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For positive downward speed, gravity and quadratic drag give $m\dot{v}=mg-\alpha v^2$, $v(0)=0$. Separation and the integral of $(1-s^2{)}^{-1}$ give the displayed approach to terminal velocity. The fallen distance is $(m/\alpha )\log \cosh (\sqrt{\alpha g/m}t)$. The limit as $\alpha \to 0$ recovers $v=gt$ and distance $g{t}^2/2$.

A spherical clean bubble with constant gas mass density and prescribed dissolution law $\dot{M}=-4\pi aD{\rho }_a$ obeys $\dot{a}=-D/a$. With negligible gas dynamic viscosity, instantaneous terminal velocity is $\dot{z}=\Delta \rho g{a}^2/(3\mu )$. Eliminating time yields the displayed radius-height relation and maximum rise $z_{\mathrm{diss}}=\Delta \rho g{a}_0^4/(12\mu D)$. This uses an instantaneous force balance; a literally constant speed during shrinkage instead gives $a^2$ linear in height. The Reynolds number must remain small for the Stokes flow drag model to apply.

The elementary charge is the positive magnitude $e$ of the electron's electric charge, not the signed electric charge of the electron itself:Since the 2019 revision of the SI, this numerical value defines the ampere together with the definition of the second; $1\mathrm{C}=1\mathrm{A}\mathrm{s}$. Thus today's numerical exactness is a choice of units, whereas measured electric charges still have experimental uncertainty. An oil drop experiment can test charge quantization and the apparatus calibration; it does not supply a new uncertain defining value of $e$ in today's SI. ^[ref].

For ordinary isolated chemical atoms, ions and charged objects whose net electric charge is counted through protons and electrons, charge quantization gives $Q=ne$ with signed integer $n$. A neutral chemical atom that loses two electrons becomes an ion of electric charge $+2e$. A toy object with $10^9$ excess electrons has $Q=-0.1602176634\mathrm{nC}$. Conversely, the scale $1\mathrm{nC}/e\simeq 6.24\times 10^9$ explains why laboratory electric charge often looks continuous: billions of microscopic increments fit into a small macroscopic amount. Averaged charge density is a continuum description, not a claim that each particle has an arbitrarily divisible electric charge.

The ordinary-matter qualifier matters. In the Standard Model, quarks have electric charges $+2e/3$ or $-e/3$, and antiquarks have the opposite signs. Confinement means these are constituents of hadrons, not isolated fractionally charged particles collected by an oil drop experiment. The usual valence quark sums give $2(2e/3)-e/3=e$ for a proton and $2e/3-2(e/3)=0$ for a neutron. Thus elementary charge should not be defined as the smallest nonzero electric charge of every constituent in particle theory. ^[ref].

The beautiful experimental idea behind the oil drop experiment is to determine a microscopic electric charge by watching a visible carrier move. In his 1913 paper, Millikan timed the descent and electrically driven ascent of individual oil drops, repeatedly changing a drop's electric charge through capture of air ions. His apparatus used parallel plates separated by $16\mathrm{mm}$, a roughly $5300\mathrm{V}$ source, and an optical scale to measure travel distances. He reported multiple relationships among a given drop's inferred electric charges. He also found that uncorrected Stokes's law gave a size-dependent inferred elementary unit, and introduced a pressure-dependent slip correction. These were measured relationships with calibration and model uncertainties, not exact readings of today's defined $e$. ^[ref].

A modern instructional implementation sprays low-volatility oil of known density through a plate aperture, illuminates the drops, and follows one through a calibrated optical scale. Measure the plate spacing $d$, electric potential difference $V$, air temperature and pressure. Time a fixed distance $L$ with the electric field off to get a terminal falling speed $v_f=L/t_f$. Then choose the field polarity to drive the same drop upward and measure $v_r=L/t_r$. Repeat timings at a fixed electric charge, and compare before and after an ion-capture event. The same drop is especially useful because its radius and mass need not be inferred anew for each charge state. ^[ref].

Here is the leading continuum model, retaining buoyancy. Assume a drop shaped as a sphere of radius $a$, oil and air densities ${\rho }_o$ and ${\rho }_a$, air dynamic viscosity $\eta$, and gravitational acceleration $g$. Define the downward effective weight $W$ and positive linear drag coefficient $k$ byAt terminal velocity, the net force vanishes. Falling with no electric field, the upward drag force balances weight minus buoyancy. Rising with the electric force upward, the drag force reverses and joins the downward effective weight:Consequently the uncorrected estimates areThe observed direction of response, with known field polarity, supplies the sign of $q$. A suspended drop gives the alternative balance $|q|E=W$, but it still requires a determination of $W$: balancing a drop alone does not determine its electric charge.

For an explicitly invented continuum example, stipulate $a=1\mu \mathrm{m}$, ${\rho }_o=900\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_a=1.2\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =1.8\times 10^{-5}\mathrm{Pa}\mathrm{s}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $q=-5e$ and a downward $E=10^5\mathrm{V}{\mathrm{m}}^{-1}$. The balances predict $W\simeq 3.69\times 10^{-14}\mathrm{N}$, $v_f\simeq 0.109\mathrm{mm}{\mathrm{s}}^{-1}$ and $v_r\simeq 0.127\mathrm{mm}{\mathrm{s}}^{-1}$. Substituting those speeds recovers $|q|=5e$ within rounding. This checks the algebra; it is not a precision prediction for real micrometre drops, which need a slip correction.

The inference has two distinct parts. Compare inferred electric charges or charge changes with a common unit, seeking $q_i\simeq n_{i}e$ within uncertainty, rather than demanding exact ratios of rounded measurements. Then establish the absolute scale through electric field, density, dynamic viscosity and distance/time calibration. If the observed steps are all even multiples, their common spacing alone cannot exclude an unobserved smaller unit. Air motion, evaporation and mistimed crossings can alter inferred electric charge without an ion-capture event. Small Reynolds number supports Stokes's law, but does not by itself justify its no-slip continuum coefficient: a significant ratio of air mean free path to drop radius requires a slip correction. Millikan's comparison of different drop sizes and pressures addressed that systematic limitation; the modern PASCO manual also includes the correction. Agreement supports charge quantization for the observed carriers within experimental resolution, rather than proving a universal statement about every possible particle.

Multiplying $e$ by the Avogadro constant gives the Faraday constant, $F=N_{A}e$, the electric charge per mole of singly charged carriers. This connects counting individual electric charges with bulk electrochemistry. ^[ref].

Robert Andrews Millikan's 1911 protocol alternated electrically driven ascent with timed descent after short-circuiting the plates; a changed ascent speed signalled ion capture.^[ref]

His 1913 refinements checked voltage before and after runs, calibrated optical distance and the clock, compared independent air-dynamic viscosity measurements, and varied pressure to test slip.^[ref] His 1924 lecture describes suppressing convection, switching the electric field, neutralizing drops, and reproducing smallest observed electrical speed increments.^[ref]

Harvey Fletcher's posthumous account recalls building the first oil apparatus and measuring with Robert Andrews Millikan; priority for suggesting oil remains uncertain.^[ref]

To interpret the readout, keep one drop and one electric field polarity. Let $u_i$ be its signed upward terminal velocity in charge state $i$, and let $v_f>0$ be its field-off falling speed. Use the linear drag coefficient $k$ already defined in elementary charge, with any required gas slip correction to Stokes drag included. Provided the drop's radius, surrounding air and electric field remain fixed, its electrical speed increment isThese relations apply while the electric force points upward in every state. A drop may still descend, with $u_i<0$, when that force is too weak to support it. A held drop has $u_i=0$ and therefore $S_i=v_f$, rather than zero electrical response. A neutral drop has $S_i=0$: its signed terminal velocity agrees with field-off descent. Thus neither raw upward speed nor the ability to hold a drop is itself a count of electrons.

The ratios cancel a fixed $k/E$, even if its value is poorly known. This is why relative charge quantization and absolute measurement calibration are different questions. A changed field-off $v_f$ warns that evaporation, air motion or another change has spoiled comparison of charge states. A stable $v_f$ is useful evidence but cannot exclude every compensating change. Repeat field-off checks around a proposed jump, allow the terminal velocity under linear drag to settle after switching, and propagate measurement uncertainty in both the field-on and field-off readings. Reusing a single estimate of $v_f$ gives the different $S_i$ a shared covariance.

At the terminal velocity, buoyancy-corrected weight balances the Stokes drag law:HenceA denser sphere falls in the direction of $\mathbf{g}$; a lighter one rises. Equal densities give no sedimentation. The calculation assumes negligible fluid Reynolds number, an unbounded fluid, and negligible transient body acceleration after terminal settling. The fluid mass density enters through buoyancy and through the small-Reynolds number condition.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

Retain the clean-bubble approximation ${\eta }_a\ll {\eta }_w$ used in parts (b) and (c), with constant mass densities and diffusion coefficient. For a spherical bubble,ConsequentlyThe buoyancy minus bubble weight is $4\pi ({\rho }_w-{\rho }_a)g{a}^3/3$. Balancing it against the clean-bubble Stokes drag gives the instantaneous terminal velocitySet $z=0$ at the initial position. Combining this with the dissolution equation gives $da/dz=-D_a/(K{a}^3)$ and hence the dissolution-limited bubble rise lawThe expression applies until the bracket vanishes. The rise speed decreases continuously as the bubble shrinks. Equivalently, $z(t)=K(a_0^{2}t-D_a{t}^2)$, providing a direct check on the elimination of time.

The phrase “constant terminal velocity” must here mean rapid adjustment to a force balance at each radius. A literally time-independent speed is incompatible with that balance because $U\propto a^2$. If instead an external mechanism maintained a fixed $U_{\ast }$, the distinct result would be $a^2(z)=a_0^2-2D_{a}z/U_{\ast }$ and $z_{\mathrm{diss}}=U_{\ast }{a}_0^2/(2D_a)$. In particular, holding the initial terminal speed fixed doubles the predicted rise distance; it is not the freely rising shrinking-bubble model.

Using the supplied values and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ givesEquality means dissolution exactly on arrival at the surface; dissolution strictly before arrival needs a greater initial depth. This is a formal low-Reynolds number estimate. The same values predict $U(a_0)\simeq 0.033\mathrm{m}{\mathrm{s}}^{-1}$ and a radius-based Reynolds number ${\rho }_{w}U(a_0)a_0/{\eta }_w\simeq 3.3$; the diameter-based value is about $6.5$. Neither is $\ll 1$. Thus the numerical input does not consistently satisfy the Stokes flow assumption initially, and the $0.82\mathrm{m}$ result should not be presented as an accurate oceanic prediction. Within the stated ideal model, bubble compression, contaminants and changes in dissolved-gas concentration are also excluded.

Set $v=\dot{z}$. The second-order position equation becomes the first-order linear ordinary differential equation $\dot{v}+\gamma v=-g$. For $\gamma >0$, multiply by the integrating factor $e^{\gamma t}$ and impose $v(0)=v_0$:The limiting terminal velocity is $-g/\gamma$, consistent with upward-positive position and downward gravity. The continuous zero-drag limit is $v_0-gt$; one must take that limit of the whole expression rather than set $\gamma =0$ in its two singular-looking terms separately.

Let $z$ measure height above the release point, and use $\mathcal{V}=4\pi a^3/3$ for the bubble volume to avoid confusing it with a speed. For constant radius, Newton's second law, the specified buoyancy and the given Stokes law giveWith positive $\alpha ,\rho ,\mu ,a$, defineThe differential equation becomes $\dot{u}+(u/\tau )=U/\tau$. Its solution with $u(0)=0$ is $u(t)=U(1-e^{-t/\tau })$. Integration using $z(0)=0$ yieldsAn arbitrary initial height is added to this expression. The effective inertia changes the relaxation time $\tau$, but not the terminal velocity obtained by balancing buoyancy and drag.

For the dissolving bubble assume $\beta >0$ and use the specified instantaneous terminal velocity, rather than the constant-radius transient formula. The bubble disappears at $t_{\ast }=a_0^2/\beta$, andIts total rise is therefore $\boxed{z(t_{\ast })=\rho g{a}_0^4/(9\mu \beta )}$. This second model neglects the initial acceleration and uses exactly the terminal-speed approximation stipulated for the shrinking bubble.

Take $0\le \gamma \le 1$ for the coefficient of restitution. Conservation of energy in the initial vacuum fall givesThe fixed plate has zero velocity, so the restitution rule makes the first upward rebound speed $u_1=\gamma u_0$. The ratio of rebound to impact kinetic energy is ${\gamma }^2$; henceGravity preserves the speed magnitude between departure from and return to the same plate height. Therefore the upward speed after the $n$th bounce is ${\gamma }^n{u}_0$. The geometric bounce sequence hasThe total travelled distance for $\gamma <1$ is the first fall plus ascent and descent for every rebound. Summing the geometric series givesThe idealized infinitely many impacts accumulate in finite time, after which the limiting state is rest on the plate; $\gamma =0$ gives rest at the first impact.

For the atmospheric fall, use upward signed velocity $u$ and height $z$. Newtonian mechanics with quadratic drag givesBefore impact $u\le 0$, so $\dot{u}=-g+(\alpha /m)u^2$. Set $v=-u\ge 0$, $v_T=\sqrt{mg/\alpha }$, and $\beta =\sqrt{\alpha g/m}=g/v_T$. Separation gives $\mathrm{artanh}(v/v_T)=\beta t$, establishing the descent from rest under quadratic drag:The speed approaches the terminal velocity $v_T$, although the formula is used only until the first impact. This occurs when $\cosh (\beta t_{\mathrm{hit}})=e^{\alpha h_0/m}$. In particular, the atmospheric impact speed is $v_T\sqrt{1-e^{-2\alpha h_0/m}}$, and both speed and height formulas recover the vacuum fall as $\alpha \to 0$.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

For $B_0\ne 0$, Newton's second law givesDefine the exponential relaxation time $\tau =mR/(B_0^2{L}^2)$ and terminal velocity $v_{\infty }={\mathcal{E}}_0/(B_{0}L)$. Solving the first-order equation for $v=\dot{s}$ and then integrating givesBoth prescribed initial values hold. The current approaches zero as the bar approaches terminal velocity. If $B_0=0$, there is no magnetic force and instead $\dot{s}=v_0$, $s=s_0+v_{0}t$; the constant current is ${\mathcal{E}}_0/R$.

Measure displacement downward from the release point and take downward speed $v\ge 0$. Newton's second law with quadratic drag givesWritingseparation of variables gives the terminal velocity solutionThe distance fallen by time $t$ is thereforeSetting $y(t)=h$ and solving for $t$ yields

For superterminal descent under quadratic drag, use $\dot{v}=vdv/dz$ in $\dot{v}=b(v_t^2-v^2)$. Equivalently,The speed approaches $v_t$ asymptotically from above. There is no finite exact distance at which it first equals the limiting speed. A practical estimate must specify an accuracy. Define “reached” as $v\le (1+ϵ)v_t$. ThenFor one-percent accuracy, $ϵ=0.01$ givesA five-percent criterion gives $z_{5\%}\simeq 0.642\mathrm{m}$. The characteristic decay length of the excess squared speed is $1/(2b)\simeq 0.0952\mathrm{m}$, but several such lengths are needed because the entry speed is so far above the limiting value. These estimates depend on the stated water-entry idealization; the source gives no finite tolerance or entry-loss model.

The air fall lastsRetain the underwater deceleration rather than replacing the whole water trajectory with motion at terminal velocity. DefineSeparation of the descent equation gives the hyperbolic cotangent solution and its integrated depth:where $t=0$ now means water entry. The derivative of the logarithm verifies $\dot{z}=v$, while $v(0)=v_t\coth \chi =u$. At water depth $H_w=5\mathrm{m}$, inversion using the inverse hyperbolic sine yieldsThe total elapsed time is thereforeFor an estimate without evaluating the exact expression, $b{H}_w\gg 1$ permits $\mathrm{arsinh}x\simeq \ln (2x)$ here, givingThis gives the same $4.43\mathrm{s}$ to the quoted precision. The simpler $H_w/v_t\simeq 4.72\mathrm{s}$ overestimates the underwater time because the pebble initially moves faster than $v_t$.

Descent equation. Take downward velocity $v$ and underwater depth $z$ as positive, with $z=0$ at the lake surface. For a sphere of radius $r$, the pebble mass is $m=4\pi r^3{\rho }_p/3$ and the frontal area is $A=\pi r^2$. Newton's second law includes downward gravitational weight, upward buoyancy and upward quadratic drag:ThusHere $b$ has units of inverse length. For arbitrary signed velocity, the last term is $-bv|v|$, so drag always opposes motion. The displayed downward form suffices for this trajectory.

Limiting speed. At terminal velocity, the net force and $\dot{v}$ vanish. For ${\rho }_p>{\rho }_w$,Use ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_w=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $r=0.020\mathrm{m}$, $C=0.7$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. ThenThe ${\rho }_p-{\rho }_w$ factor is essential: omitting buoyancy would overestimate the limiting speed.

For the numerical descent, assume the pebble is released from rest, neglect air drag, take $g$ and the water mass density as constant, and apply the given constant drag coefficient once it is submerged. Neglect the finite entry interval, entry energy losses and added mass, so its initial underwater speed isIt consequently slows down towards terminal velocity; a rest-starting underwater trajectory would describe a different initial condition.

Take upward velocity as positive. The linear drag force is $-mkv$, so it opposes the motion both during ascent and descent. Newton's second law gives $\dot{v}+kv=-g$. With $v(0)=u$, the integrating factor $e^{kt}$ yieldsThe terminal velocity is $-g/k$. The limit $k\to 0$ at fixed $t$ recovers $u-gt$; expanding the exponential is necessary to see the cancellation of the apparently singular terms.

For positive linear drag, $v_x$ decays exponentially toward zero. The vertical velocity decreases monotonically toward $-mg/\alpha$, with an initially steeper gradient $-g-(\alpha /m)v_0\sin \theta$ and positive second time derivative for an upward launch. It crosses zero atThe right panel shows the two components and the formal terminal velocity. To identify the end of the physical flight, integrate the vertical velocity once:The first positive root of $y(t)=0$ is ground contact. The terminal velocity need not be reached before that event; dashed portions in the sketch are mathematical extensions without a ground surface.

Take $x$ horizontal, $y$ upward and $g>0$. Resolve Newton's second law into components for projectile motion with linear drag:For $\alpha >0$, let ${\tau }_d=m/\alpha$. An integrating factor $e^{t/{\tau }_d}$ givesThe relaxation time is ${\tau }_d$, and the terminal velocity is $(0,-g{\tau }_d)$. For fixed finite $t$, expanding the exponential function as $\alpha \to 0$ gives the continuous limit $v_x=v_0\cos \theta$ and $v_y=v_0\sin \theta -gt$; the two terms proportional to ${\tau }_d$ cancel before taking the limit.

Write the linear drag as ${\mathbf{F}}_d=-b\mathbf{v}$, with $b>0$, and let $m$ be the projectile's mass. Taking upward as positive, Newton's second law givesThe downward terminal velocity is $-v_t$, whereEquivalently, if the drag acceleration is specified as $-k\mathbf{v}$, then $v_t=g/k$. Solving the first-order equation,The highest point occurs at $v_y=0$, soFor an upward launch this root is positive and ${\dot{v}}_y=-g$ there, confirming a maximum height. As $b\to 0$, expanding the logarithm gives the drag-free result $t_{\mathrm{top}}\to u_0\sin \alpha /g$. A horizontal launch has its highest point at launch; a downward launch from the ground does not have a later airborne maximum.