kinetic energy. A drop accelerating at $g/7$ plainly does not: its speed is far below what free fall would give at the same height. The rest goes somewhere.

It goes into the gathering. Each droplet absorbed is at rest and is brought instantly to the drop’s speed in a perfectly inelastic collision, and such a collision always turns kinetic energy into heat: the drop loses momentum to the droplet, and the kinetic energy of the combined body is less than the drop’s was. Adding up those losses over a fall at $g/7$ gives exactly three-sevenths of the work gravity has done. The figure follows the shares for a drop starting at a fifth of a millimetre: at first it is falling freely and keeps everything; as the gathered mass takes over, its kinetic energy settles to four-sevenths of the work and the losses to three.

That is a statement nobody would guess from the equation of motion, which contains no friction and no heat. The drop’s centre of mass, together with the mist it has gathered, still obeys the point that keeps moving: the momentum of the whole system grows at exactly the rate gravity supplies. What is not conserved is the kinetic energy, which momentum conservation never promised. It is the same accounting that made the chain land heavier than it weighs. There each arriving link was stopped by the pile, with all its kinetic energy lost, and the force of stopping it doubled the reading. Here each gathered droplet is started by the drop, and the loss in starting it is what slows the fall. The energy that depends on the observer found that kinetic energy is frame-dependent while energy conservation holds in every frame; the dissipated three-sevenths is the part that is the same in all of them.

A family of falls #

The arithmetic generalises immediately, and the generalisation is where the seventh comes from. Suppose a falling body’s moving mass grows as some power of the distance it has fallen, $m \propto x^n$, by picking up material at rest. Trying $v^2$ proportional to $x$ in the momentum equation gives an acceleration of $g/(2n+1)$, and the share of gravity’s work lost in the pick-up is $n/(2n+1)$.

A stone gathers nothing: $n = 0$, acceleration $g$, nothing lost — the fall that does not depend on what is falling, which is the special case in which nothing is gathered. A chain lying in a heap on a table, with one end hanging through a hole, pulls out more links as it falls, and the moving length is just the distance fallen: $n = 1$, acceleration $g/3$, a third of the work lost. Cayley found that in 1857, in one of the first published problems about a body of varying mass. A raindrop’s mass goes as the cube of its radius, and its radius as the distance fallen: $n = 3$, a seventh of $g$, three-sevenths lost.

As $n$ grows the acceleration falls towards zero and the loss climbs towards one half — the share of the energy a perfectly inelastic pick-up wastes when the body doing the picking up is far heavier than what it picks up, as five balls and the law that does not choose found the equations of collision leave open and the materials decide. The seventh is not a coincidence of water. It is the $n = 3$ member of a family in which the only input is how steeply the moving mass grows.

The chain that does not leave its heap quietly #

Cayley’s chain shows the lost energy in a form that can be heard. A chain piled in a loose heap and pulled out through a hole by the weight of the part already hanging snatches each link from rest, and the snatch is a small collision: a link at rest is jerked to the speed of the moving chain by the tension in the link ahead. The jerks make the rattle a chain makes running off a heap, and the energy that goes into them — a third of the work gravity does — ends up as sound and heat in the heap. The falling end accelerates at $g/3$, and its speed after falling a distance $x$ is $\sqrt{2gx/3}$, well below free fall.

The assumption hidden in that result is that each link is picked up from rest with no help from the heap. In 2013 a demonstration showed what happens when that assumption fails. A long chain of beads pulled out of a beaker and over its rim towards the floor does not slide over the rim; it leaps up from the beaker in a fountain that can stand tens of centimetres above it. The explanation, worked out the following year, is that a link lifted from the heap by one end acts as a lever: the heap pushes up on its other end, so the heap supplies part of the momentum the link needs and the chain is pulled up faster than the tension alone would pull it. The heap becomes a source of momentum, the energy lost in the pick-up is less than Cayley’s third, and the excess lifts the chain into the air.

The raindrop has no such help. A droplet of mist is a free body at rest, nothing pushes it towards the drop, and the full inelastic cost of starting it is paid. That is why its three-sevenths can be computed exactly, while a chain’s losses depend on how the heap is piled.

What air does to it #

None of this describes the rain that falls. A real drop falls through air, and air resistance is far stronger than the drag of gathering droplets.

A drop a quarter of a millimetre across reaches its terminal speed, where the drag of the air balances its weight, within a fraction of a second; a millimetre drop in about a second, at six or seven metres a second. Once there it stays at terminal speed, which rises only as the drop slowly grows. Over four seconds of gathering in a real cloud the radii in the figure grow by less than one per cent, and the drag of the gathered droplets is a tiny correction to the drag of the air. In real rain the speed is set by the air, and the sweeping matters only for how big the drop gets. The air’s own effect on a falling body is the subject of the angle that drag moves, where a thrown ball’s best launch angle falls below forty-five degrees once drag is included; here the drag is so strong compared with the gathering that it decides everything about the speed.

So the seventh of $g$ is a property of an idealisation: a drop in a vacuum filled with motionless mist. It is still worth having, for the same reason the bounces that add up to a stop was worth having for a ball that is not quite elastic. It isolates one effect — the cost of sharing momentum with what is gathered — and shows its size and its universality before the other effects are put back.

Snowploughs, planets and dust #

Gathering mass from rest is common outside meteorology, and the same accounting runs through all of it. A snowplough pushing snow ahead of it at constant speed has to supply a force equal to the rate at which it adds momentum to the snow, $\rho A v^2$, which rises as the square of its speed, and half the work it does goes into heat in the snow. A shock wave sweeping up interstellar gas — the “snowplough” phase of a supernova remnant — slows as it gathers, conserving momentum rather than energy, because the inelastic sweeping radiates the difference away. A planet forming by gathering dust and small bodies in its orbit, or a dust grain falling through a gas and collecting molecules, obeys the same equation with different powers. In each case momentum is shared and energy is lost, and the share lost is fixed by how fast the gathered mass grows.

The structure is the reverse of the push that needs nothing to push against. A rocket throws mass away and is pushed forward; a raindrop takes mass in and is held back. Both are momentum balances for a body whose mass changes, and both turn out to depend on the relative velocity of the mass entering or leaving: here zero for the mist, so all of the drop’s speed must be given to each droplet.

What the pictures cannot show #

The no-air figures assume a drop that captures every droplet in its path, a mist that is uniform and at rest, and a drop that stays spherical; real collection efficiencies are well below one for small droplets, clouds are turbulent and patchy, and large drops flatten and break up. The air-resistance figure uses a fixed drag coefficient of 0.5, a reasonable value for small drops but not an exact one; measured terminal speeds are close to those drawn for millimetre drops and lower for the largest, which distort. The growth figure ignores the updraughts that carry real drops through the same cloud several times. And none of the figures shows what happens to the droplets that a drop pushes aside rather than absorbs, which in real clouds is most of them.

Still open: how rain forms fast enough #

The growth law says that collecting droplets from a cloud is slow, and the observation that rain often forms within twenty minutes of a cumulus cloud appearing has long been hard to reconcile with it. The bottleneck is the first step: droplets near ten to twenty micrometres grow too slowly by condensation to become large enough to collide efficiently, and too slowly by collision because they are all falling at nearly the same speed. Turbulence in the cloud, which brings droplets together and clusters them, giant salt particles that seed a few large drops early, and the electrical charges droplets carry have all been proposed to bridge the gap. How much each contributes is measured in cloud chambers, in aircraft and in simulations, and is not yet settled.

The habit worth carrying away is to ask what a moving body must share its momentum with. A falling body that gathers mass at rest must bring each piece up to its own speed, so it accelerates more slowly than g and loses energy in every pick-up: a raindrop sweeping mist settles at g/7 and loses three-sevenths of gravity’s work, and in general a body whose moving mass grows as distanceⁿ falls at g/(2n + 1) and loses n/(2n + 1). The mist’s density and the drop’s size cancel; only the power survives.