Drag Forces and Terminal Velocity

Step out of an aeroplane and gravity's promise of ever-increasing speed lasts about twelve seconds. After that you fall at a steady 180 km/h — fast, but constant: velocity flat, acceleration zero, the falling equivalent of cruise control. Open the parachute and the same physics renegotiates the contract down to a survivable 5 m/s. Meanwhile a fog droplet performs the opposite miracle, falling so slowly — about a centimetre per second — that a cloud is effectively a suspended ocean. Same planet, same gravity, terminal speeds five thousand times apart. The negotiator in every case is drag, the resistive force a fluid exerts on anything moving through it — and the first honest thing to learn about drag is that there is no single drag law. Nature runs two regimes. Small and slow things (dust, droplets, bacteria) feel a linear drag, proportional to speed, ruled by the fluid's viscosity: Stokes' world. Big and fast things (skydivers, raindrops, cars) feel a quadratic drag, proportional to speed squared, ruled by the inertia of the air being shoved aside: Newton's world. A single dimensionless number — the Reynolds number — tells you which world you are in, and using the wrong law produces answers that are wrong by orders of magnitude, not percentages. This article builds both laws, solves both equations of motion, extracts the terminal velocity each predicts, and — the part most treatments skip — gives you the Reynolds honesty test for deciding which law applies. Your friction toolkit gains its missing member: dry surfaces you have done; fluids start here.

Key takeaways

Drag opposes motion through a fluid and grows with speed — so a falling body accelerates until drag balances weight: terminal velocity, the zero-net-force falling state.
Two regimes, two laws: linear (Stokes) drag $F = bv$ ( $b = 6\pi\eta r$ for a sphere) when viscosity dominates; quadratic drag $F = \tfrac12 \rho C_d A v^2$ when the inertia of displaced fluid dominates.
The Reynolds number $\mathrm{Re} = \rho v L/\eta$ decides the regime: $\mathrm{Re} \ll 1$ linear, $\mathrm{Re} \gtrsim 10^3$ quadratic, in between neither is exact.
Linear terminal velocity: $v_t = mg/b \propto r^2$ (fog falls at cm/s); quadratic: $v_t = \sqrt{2mg/\rho C_d A}$ (skydiver ≈ 50 m/s belly-down, ≈ 5.5 m/s under canopy).
Approach to terminal speed: exponential $v = v_t(1 - e^{-t/\tau})$ with $\tau = v_t/g$ in the linear regime; $v = v_t\tanh(gt/v_t)$ in the quadratic — same timescale $v_t/g$ , gentler saturation.
Terminal velocity scales with size: $v_t \propto \sqrt{L}$ for geometrically similar bodies in the quadratic regime — why hail hurts, rain doesn't, and insects survive any fall.
Key idea: drag grows with speed until it cancels weight; which drag law applies — linear in viscosity or quadratic in inertia — is decided by the Reynolds number, and guessing wrong costs orders of magnitude.

The core insight

A body falling through fluid accelerates only until the speed-dependent drag force grows to cancel its weight, leaving it at a terminal velocity set by the balance $mg = F_{\rm drag}(v_t)$ — with $F = 6\pi\eta r v$ (Stokes, $\mathrm{Re} \ll 1$ ) giving $v_t \propto r^2$ , and $F = \tfrac12\rho C_d A v^2$ (Newtonian, $\mathrm{Re} \gg 1$ ) giving $v_t = \sqrt{2mg/\rho C_d A}$ .

Where this lives today

The two drag laws are daily infrastructure at both ends of the size spectrum. At the small end, Stokes' law is the workhorse of aerosol science: the settling of pollution particles (PM2.5 stays airborne for days precisely because $v_t \propto r^2$ makes micron-scale particles nearly buoyant), the physics of respiratory droplets and airborne disease transmission, the design of inhalers and cleanroom air handling, and every differential centrifuge in molecular biology — where "spinning down" a sample is Stokes sedimentation with $g$ replaced by $10^5 g$ and the protocols written directly from equation (2). At the big end, the quadratic law runs vehicle engineering: the $C_d$ of a production car (0.25–0.3, and heavily advertised) sets its highway fuel burn, since aerodynamic power grows as $\tfrac12\rho C_d A v^3$ — the cube law behind why 120 km/h costs disproportionately more fuel than 100, and why cycling's hour record is essentially a drag-optimization contest. Sport and weather live in the honesty band between regimes. Ball games are Reynolds-number theatre: a football's knuckling flight, a cricket ball's swing, and golf's dimples (which deliberately trip the boundary layer to trigger the drag crisis early, nearly halving $C_d$ ) all exploit the measured $C_d(\mathrm{Re})$ curve where no simple law holds. Meteorology integrates raindrop and hailstone terminal velocities — this article's worked example 2 industrialized — inside every precipitation model, and planetary scientists redo the calculation with other atmospheres: on Titan, methane raindrops in thick air and low gravity drift down at ~1.5 m/s, snowflake-slow rain on an alien coast, computed with exactly the equations above.

The physical picture

Free fall is a lie that lasts a few seconds. The moment a body moves through air, the air pushes back, and — the crucial feature — pushes back harder the faster you go. That feature makes terminal velocity inevitable: start from rest and gravity wins at first ( $v$ small, drag small, acceleration $\approx g$ ); as speed builds, drag climbs; the net force shrinks; and the process self-terminates at the speed $v_t$ where drag exactly cancels weight. Beyond that no acceleration is available — fall faster than $v_t$ (say, after a parachute opens) and drag exceeds weight, braking you back down to it. Terminal velocity is an attractor: the fall forgets its history and converges to $v_t$ from either side.

Figure

The self-terminating fall — velocity converges to the terminal attractor

But how does drag grow with speed? Here the physics forks, and the fork is the heart of this article. Watch the fluid do two different jobs. A large, fast body — your hand out a car window — must shove the air out of its way: every second it sweeps out a tube of air and hurls that mass aside at roughly its own speed. The momentum bill for that shoving is a force proportional to the density of the air, the area being swept, and the square of the speed (one factor of $v$ for how much air per second, one for how fast it must be thrown). That is quadratic drag — an inertial, momentum-transfer effect, almost independent of how sticky the fluid is. A tiny, slow body — a fog droplet — barely disturbs the air at all; instead the air's viscosity matters: fluid layers must slide over one another to let the sphere through, each layer dragging on the next like sheets of wet paper. Shearing fluid at speed $v$ across a body of size $r$ produces a force proportional to $\eta r v$ — linear in speed, blind to density. That is Stokes drag. Which job dominates? Compare the two forces and a single dimensionless group emerges: the Reynolds number, the ratio of inertial shoving to viscous sliding. Small $\mathrm{Re}$ : viscosity rules, drag is linear, the world feels like honey (a bacterium swimming in water lives at $\mathrm{Re} \sim 10^{-5}$ and stops dead in a microsecond when it stops swimming). Large $\mathrm{Re}$ : inertia rules, drag is quadratic, the world feels like wind (a skydiver falls at $\mathrm{Re} \sim 10^5$ ). The two regimes are not rival theories but different limits of one fluid mechanics — and the honest physicist's first move in any drag problem is to estimate $\mathrm{Re}$ before choosing a formula.

The math: two drag laws and the number that arbitrates

The linear law first. For a sphere of radius $r$ creeping through a fluid of viscosity $\eta$ , George Gabriel Stokes solved the full slow-flow equations in 1851 and obtained one of the most used results in physics:

F_{\rm Stokes} = 6\pi\,\eta\, r\, v \qquad (1)

where: $\eta$ — the dynamic viscosity (air: $1.8\times10^{-5}$ Pa·s; water: $1.0\times10^{-3}$ Pa·s), $r$ — the sphere's radius, $v$ — its speed through the fluid. Read the structure: density does not appear — the fluid's mass is irrelevant when nothing is being thrown aside — and the force is linear in both size and speed, the signature of shear. Equation (1) is exact only in the creeping limit $\mathrm{Re} \ll 1$ , and its derivation from the fluid equations belongs to the viscous-flow article at the end of this track; here we take it as Stokes' gift and put it to work. Its terminal velocity follows from $mg = 6\pi\eta r v_t$ : for a solid sphere of density $\rho_p$ , using $m = \tfrac43\pi r^3 \rho_p$ ,

v_t^{\rm lin} = \frac{mg}{6\pi\eta r} = \frac{2\,\rho_p\, r^2 g}{9\,\eta} \qquad (2)

where: the celebrated $r^2$ scaling does the talking — halve a droplet's radius and it falls four times slower. This is why clouds float (10 μm droplets: ~1 cm/s), why fine dust hangs in sunbeams for minutes, and why sediment grading by settling speed works in geology labs. The quadratic law is written with an engineering-standard bookkeeping factor:

F_{\rm drag} = \tfrac12\,\rho\, C_d\, A\, v^2 \qquad (3)

where: $\rho$ — the fluid density (air: 1.2 kg/m³), $A$ — the body's cross-sectional area facing the flow, and $C_d$ — the dimensionless drag coefficient that absorbs the shape's aerodynamic character: about 0.5 for a smooth sphere at high $\mathrm{Re}$ , ≈1.0 for a flat-ish sprawled human, 0.25–0.3 for a modern car, over 1.5 for an open parachute canopy. Now viscosity is the silent partner: the force is pure momentum flux, $\rho A v \cdot v$ , dressed with a shape factor. Setting (3) equal to weight gives the quadratic terminal velocity

v_t = \sqrt{\frac{2mg}{\rho\, C_d A}} \qquad (4)

where: the square root tames everything — to fall twice as fast you must quadruple your weight-to-area ratio, which is why a skydiver can double her speed by going head-down (slashing $C_dA$ by a factor ~3.5) and why a parachute's fifty-fold area increase cuts speed only ten-fold. The arbiter between (1) and (3) is the Reynolds number,

\mathrm{Re} = \frac{\rho\, v\, L}{\eta} \qquad (5)

where: $L$ is the body's size (diameter, for spheres). $\mathrm{Re}$ is the ratio of the two drag mechanisms' strengths — derived honestly below — and the empirical map for spheres reads: $\mathrm{Re} \lesssim 1$ : Stokes' law (1) holds (equivalently $C_d = 24/\mathrm{Re}$ , a useful translation). $1 \lesssim \mathrm{Re} \lesssim 10^3$ : neither law is right; the measured $C_d(\mathrm{Re})$ curve must be used. $10^3 \lesssim \mathrm{Re} \lesssim 2\times10^5$ : quadratic law with $C_d \approx 0.5$ for spheres. (Beyond that lurks the "drag crisis," a sudden $C_d$ drop when the boundary layer goes turbulent — the reason golf balls have dimples, and a story for the fluid-dynamics articles.)

Derivation ALinear drag: the exponential approach to terminal velocity

We are solving the equation of motion for a body falling with linear drag — $m\,dv/dt = mg - bv$ with $b = 6\pi\eta r$ for a Stokes sphere — to obtain the full velocity history, the terminal velocity, and the relaxation time $\tau$ that tells you how long terminal-velocity thinking takes to become valid. The strategy: recognize a first-order linear ODE, identify its fixed point (that is $v_t$ ), and watch deviations from the fixed point die exponentially. The physical payoff to watch for: $\tau$ is also $v_t/g$ , a quantity you can estimate in your head.

Step 1. Write Newton's second law along the fall, taking down as positive: $m\dfrac{dv}{dt} = mg - bv$ . Why the signs: gravity aids the fall, drag opposes the velocity; for a straight downward fall from rest the drag force points up throughout.

Step 2. Find the fixed point by setting $dv/dt = 0$ : $v_t = mg/b$ . Define the deviation $u = v - v_t$ , so $m\,du/dt = -bu$ . Why change variables: the equation for $u$ has the forcing removed — it says deviations from terminal velocity decay on their own, at a rate set by $b/m$ , regardless of sign. This is the "attractor" property in algebra.

Step 3. Integrate: $u(t) = u(0)\,e^{-t/\tau}$ with $\tau = m/b$ . Starting from rest ( $u(0) = -v_t$ ):

v(t) = v_t\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{m}{b} = \frac{v_t}{g} \qquad (6)

Why $\tau = v_t/g$ is worth memorizing: it says the time to reach terminal velocity is roughly the time free fall would need to reach that speed — a fog droplet with $v_t = 1.2$ cm/s equilibrates in $\tau \approx 1.2$ ms (it is born at terminal velocity, for all practical purposes), while a skydiver with $v_t \approx 50$ m/s needs of order five seconds. One quantity, instant intuition.

Step 4. Integrate once more for the distance fallen: $x(t) = v_t t - v_t\tau(1 - e^{-t/\tau})$ , which at late times is simply $v_t(t - \tau)$ — the fall proceeds as if it had moved at terminal velocity all along but started $\tau$ seconds late. Why this reading helps: it converts the transient into a single "lost time" correction, the standard trick for estimating fall durations without solving anything. *Three quick checks on (6): dimensionally $[b] =$ N·s/m so $m/b$ is indeed a time; taking $t \to \infty$ returns $v_t = mg/b$ ; expanding for $t \ll \tau$ gives $v \approx gt$ , recovering free fall — all consistent.*

Go deeper · Derivation B · 2 minQuadratic drag: momentum flux, the tanh fall, and the skydiver timetable

Derivation BQuadratic drag: momentum flux, the tanh fall, and the skydiver timetable

We are doing two jobs: first, deriving the form of the high-Reynolds drag law (3) from momentum conservation — no fluid mechanics beyond Newton required — and second, solving the resulting equation of fall $m\,dv/dt = mg - kv^2$ exactly, to get the $\tanh$ velocity profile and the practical timetable of a skydive. Watch how the momentum argument produces everything in (3) except the numerical value of $C_d$ — that honesty about what dimensional physics can and cannot deliver is the derivation's second lesson.

Step 1. A body of cross-section $A$ moving at speed $v$ sweeps through a tube of fluid: in time $dt$ it encounters the fluid in volume $A\,v\,dt$ , of mass $dm_f = \rho A v\,dt$ . To pass, it must push this fluid out of the way, giving it a sideways/forward speed of order $v$ . Momentum delivered to the fluid per unit time: $\dot p \sim (\rho A v)\,v = \rho A v^2$ . By Newton's third law the fluid pushes back on the body with the same magnitude — hence $F = \tfrac12 C_d\,\rho A v^2$ , where the order-one constant $\tfrac12 C_d$ collects everything the estimate cannot fix (how completely the fluid is deflected, how cleanly it rejoins behind the body). Why this argument earns trust: it predicts the exact dependence on $\rho$ , $A$ , and $v$ , and predicts that viscosity should NOT appear — all confirmed by experiment at high $\mathrm{Re}$ . The price is $C_d$ , an empirical shape-dependent number.

Step 2. Write the fall equation with $k \equiv \tfrac12\rho C_d A$ : $\ m\dfrac{dv}{dt} = mg - kv^2$ , and read off the terminal velocity from the zero of the right side: $v_t = \sqrt{mg/k}$ — equation (4). Why the balance is stable: below $v_t$ the right side is positive (speeding up), above $v_t$ negative (slowing down); the fall is funnelled toward $v_t$ from both sides, exactly like the linear case but with a stiffer wall — drag grows quadratically, so overshoots are punished harder.

Step 3. Separate variables using $v_t$ : $\dfrac{dv}{1 - v^2/v_t^2} = g\,dt$ . The left side integrates to an inverse hyperbolic tangent, giving, from rest,

v(t) = v_t \tanh\!\left(\frac{g\,t}{v_t}\right) \qquad (7)

Why tanh: the equation is the textbook logistic-type saturation — linear growth $v \approx gt$ at early times (the tanh's straight start), then a smooth roll-over onto the plateau $v_t$ . The characteristic time is again $v_t/g$ , but the approach is gentler than an exponential: 76% of $v_t$ at $t = v_t/g$ , 90% at $t \approx 1.5\,v_t/g$ .

Step 4. Integrate again for the altitude ledger: $x(t) = \dfrac{v_t^2}{g}\ln\cosh\!\left(\dfrac{gt}{v_t}\right)$ . For the skydiver of the figure ( $v_t = 51$ m/s): 90% of terminal velocity arrives at $t = 1.47\,v_t/g = 7.7$ s, after $x = 221$ m of fall; essentially full terminal velocity by ~12 s and ~400 m. Why jumpers care: this transient is the "hill" — the seconds of true acceleration a skydiver feels before the airflow steadies; freefall photographers use it to time when relative speeds within a formation stabilize. *Three checks on (7): direct differentiation using $\tanh' = 1 - \tanh^2$ confirms it satisfies the ODE; expanding at early times gives $v \approx gt$ ; at late times $v \to v_t$ — all consistent.*

Go deeper · Derivation C · 2 minThe Reynolds number: one ratio to rule both regimes

Derivation CThe Reynolds number: one ratio to rule both regimes

We are deriving the criterion that decides between Stokes' law and the quadratic law — by directly forming the ratio of the two drag estimates — and then locating the crossover size for falling water drops in air, the worked boundary between cloud physics and rain physics. The strategy is pure scaling: no coefficients, just the two force laws face to face. Watch for the punchline $C_d = 24/\mathrm{Re}$ , which shows Stokes' law is not a rival of the drag-coefficient language but a special case of it.

Step 1. Form the ratio of inertial to viscous drag on a body of size $L$ : $\dfrac{F_{\rm quad}}{F_{\rm visc}} \sim \dfrac{\rho A v^2}{\eta L v} \sim \dfrac{\rho L^2 v^2}{\eta L v} = \dfrac{\rho v L}{\eta} \equiv \mathrm{Re}$ . Why this defines the regime: when $\mathrm{Re} \ll 1$ the viscous force dwarfs the inertial one and Stokes physics rules; when $\mathrm{Re} \gg 1$ the momentum-flux force dominates and viscosity retreats into thin boundary layers. The Reynolds number IS the drag-law selector — the same group that will later decide laminar versus turbulent flow.

Step 2. Translate Stokes' law into drag-coefficient language by equating (1) with (3): $6\pi\eta r v = \tfrac12 \rho C_d (\pi r^2) v^2$ gives $C_d = \dfrac{12\eta}{\rho r v} = \dfrac{24}{\mathrm{Re}}$ (with $\mathrm{Re}$ built on diameter $2r$ ). Why this line matters: it unifies the article — one $C_d(\mathrm{Re})$ curve describes all sphere drag, falling as $24/\mathrm{Re}$ on the left (Stokes), flattening to $\approx 0.5$ on the right (Newtonian), with the awkward but measured bridge in between. Using "the" quadratic law at $\mathrm{Re} = 10$ or Stokes at $\mathrm{Re} = 100$ means reading the wrong end of this curve.

Step 3. Find where falling water drops in air leave the Stokes regime. A drop falling at its Stokes terminal velocity (2) has $\mathrm{Re}(v_t) = \dfrac{\rho_{\rm air}\, (2r)\, v_t}{\eta} = \dfrac{4\rho_{\rm air}\rho_w g\, r^3}{9\eta^2}$ . Setting $\mathrm{Re} = 1$ and inserting numbers ( $\rho_{\rm air} = 1.2$ , $\rho_w = 1000$ kg/m³, $\eta = 1.8\times10^{-5}$ Pa·s): $r^3 = \dfrac{9\eta^2}{4\rho_{\rm air}\rho_w g} = 6.2\times10^{-14}\ \mathrm{m}^3$ , so $r \approx 40\ \mu\mathrm{m}$ . Why this single number organizes a lot of weather: droplets below ~40 μm (cloud and fog) are honest Stokes fallers — slow, suspended, cloud-forming; above it drops accelerate into the transitional and then quadratic regimes, and by millimetre size they are rain, falling at metres per second under Newtonian drag. The crossover radius is the physical boundary between "cloud" and "rain". *The crossover radius is self-consistent: numerically $9\times(1.8\times10^{-5})^2 = 2.92\times10^{-9}$ and $4\times1.2\times1000\times9.8 = 4.7\times10^{4}$ , giving $r^3 = 6.2\times10^{-14}$ m³ and $r = 3.96\times10^{-5}$ m — confirming $r \approx 40\ \mu$ m. The regime assignments are also self-consistent: a fog droplet with $r = 10\ \mu$ m has $\mathrm{Re} = 0.016 \ll 1$ (Stokes legitimate), while a 1-mm raindrop has $\mathrm{Re} \approx 900$ (just under the $10^3$ line, already on the near-plateau of the measured curve).*

Which equation answers which question

you want	use	notes
which regime am I in?	$\mathrm{Re} = \rho v L/\eta$ , eq. (5)	FIRST question, always; $<1$ linear, $>10^3$ quadratic
drag on something small/slow	Stokes, eq. (1)	spheres, $\mathrm{Re}\lesssim 1$ ; $C_d = 24/\mathrm{Re}$ translation
terminal speed, small sphere	$v_t = 2\rho_p r^2 g/9\eta$ , eq. (2)	the $r^2$ law: fog, dust, sediment
drag on something big/fast	$\tfrac12\rho C_d A v^2$ , eq. (3)	need $C_d$ from tables: sphere 0.5, human ~1, car 0.25–0.3
terminal speed, big body	$v_t = \sqrt{2mg/\rho C_d A}$ , eq. (4)	skydiver, raindrop, hailstone
speed during the fall	eq. (6) linear · eq. (7) quadratic	both saturate on timescale $v_t/g$
how long till terminal?	$t_{90\%} \approx 1.5\,v_t/g$ (quadratic)	skydiver ~8 s; droplet: milliseconds

Limiting cases

Case	Result	Meaning
$t \ll v_t/g$	$v \approx gt$ in both regimes	early fall is always free fall — drag needs speed to act
$t \gg v_t/g$	$v \to v_t$ , $a \to 0$	the terminal attractor; history forgotten
$\eta \to 0$ at fixed $\mathrm{Re} \ll 1$ setup	$\mathrm{Re} \to \infty$ : regime change	Stokes' law cannot survive vanishing viscosity
vacuum ( $\rho \to 0$ , $\eta \to 0$ )	$v_t \to \infty$ : no terminal velocity	Apollo 15's hammer and feather land together
$r \to 0$ (Stokes)	$v_t \propto r^2 \to 0$	small things effectively float: fog, aerosol, smoke
geometric scaling, quadratic	$v_t \propto \sqrt{L}$ ( $m \propto L^3$ , $A \propto L^2$ )	mouse survives, horse doesn't — Haldane's law
start above $v_t$ (canopy opens)	drag > weight: deceleration to $v_t$	terminal velocity approached from above too
$\mathrm{Re} \sim 1$ – $10^3$	neither law; empirical $C_d(\mathrm{Re})$	the honesty band — interpolate, don't extrapolate

Figure

One curve, two regimes — the sphere drag coefficient across eight decades of Reynolds number

Common trapHeavier objects fall faster — Galileo settled this, right?

In vacuum Galileo wins absolutely: all bodies fall identically, as Apollo 15's hammer-and-feather demonstration made theatrical. But in air the drag force cares about size and shape while gravity cares about mass, so the ratio — which is what sets terminal velocity — differs between bodies, and "heavier falls faster" becomes conditionally true in a precise way: at equal size and shape, the denser body has the higher $v_t$ (a steel ball outruns a styrofoam one of the same diameter); at equal density, the bigger body falls faster ( $v_t \propto \sqrt{L}$ in the quadratic regime, $\propto r^2$ in Stokes), which is why drizzle drifts and hail hammers. The classroom error is applying either intuition to short falls: over the first metre or two of drop, drag has had no time to act ( $v \ll v_t$ , both regimes reduce to $v \approx gt$ ) and everything really does fall together — Galileo's Pisa experiment works fine with dense objects precisely because the transient is nearly drag-free. Terminal-velocity reasoning belongs to long falls, $t \gtrsim v_t/g$ . And one more Galilean rescue: the independence of $v_t$ from fall height means a cat from the 6th floor and from the 20th arrive at the same speed — the famous (and real) feline high-rise statistics.

Worked examples

1 · The skydive, in numbers (confidence builder)

A skydiver of mass $m = 80$ kg falls belly-to-earth with $C_d A = 0.5\ \mathrm{m}^2$ in air of density $\rho = 1.2\ \mathrm{kg/m^3}$ . Find (a) her terminal velocity, (b) the time and altitude cost to reach 90% of it, (c) her landing speed under a canopy of $C_d A = 44\ \mathrm{m}^2$ . (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Regime check first, as the law requires: guessing $v \sim 50$ m/s and $L \sim 0.5$ m, $\mathrm{Re} = 1.2 \times 50 \times 0.5 / 1.8\times10^{-5} \approx 1.7\times10^6$ — deep quadratic territory. Equation (4) applies.

Step 2. Terminal velocity: $v_t = \sqrt{\dfrac{2 \times 80 \times 9.8}{1.2 \times 0.5}} = \sqrt{\dfrac{1568}{0.6}} = \sqrt{2613} = \boxed{51\ \mathrm{m/s}}$ — about 184 km/h, matching the standard belly-flight figure. (Head-down, $C_dA \approx 0.14$ : $v_t = \sqrt{1568/0.168} = 97$ m/s ≈ 350 km/h — the competitive speed-skydiving regime.)

Step 3. The timetable, from the tanh solution (7): 90% of $v_t$ when $\tanh(gt/v_t) = 0.9$ , i.e. $gt/v_t = 1.47$ , so $t = 1.47 \times 51/9.8 = \boxed{7.7\ \mathrm{s}}$ . Altitude spent: $x = \dfrac{v_t^2}{g}\ln\cosh(1.47) = 266 \times 0.83 = \boxed{\approx 220\ \mathrm{m}}$ . From a 4000 m jump, the acceleration phase is a small opening tax; the remaining ~50 s of freefall happen at essentially constant speed.

Step 4. Canopy: $v_t = \sqrt{\dfrac{1568}{1.2 \times 44}} = \sqrt{29.7} = \boxed{5.5\ \mathrm{m/s}}$ — the impact of a jump from $h = v_t^2/2g = 1.5$ m. A parachute does not stop you; it renegotiates $v_t$ down to a hop off a garden wall.

2 · Cloud physics on the back of an envelope (exam level)

(a) Find the terminal velocity of a fog droplet of radius $10\ \mu$ m and verify the Stokes regime is legitimate. (b) A drizzle drop has grown to $r = 0.1$ mm; what does Stokes predict, and can you trust it? (c) A raindrop of $r = 1$ mm falls — get its terminal velocity honestly. ( $\rho_w = 1000$ , $\rho_{\rm air} = 1.2\ \mathrm{kg/m^3}$ , $\eta = 1.8\times10^{-5}$ Pa·s, $C_d \approx 0.5$ .)

Step 1. Fog droplet, Stokes formula (2): $v_t = \dfrac{2 \times 1000 \times (10^{-5})^2 \times 9.8}{9 \times 1.8\times10^{-5}} = \dfrac{1.96\times10^{-6}}{1.62\times10^{-4}} = \boxed{1.2\ \mathrm{cm/s}}$ . Honesty check: $\mathrm{Re} = \dfrac{1.2 \times 0.012 \times 2\times10^{-5}}{1.8\times10^{-5}} = 0.016 \ll 1$ ✓ — Stokes fully earned. At 1.2 cm/s a droplet needs a day to fall 1 km: clouds persist because their water is aerodynamically suspended. Relaxation time $\tau = v_t/g = 1.2$ ms — droplets live at terminal velocity.

Step 2. Drizzle, $r = 10^{-4}$ m: Stokes (2) scales as $r^2$ , predicting $v_t = 1.2\ \mathrm{cm/s} \times 100 = 1.2$ m/s. Check the licence: $\mathrm{Re} = \dfrac{1.2 \times 1.2 \times 2\times10^{-4}}{1.8\times10^{-5}} = 16$ — outside the Stokes regime. The prediction overshoots (real drizzle of this size falls at ~0.7–0.8 m/s); at $\mathrm{Re} \sim 16$ the measured $C_d$ exceeds $24/\mathrm{Re}$ noticeably. Answer: about 1 m/s, but only the measured $C_d(\mathrm{Re})$ curve gives it accurately — Stokes has expired.

Step 3. Raindrop, $r = 10^{-3}$ m, quadratic law: mass $m = \tfrac43\pi r^3\rho_w = 4.19\times10^{-6}$ kg, area $A = \pi r^2 = 3.14\times10^{-6}$ m². Equation (4): $v_t = \sqrt{\dfrac{2 \times 4.19\times10^{-6} \times 9.8}{1.2 \times 0.5 \times 3.14\times10^{-6}}} = \sqrt{\dfrac{8.21\times10^{-5}}{1.88\times10^{-6}}} = \sqrt{43.6} = \boxed{6.6\ \mathrm{m/s}}$ — matching measured 1-mm raindrop speeds (~6.5 m/s). Consistency: $\mathrm{Re} = \dfrac{1.2 \times 6.6 \times 2\times10^{-3}}{1.8\times10^{-5}} \approx 880$ — just below the $10^3$ boundary, at the top of the transition band where the measured $C_d$ has already flattened onto its $\approx 0.5$ plateau, so the quadratic estimate is a good approximation here ✓.

Step 4. Extract the moral: three water drops, three regimes, three different formulas — and the Reynolds check was what kept each answer honest. Had we used Stokes for the raindrop it would have predicted $v_t = 121$ m/s, wrong by a factor of eighteen; the quadratic law applied to the fog droplet errs by a similar factor the other way. In drag problems the regime check IS the physics.

Origins · 1 minGeorge Gabriel Stokes and how this was discovered

George Gabriel Stokes

1819–1903 · Cambridge

Stokes held the Lucasian Chair — Newton's chair — for 54 years, and filled it with the physics of continuous media: the Navier–Stokes equations bear his name, and his 1851 paper on pendulums swinging in viscous fluids contains the drag law of this article, derived by solving the creeping-flow equations exactly for a sphere — one of the first complete solutions of a fluid-resistance problem in history, produced to explain why real pendulums (then the world's timekeepers and gravity-meters) damped as they did. The formula's afterlife has been extraordinary: Millikan hung his oil drops on it to weigh the electron, Einstein built it into his 1905 Brownian-motion theory (whence Perrin counted atoms), and it remains the working equation of aerosol science, sedimentation analysis, and every centrifuge protocol in a biology lab. Stokes was also the man who explained fluorescence, established how X-rays would later be understood to polarize, and served simultaneously as physicist, mathematician, and long-time secretary of the Royal Society — a scientific civil servant of ferocious usefulness whose least glamorous formula, six-pi-eta-r-v, may be his most used.

History

1687

— Newton's Principia treats resistance proportional to

\rho A v^2

— the quadratic law is as old as mechanics itself; the famous vertical-fall tests, dropping globes inside St Paul's Cathedral, came later (June 1710) and were reported in the second edition of 1713.

1851

— Stokes solves creeping flow around a sphere:

F = 6\pi\eta r v

, the linear law, born from the practical problem of pendulum damping.

1883

— Osborne Reynolds' pipe-flow experiments isolate the dimensionless group

\rho v L/\eta

that now arbitrates every drag problem (and much else in fluid mechanics).

1910

— Millikan's oil-drop experiment uses Stokes terminal velocities to weigh single electron charges — tabletop drag physics delivers a fundamental constant.

1912

— Eiffel measures the drag crisis (the sudden

C_d

collapse near

\mathrm{Re} \sim 3\times10^5

), opening the boundary-layer era that Prandtl's theory would organize.

Going further

Texts. Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 3 (drag problems) · Morin, Introduction to Classical Mechanics, Ch. 3 · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 6 · Purcell, "Life at Low Reynolds Number," Am. J. Phys. 45, 3 (1977) — the classic tour of the Stokes world · Batchelor, An Introduction to Fluid Dynamics, for the honest derivation of Stokes' law. Key numerical anchors: skydiver $v_t = \sqrt{1568/0.6} = 51$ m/s; $t_{90\%} = 1.47 \times 51/9.8 = 7.7$ s; $x = 266\ln\cosh(1.47) \approx 220$ m; canopy $v_t = \sqrt{1568/52.8} = 5.5$ m/s. Fog droplet: $v_t = 1.96\times10^{-6}/1.62\times10^{-4} = 1.2$ cm/s, $\mathrm{Re} = 0.016$ . Raindrop: $v_t = \sqrt{43.6} = 6.6$ m/s, $\mathrm{Re} \approx 880$ . Stokes-validity crossover: $r \approx 40\ \mu$ m from $\mathrm{Re}(v_t) = 1$ . Stokes–quadratic bridge: $C_d = 24/\mathrm{Re}$ .

Builds on: Newton's Second Law · Friction Forces · One-Dimensional Kinematics · Conservation of Mechanical Energy. Leads to: Damped Harmonic Oscillator · Ideal Fluid Kinematics and the Continuity Equation · The Navier–Stokes Equations and Viscous Flow.

Why does drag grow with speed at all? Friction between solids doesn't care about speed.

Because the fluid's resistance is built from processes that themselves scale with speed. In the viscous regime, drag comes from shearing fluid layers past one another, and the shear rate — how fast adjacent layers slide — is proportional to the body's speed: double $v$ , double the shear, double the force (linear law). In the inertial regime, drag is a momentum bill: the body sweeps up fluid at a rate $\rho A v$ (one factor of $v$ ) and must fling it aside at a speed proportional to $v$ (the second factor), giving $F \propto v^2$ . Dry solid-on-solid friction is genuinely different — its microscopic origin (contact-asperity adhesion) supplies a force set by the normal load, nearly independent of sliding speed, which is why the friction article's $\mu N$ has no $v$ in it. The speed dependence is precisely what gives fluid drag its signature phenomenon: a self-limiting fall. Constant-force friction could never produce a terminal velocity.

Is terminal velocity ever actually reached, mathematically?

Strictly never — both solutions (6) and (7) approach $v_t$ asymptotically and touch it only at $t = \infty$ . Physically the distinction is empty: the approach is exponential-grade fast, so the gap below terminal velocity halves every fraction of the timescale $v_t/g$ , and within a handful of those timescales the difference drops beneath anything measurable (for the skydiver, under 1% by ~16 s; for a fog droplet, immeasurably small after a few milliseconds). The practical vocabulary is the relaxation time $\tau = v_t/g$ : fall durations long compared to $\tau$ mean "at terminal velocity" for every purpose, durations short compared to $\tau$ mean drag hasn't yet mattered and $v \approx gt$ . What IS worth care: in variable conditions — air density rising as a skydiver descends, a hailstone growing as it falls — $v_t$ itself drifts, and the body continuously chases its slowly moving target, staying near quasi-equilibrium whenever $\tau$ is short compared to the rate of change. That chasing picture is how atmospheric physicists actually use terminal velocity.

Where does the ½ in ½ρC_dAv² come from?

Pure convention — and knowing that spares you factor-of-two anxiety. The momentum-flux argument delivers $F \sim \rho A v^2$ with an undetermined order-one constant; engineers chose to write the constant as $\tfrac12 C_d$ so that the reference quantity $\tfrac12\rho v^2$ is the dynamic pressure — the kinetic energy per unit volume of the oncoming stream, which is also the pressure scale in Bernoulli's equation. With that convention, $C_d$ has a clean meaning: the drag force per unit area, measured in units of the natural pressure of the flow. A flat plate held square to the wind has $C_d$ near 1.2 (it converts nearly the full dynamic pressure into drag over its whole face); a streamlined teardrop achieves 0.04 (it lets the flow close smoothly behind, recovering almost all the pressure); a sphere's 0.5 sits between. Any other convention would just rescale every tabulated $C_d$ — the physics lives in the $\rho A v^2$ , the $\tfrac12$ lives in the handbook.

Why doesn't the fluid density appear in Stokes' law? Surely denser fluid resists more.

In the creeping regime, no — and that absence is a deep statement about what low Reynolds number means. Density enters fluid resistance only through inertia: the cost of accelerating fluid out of the body's path. At $\mathrm{Re} \ll 1$ that cost is negligible compared to the cost of shearing the fluid, which is set entirely by viscosity $\eta$ . The fluid at low Re behaves as if massless but sticky — flow patterns adjust instantaneously, momentum plays no role, and the drag can only be built from $\eta$ , $r$ , and $v$ (which by dimensions forces $F \propto \eta r v$ , Stokes' form, before any hard calculation). This is also why low-Re locomotion is so alien: a bacterium that stops swimming halts within a body length in microseconds — no coasting, because coasting is an inertial luxury. Purcell's famous "scallop theorem" (reciprocal flapping gets you nowhere in Stokes flow) follows from the same physics. Density sneaks back only through buoyancy — in liquids you must replace $\rho_p$ by $\rho_p - \rho_{\rm fluid}$ in the terminal-velocity formula (2) — but that is Archimedes' statics, not drag.

My textbook writes linear drag as F = −bv for oscillators too. Is that the same b as Stokes?

Same mathematical form, and for genuinely small, slow oscillators (a pendulum bob in air at low Re, a colloidal particle in an optical trap) the coefficient really is Stokes' $b = 6\pi\eta r$ . But in the damped-oscillator articles the linear term earns its place mostly by a different argument: for small velocities, whatever the true drag law, a Taylor expansion around $v = 0$ starts with the linear term, and linearity is what keeps the oscillator ODE solvable — so $-bv$ is often a modelling choice rather than a Stokes fact. The honesty test is the same Reynolds number as ever: a lab pendulum at $v \sim 1$ m/s with a 2-cm bob has $\mathrm{Re} \sim 10^3$ , meaning its damping is actually closer to quadratic and the exponential decay is an approximation (a decent one over a few periods, since velocities average down). The clean rule: check Re; below 1 the linear coefficient is physical and computable from (1); above, linear damping is a small-velocity effective model whose $b$ you calibrate, not derive.

How can a cat survive a fall from any height above six storeys?

Because a fall's damage is set by landing speed, and landing speed saturates at $v_t$ — after which extra height adds nothing but time. A cat's numbers are friendly: mass ~4 kg, spread-eagled $C_dA \approx 0.06$ –0.08 m², giving $v_t = \sqrt{2mg/\rho C_dA} \approx \sqrt{78.4/0.084} \approx 30$ m/s, notably below a human's, and reached after roughly $1.5\,v_t/g \approx 4.5$ s — about twenty-five storeys of fall. The (real) veterinary study behind the legend found injuries rising with height up to about the seventh storey, then plateauing or improving — consistent with the cat reaching terminal velocity, relaxing out of its startle posture, spreading out (raising $C_dA$ , lowering $v_t$ ), and preparing its legs. The physics core is solid and human-relevant too: it is why BASE jumps from modest cliffs are the dangerous ones (no altitude budget to deploy at $v_t$ ), and why "the fall doesn't get worse forever" is literally true in any atmosphere. (Do not test on any cat.)

Does terminal velocity depend on altitude?

Yes — through the air density in equation (4), and the effect is large enough that skydiving computers correct for it. Air density at 4000 m is about two-thirds of sea-level density, so the same jumper's terminal velocity is higher aloft by $\sqrt{1/0.67} \approx 1.22$ — she genuinely slows down as she descends into thicker air, drag exceeding weight the whole way, no parachute involved. Felix Baumgartner's stratospheric jump is the extreme demonstration: leaping from 39 km, where the air is ~0.36% of sea-level density, his early-fall "terminal velocity" was so high he broke the sound barrier (1357 km/h) — then thickening air dragged him back to an ordinary ~180 km/h by canopy altitude. The dependence also runs the other way in engineering: high-altitude airports mean higher true landing speeds (same dynamic pressure needs more speed in thin air), and Mars landers face the problem in the extreme — Martian air is so thin that a parachute alone cannot bring $v_t$ below ~60 m/s, which is why every Mars landing ends with retrorockets, airbags, or a sky crane.

Why do all the classic drag problems assume spheres?

Because the sphere is the only shape with exact answers at both ends — Stokes' closed-form $6\pi\eta rv$ at low Re, a single well-measured $C_d(\mathrm{Re})$ curve at high Re — and because a huge fraction of naturally falling objects (droplets, hailstones before lobing, sediment grains, aerosol particles) are spheres or near-spheres, since surface tension and abrasion round things off. For other shapes the framework is unchanged and only the coefficients move: any body has a Stokes-regime drag linear in $\eta L v$ with a shape-dependent prefactor (computable numerically; tabulated for ellipsoids, rods, disks), and a high-Re drag $\tfrac12\rho C_dAv^2$ with $C_d$ from wind-tunnel tables. Two genuinely new behaviours appear beyond spheres, both worth flagging: orientation matters (a falling rod turns broadside-on, a disk flutters or tumbles — the drag can depend on an angle the sphere doesn't have), and lift appears (a non-spherical or spinning body can generate force perpendicular to its motion — whence curveballs, falling-leaf flutter, and the entire subject of flight). The sphere is the honest baseline; asymmetry is where aerodynamics begins.