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Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:The geometric factors cancel, givingHere $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $B=1\mathrm{T}$, ${\rho }_m=8960\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_c=1.7\times 10^{-8}\mathrm{\Omega }\mathrm{m}$. As numerical checks, the specified $a=0.08\mathrm{m}$ and $d_w=0.001\mathrm{m}$ give $M=2.252\times 10^{-3}\mathrm{kg}$, $R=6.926\times 10^{-3}\mathrm{\Omega }$ and $I_s=0.276\mathrm{A}$.

The gravitational power lost, $Mg{v}_s$, equals the Joule heating $I_s^{2}R$, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

At the terminal velocity, the acceleration in iii vanishes. ThusFor $0<\varphi <\pi /2$ and nonzero $B$, the magnetic drag increases linearly with speed and this terminal state is stable. Horizontal rails have no gravitational drive. Vertical rails have no linked vertical magnetic flux change, so the model gives free fall rather than a finite terminal velocity.

A Newtonian fluid exerts viscous shear stress proportional to the rate of strain. For an incompressible fluid of constant dynamic viscosity $\eta$,The local momentum balance is the Navier-Stokes equation:The left side is density times material acceleration. A steady velocity field can still have convective acceleration when a moving parcel changes speed or direction; setting $\partial \mathbf{u}/\partial t=0$ does not eliminate it.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low Reynolds number, the fluid's own inertia can often be neglected, giving Stokes flow. This does not say that every immersed body has zero acceleration: its mass and the net pressure, viscous and body forces still enter its own equation of motion.

For a small rigid sphere moving through a slowly varying viscous background, Stokes drag is $-6\pi \eta a(\mathbf{v}-\mathbf{u})$. In still fluid, a simple settling model includes weight, buoyancy and drag:Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi \eta a)$ and terminal velocityAt the terminal velocity the net force and acceleration vanish, while viscous drag and gravity remain individually nonzero. The dissipated power $6\pi \eta a{v}_t^2$ is supplied by the loss of gravitational potential energy relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

An upward drag force $F_D\ge 0$ changes the variable-mass system equation toFor an ordinary drag force increasing with positive speed, the new curve lies below the no-drag curve. It still starts with slope $g$ if $F_D=0$ at rest. The dashed curve illustrates quadratic drag on a sphere with constant radial growth, using $F_D=\frac{1}{2}{C}_D{\rho }_{\mathrm{air}}\pi r^2{v}^2$ with constant drag coefficient $C_D$ and air mass density ${\rho }_{\mathrm{air}}$. The equation for these dimensionless quantities isThe figure uses $\eta =2$ solely to show the shape; it is not an additional datum of the paper.

There is a limitation to the usual terminal velocity sketch here. The examiner's commentary describes a curve approaching terminal velocity, but the paper maintains unbounded radius growth and specifies no drag law. A horizontal limit is justified if one additionally treats the radius as approximately fixed, or otherwise assumes a drag-to-mass ratio with a fixed limiting balance. With the growing radius and constant-$C_D$ quadratic drag just written, the late balance instead givesThis curve flattens in the sense that its slope tends to zero, but its speed remains unbounded. Thus a lower, more strongly flattening curve is warranted; a finite terminal speed requires an additional assumption. The no-growth approximation gives the familiar finite value $v_T=\sqrt{8\rho gr/(3C_D{\rho }_{\mathrm{air}})}$ for a specified fixed radius.^[ref]

For an initially monodisperse cloud the relative settling speed vanishes, so there is no gravitational collection term in this model. Set $D=(S-1)/(F_k+F_d)$ and use $M=4\pi {\rho }_l{r}^3/3$. Condensational growth givesTaking upward height from cloud base as positive, the droplet speed relative to the ground is the updraft minus its downward terminal velocity:For a launch at $h_r(0)=0$,This assumes the prescribed uniform updraft and supersaturation persist along the trajectory, and that the Stokes-type radius-squared fall law remains applicable.

Quadratic drag has magnitude proportional to the square of the relative speed and points opposite to the relative velocity.

For an isotropic model of a body of fixed shape, let $\mathbf{U}={\mathbf{v}}_{\mathrm{body}}-{\mathbf{u}}_{\infty }$ denote its relative velocity through the distant fluid, with $U=|\mathbf{U}|$. A constant positive resistance $k$ givesHere density $\rho$, reference area $A$ and drag coefficient $C_D$ are held fixed over the modeled range of relative speeds. The vector formula makes the direction explicit: the drag force on the body opposes its motion relative to the fluid, and vanishes at $U=0$. For a general body, orientation and other flow conditions can affect the drag coefficient; treating it as a constant requires a regime in which that is an adequate approximation. ^[ref]

Compare two distinct models at a chosen relative speed $U_0>0$. A linear drag model has $F_D(U)/F_D(U_0)=U/U_0$, while a constant-coefficient quadratic drag model has $F_D(U)/F_D(U_0)=(U/U_0{)}^2$. At $U=2U_0$ their drag force ratios are two and four respectively.

This distinction matters when estimating terminal velocity or trajectories such as descent from rest under quadratic drag: doubling relative speed increases the resistance much more strongly in the quadratic drag model. An approximately constant drag coefficient can occur in another Reynolds number regime; its use must be justified there. Conversely, substituting the creeping-flow sphere coefficient $C_D=24/{\mathrm{Re}}_d$ into $F_D=\frac{1}{2}\rho U^{2}A{C}_D$ recovers linear drag, so writing this general formula does not make the resistance quadratic. Neither approximation should be extended across all Reynolds numbers.

The distinction already appears in the original investigation underlying the Stokes drag law, read on 9 December 1850: article 42 identifies resistance proportional to relative speed and distinguishes it from a contribution proportional to the square of relative speed. ^[ref]

For surface-fed growth of an accreting sphere with $\mu =\dot{r}>0$, stationary incoming material, and quadratic drag $F_D=C_D{\rho }_a\pi r^2{v}^2/2$, the variable-mass system obeys $\dot{v}+3\mu v/r=g-3C_D{\rho }_a{v}^2/(8\rho r)$. For constant positive parameters, $r\sim \mu t$ and the late balance gives the displayed speed. The omitted acceleration and accretion terms are both $O(t^{-1/2})$, consistent with the balance. Thus this growing sphere has unbounded speed with vanishing acceleration, rather than a finite terminal velocity. This model has a different growth law from the existing accreting raindrop with quadratic drag, whose growth rate is proportional to speed.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

Take downward as positive and let gravity have magnitude $g$. For a rigid sphere of density ${\rho }_p$ in otherwise stationary fluid of density ${\rho }_f$, its weight minus buoyancy is $\frac{4}{3}\pi a^3({\rho }_p-{\rho }_f)g$. At terminal velocity, this balances the Stokes drag law:This signed downward settling velocity is negative for a buoyantly rising sphere and zero for equal densities. For fixed density contrast and dynamic viscosity, the submerged weight grows as $a^3$ while linear drag at fixed speed grows as $a$; their balance gives the settling-velocity radius scaling $U_s\propto a^2$. Meanwhile,Thus doubling the radius multiplies the predicted Stokes settling speed by four and the particle Reynolds number by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.

For a computed water example, take ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and $a=10.0\mu \mathrm{m}$. These are specified rounded inputs rather than a claim of exact water properties at every temperature. They giveThe diameter is $20.0\mu \mathrm{m}$, and both the net downward weight and the upward drag force are $6.16\times 10^{-11}\mathrm{N}$. At this terminal velocity, falling $1.00\mathrm{mm}$ takes $3.06\mathrm{s}$. At $a=20.0\mu \mathrm{m}$ the prediction is $1.308\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.05232$; at $a=50.0\mu \mathrm{m}$ it is $8.175\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.8175$, which is no longer a convincing small-Reynolds number case. The figure labels these as predictions of the Stokes settling velocity formula and shows where its inertia check fails.

The calculation describes the eventual terminal velocity, not the release transient governed by particle acceleration and possibly the Unsteady Stokes equation. Walls, hydrodynamic interactions, departures from a rigid sphere, or a different surface condition require a different resistance. For sufficiently small particles, Brownian motion can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of ^[ref].

Let downward velocity $v$ obey $\dot{v}=a-b{v}^2$, with constants $a,b>0$, terminal velocity $v_t=\sqrt{a/b}$ and initial speed $u>v_t$. Gravity minus buoyancy can supply the constant downward force represented by $a$. Since $\dot{z}=v$, the equation for $v^2$ is linear in depth: $d(v^2)/dz=-2b(v^2-v_t^2)$, giving the displayed decay. Exact terminal velocity is approached only asymptotically. A practical tolerance $v\le (1+ϵ)v_t$ is attained at $z_{ϵ}=(2b{)}^{-1}\ln [(u^2/v_t^2-1)/(2ϵ+{ϵ}^2)]$, provided the initial excess exceeds that tolerance. Its time-domain form is $v=v_t\coth (b{v}_{t}t+\chi )$ with $\chi =\mathrm{arctanh}(v_t/u)$, in contrast to the hyperbolic tangent for descent from rest under quadratic drag.

Terminal velocity is a constant limiting velocity approached when a velocity-dependent resistance balances the other applied forces. A falling body may keep moving while its net force and acceleration tend to zero. For downward motion in a stationary medium, its limiting speed is set by balancing gravitational weight against drag force, including buoyancy if appreciable. The balance, rather than a universal numerical value, is the defining idea. ^[ref]

The terminal velocity under linear drag gives a simple example with an explicit approach to the limit. Real falling objects need an appropriate resistance model: linear drag is useful for slow motion of a small sphere in Stokes flow, whereas many faster falling objects require quadratic drag. Changing mass also changes the question: the existing accreting raindrop with quadratic drag has a speed-radius attractor of an accreting raindrop, rather than the fixed-mass balance used below.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]