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For a clean inviscid bubble a distance $d\gg a$ below a gravity-restored approximately flat free surface, the surface displacement scales as $F/(\rho g{d}^2)=O(a^3/d^2)$. Zero normal velocity and zero tangential traction are imposed to leading order by an opposite normal image Stokeslet at distance $2d$ from the bubble. Its incident velocity is $-F/(8\pi \mu d)$, which gives the displayed correction at fixed buoyancy. The image can be represented by a same-radius inviscid drop of density $2\rho$ in liquid of density $\rho$. Its deformation-producing strain is $O(F/(\mu d^2))$, so the leading fractional capillary deformation is $O(\mathrm{Bo}(a/d{)}^2)$; small Bond number is the conventional stronger sufficient criterion.

The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

The gravity-capillary drop with a meridional curvature approximation uses the stipulated $\kappa =-h^{\prime \prime }$ for this calculation. At equilibrium the thin-film equation givesRegularity at $r=0$ forces $C=0$. In the wetted interior $h>0$, the equilibrium pressure gradient therefore vanishes, givingThis is the capillary length. Integrating the constant-coefficient ordinary differential equation givesThe central boundary condition $h^{\prime }(0)=0$ sets $C_2=0$, and contact at $r=R$ gives $C_0=-C_1\cosh (R/l)$. Write $C_1=-\alpha l$. The prescribed contact angle fixesHere $\alpha$ is the dimensionless coefficient of this drop solution, unrelated to the hinge angle in question 1.

The axisymmetric volume integral is $V=2\pi {\int }_0^{R}rh(r)dr$. Substituting $\xi =r/l$ givesFor numerical evaluation, with $s=R/l$, the integral isTogether with $\alpha =\tan {\theta }_c/\sinh s$, this determines $R$ for fixed $V$ and ${\theta }_c$.

The Bond number $R^2/l^2=\rho g{R}^2/\gamma$ distinguishes the dominant balances. For $R\ll l$, surface tension dominates gravity and the surface is nearly a small-slope constant-curvature cap:For $R\gg l$, gravity flattens the interior, while surface tension bends the edge over the capillary length. The stipulated model givesup to relative corrections of order $l/R$. The contact angle fixes the edge slope and therefore the height scale; conservation of mass then fixes the lateral extent. Consistency of lubrication theory requires small slopes, in particular $\tan {\theta }_c\ll 1$.

The stated approximation differs from axisymmetric free-surface curvature. For an axisymmetric graph with the same sign convention, the sum of principal curvatures isThe omitted $h^{\prime }/r$ term is generally of the same order as $h^{\prime \prime }$. For example, $h=H(1-r^2/R^2)$ gives stipulated $\kappa =2H/R^2$ but axisymmetric small-slope $\kappa =4H/R^2$. Thus the required hyperbolic-cosine profile solves the explicitly prescribed reduced model, not the full axisymmetric capillary equilibrium. Keeping both small-slope terms instead giveswhere $I_0,I_1$ are modified Bessel functions. The qualitative capillary length and Bond number balances remain valid.

Let $z$ measure height above the substrate, and write $\mu =\rho \nu$. In the lubrication approximation, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:The no-slip boundary condition gives $u(0)=0$; negligible air viscosity and constant surface tension give $u_z(h)=0$. Integrating twice and then through the depth yieldsThe free-surface kinematic boundary condition and incompressibility imply $h_t+q_x=0$. ThereforeThe cross-sectional area $A=\int hdx$ is conserved when the endpoint volume flux vanishes.

The relevant small groups follow from the same balances. Put ${ϵ}_h=H_0/\ell$ and $U_{\ast }=g{H}_0^3/(3\nu \ell )$. Small ${ϵ}_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O({ϵ}_h^2)$. The ratio of horizontal inertia to vertical viscous stress isThus a small full-length Reynolds number is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_{\ast })$ must likewise be small; on the spreading timescale $t_{\ast }=\ell /U_{\ast }$ it equals the displayed group. If surface tension is retained, its pressure-gradient correction relative to gravity isThe Bond number here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/{\ell }^2)$ whereas the hydrostatic pressure is $O(\rho g{H}_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant kinematic viscosity.

To find the similarity solution, put $K=g/(3\nu )$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the time derivative against the nonlinear diffusion gives ${\ell }^5\sim K{A}^{3}t$. A normalized ansatz isSubstitution gives $(f^3{f}^{\prime }{)}^{\prime }=-(f+\xi f^{\prime })/5$. For the symmetric zero-flux profile, one integration gives $f^3{f}^{\prime }=-\xi f/5$, and on its positive support a second integration givesThus the solution is a compactly supported Barenblatt solution of the porous medium equation. Let $\ell (t)$ denote its half-width. In dimensional variables,Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell \sin \theta$,Writing the remaining integral as $I$, the final widths areThe height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The volume flux tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported weak solution of the reduced equation. The diverging edge slope means the lubrication approximation is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.

The force scale is $F\sim \rho g{a}^3$. At the distant free surface, the Stokeslet's pressure and viscous normal-stress scales are $F/d^2$. Gravity restores a displaced surface by a pressure change $\rho gh$, soThus $h/d=O((a/d{)}^3)\ll 1$, consistent with treating the boundary as approximately flat and impermeable at leading order. This uses the stipulated gravity-dominated restoring balance.

The liquid-air interface has negligible tangential traction. The stress-free image of a normal Stokeslet places an opposite point force at the reflection point, a distance $2d$ above the bubble centre. Across the flat surface the combined normal velocity is odd and the tangential velocity is even, so the normal velocity and tangential stress vanish there. A same-radius inviscid drop of density $2\rho$ has downward excess weight $4\pi \rho g{a}^3/3$, exactly opposite the bubble's upward buoyancy. Its exterior Stokeslet therefore supplies this image, giving the asserted full-space comparison to leading order in $a/d$.

The image's incident axial velocity at the bubble isTo first reflection, the bubble translates with this locally uniform incident velocity plus its isolated rise speed $F/(4\pi \mu a)$. Hence the bubble rise near a distant stress-free free surface isOnly the first relative correction is retained; the image's spatial variation and additional reflections affect higher terms.

The customary capillary-controlled small-deformation condition is a small capillary number,equivalently a small Bond number $\mathrm{Bo}=\rho g{a}^2/\gamma$. This is a sufficient condition, not a necessary condition for the special isolated spherical solution in part (b). More precisely, the image's strain is $O(F/(\mu d^2))$, so its deformation-producing normal stress compared with $\gamma /a$ is $O(\mathrm{Bo}(a/d{)}^2)$. When exploiting the exact isolated solution, that smaller combination controls the leading boundary-induced distortion. If capillarity at the distant free surface is also retained, the gravity-dominated deflection scaling requires $\gamma /(\rho g{d}^2)\ll 1$; $a\ll \sqrt{\gamma /(\rho g)}\ll d$ allows both the usual spherical-bubble criterion and gravity-dominated surface restoration.

Treat the two film surfaces as an effective interfacial tension $\Gamma =2\gamma$. The relevant Bond number for the gas-density difference is $\mathrm{Bo}=\Delta \rho g{r}^2/\Gamma$. Substituting the Young–Laplace equation and the preceding density relation givesThus the formal crossover $\mathrm{Bo}>1$ gives $r>h_0/2\simeq 4.25\mathrm{km}$. This compares gravity on the excess gas with capillary restoration and neglects film weight. It does not say gravity-driven drainage of a real water film is negligible below kilometre scales. At this enormous radius, treating ambient pressure and density as uniform is itself no longer quantitatively reliable.

On Teflon, the water drop has an obtuse contact angle and beads up, whereas the oil drop has an acute contact angle and spreads into a shallower patch. For small drops with negligible Bond number, both can be drawn as spherical caps.

To compare shape at equal volume $V$, let $R$ denote the generating sphere radius. The cap height and footprint radius areFor a fixed $V$, these relations give a taller, narrower drop for the larger contact angle. The angles are measured inside the liquid, including the obtuse angle of the water drop.

The relevant comparison is between gravity and surface tension. With the mass density difference $\Delta \rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, the Bond number and capillary length areThus the drop is small compared with the capillary length; hydrostatic pressure variations only weakly perturb its shape. The Young–Laplace equation then gives approximately constant interfacial mean curvature, and an axisymmetric drop on a uniform flat substrate is a spherical cap. The substrate selects the contact angle, measured through the water.

Write $R$ for the radius of the generating sphere, $s$ for the contact-circle radius, and $h$ for the apex height. Elementary trigonometric identities giveTaking the stated droplet radius to mean its radius of curvature, $R=r$, givesThere is no diagram or explicit radius convention in the original PDF. If $r$ instead denotes the contact-circle radius, the answer is $h=r\tan 27.5^{\circ }=0.260\mathrm{mm}$. If it denotes the radius of an equal-volume free sphere, use the volume of a spherical cap, $\pi h^2(3R-h)/3=4\pi r^3/3$, to obtain $h=r[4(1-\cos \theta )/(2+\cos \theta ){]}^{1/3}=0.436\mathrm{mm}$. The spherical cap approximation is justified in all three conventions; a unique numerical height requires specifying which radius is meant.