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John F. Lindner

November 11, 2026

Last weekend the twin ESCAPADE spacecraft (Blue and Gold) performed powered flybys of Earth, combining a gravitational “slingshot” and theOberth effect to accelerate their journeys to Mars. (The former name is a science popularization that evokes using a sling to whip a stone in a tight curve before releasing it along a fast tangent, and the latter name is that of its discoverer, physicist and rocket pioneer Hermann Oberth.)

“Slingshot”

A spacecraft passes behind a planet’s motion allowing its gravity to bend its trajectory toward its velocity, thereby gaining speed and kinetic energy relative to its star without expending fuel. Assuming a nearly $180^\circ$ bend, the motion is quasi-one-dimensional. If $v_{ab}$ is the velocity of $a$ relative to $b$ , and indices $r, p, s$ denote “rocket”, “planet”, “star”, then before the flyby
$v_{rs}=v_{rp}+v_{ps}$
and after the flyby
$v_{rp}^\prime=-v_{rp}.$
But $m_p \gg m_r$ implies $v_{ps}^\prime\approx v_{ps}$ , so
$\begin{aligned}v_{rs}^\prime&=v_{rp}^\prime+v_{ps}^\prime\\&\approx -v_{rp}+v_{ps}\\&=-(v_{rs}-v_{ps})+v_{ps}\\&=-v_{rs}+2v_{ps},\end{aligned}$
and the rocket’s return speed is twice the planet’s speed relative to the star!

Oberth Effect

If a spacecraft’s rocket engine can cause a speed change $\Delta v$ , the corresponding kinetic energy change
$\begin{aligned}\Delta K &= \frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2\\&=mv\Delta v+\frac{1}{2}m(\Delta v)^2,\end{aligned}$
where the first term is proportional to $v$ , so the larger the speed, the larger the kinetic energy change, and hence the larger speed at infinity. Indeed, if the spacecraft exhausts its burned fuel backward near perigee where it moves fastest, it leaves the burned fuel at minimum orbital speed relative to the stars, stripping it of its kinetic energy and transferring that energy to the spacecraft itself.
3D Ein Stein

John F. Lindner

October 17, 2026

In yet another recent exciting advance in mathematical tiling, last month Ioannis Tsiokos and the reasoning model GPT-6 Astra announced the discovery of a 3D aperiodic mono tile, an ein stein (or “one-stone”), a single shape that forces a non-periodic tiling of space, and reported formal verification by the Lean 4 theorem-prover software. Independently Felix Flicker and Chaim Goodman-Strauss soon simplified and elucidated the result.

The shape is a decorated 3D chair: a cube missing one of its corners, as in the figure below. The undecorated chair can tile space either periodically or nonperiodically, but the decorations are chosen to prohibit periodicity and force a hierarchical tiling of chairs into inflated superchairs with the same matching rules at twice the length scale. Thus, if the pattern repeats through a translation $\vec d$ , it must repeat through $2^n\vec d$ , but $\vec d\neq 2^n \vec d$ for $\vec d \neq 0$ . (The inflation can push the missing octant to infinity or, alternately, the octant can be recursively filled with deflated chairs.)

In Goodman-Strauss’s scheme, inflate the chair by arranging copies (left) so green arrows match red arrows and blue arrows match blue arrows to form a similar superchair (right), and repeat infinitely.

Freely downloadable computer files are already available from Bob MacKay to 3D-print copies of the chair, where patterns of bumps and dents enforce the matching rules (and cause the copies to “stick” together).

MacKay’s 3D-printed decorated chairs.
n-Ball Volumes

John F. Lindner

October 6, 2026

The way a ball’s volume varies with dimension is both fascinating and practically important (in statistical physics, data science, and elsewhere). Here I offer an accessible derivation.

Recursion

Let $V_n(R)$ be the volume of a ball of radius $R$ in dimension $n$ . Note that $V_1(R)=2R$ , $V_2(R)=\pi R^2$ , $V_3(R)=4\pi R^3/3$ , and $V_n(R)=R^nV_n(1)$ . The volume of a unit $n$ -ball
$\begin{aligned}V_n(1)&=\mathop{\int \cdots \int}\limits_{x_1^2 + \cdots + x_n^2 \le 1} \mathrm{d}x_1 \cdots \mathrm{d}x_n\\&=\mathop{\int\int}\limits_{x_1^2 + x_2^2 \le 1} \mathrm{d}x_1\mathrm{d}x_2\left(\mathop{\int \cdots \int}\limits_{x_3^2 + \cdots + x_n^2 \le 1-x_1^2-x_2^2} \mathrm{d}x_3 \cdots \mathrm{d}x_n \right)\\&=\mathop{\int\int}\limits_{x_1^2 + x_2^2 \le 1} \mathrm{d}x_1\mathrm{d}x_2 V_{n-2}\left(\sqrt{1-x_1^2-x_2^2}\right)\\&=\mathop{\int\int}\limits_{x_1^2 + x_2^2 \le 1} \mathrm{d}x_1\mathrm{d}x_2 \left(\sqrt{1-x_1^2-x_2^2}\right)^{n-2}V_{n-2}(1)\\&=V_{n-2}(1)\mathop{\int\int}\limits_{\substack{0\le r\le 1\\ 0\le \theta\le 2\pi}} r\mathrm{d}\theta\mathrm{d}r \left(1-r^2\right)^{n/2-1}\\&=V_{n-2}(1)\int_0^{2\pi} \!\mathrm{d}\theta\int_0^{1} \mathrm{d}r\,\left(1-r^2\right)^{n/2-1}r \\&=V_{n-2}(1)\,\theta\bigg|_0^{2\pi}\, \frac{(1-r^2)^{n/2}}{n/2}\left(\frac{1}{-2}\right)\bigg|_0^{1}\\&=V_{n-2}(1)\,\frac{2\pi}{n},\end{aligned}$
and so
$\color{red}R^n\color{black}V_n(1)=\color{red}R^{n-2}\color{black}V_{n-2}(1)\,\frac{2\pi \color{red}R^2\color{black}}{n}$
implies the volume of a generic $n$ -ball
$V_n(R)=V_{n-2}(R)\,\frac{2\pi R^2}{n}.$

Solution

This recursion has the beautiful solution
$\boxed{V_n(R)=\frac{\pi^{n/2}}{(n/2)!} R^n},$
where
$\begin{aligned}n!&=\int_0^\infty \!\!\!\!\mathrm{d}x\, e^{-x} x^n\\&= -e^{-x}x^n \bigg|_0^\infty + \int_0^\infty \!\!\!\!\mathrm{d}x\, e^{-x} n x^{n-1}\\&=n(n-1)!\end{aligned}$
is itself recursive with
$0!=\int_0^\infty \!\!\!\!\mathrm{d}x\, e^{-x} = -e^{-x} \bigg|_0^\infty = 1,$
(so, for example, $4! = 4\cdot 3\cdot 2\cdot 1 \cdot 0! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ ) and
$\begin{aligned}\left(-\frac{1}{2}\right)!&=\int_0^\infty \!\!\!\!\mathrm{d}x\, e^{-x} x^{-1/2}\\&=\int_0^\infty \!\!\!\!2z \mathrm{d}z\, e^{-z^2}z^{-1}\\&=\int_{-\infty}^\infty \!\!\!\! \mathrm{d}z\, e^{-z^2}\\&\color{red}=\sqrt{\int \mathrm{d}x\, e^{-x^2}\int \mathrm{d}y\, e^{-y^2}}\\&\color{red}=\sqrt{\iint \mathrm{d}x\mathrm{d}y\, e^{-(x^2+y^2)}}\\&\color{red}=\sqrt{\iint r \mathrm{d}\theta\mathrm{d}r\, e^{-r^2}}\\&\color{red}=\sqrt{\int_0^{2\pi} \!\!\!\!\mathrm{d}\theta\int_0^\infty \!\!\!\!\mathrm{d}r\, re^{-r^2}}\\&\color{red}=\sqrt{\theta\bigg|_0^{2\pi}\frac{e^{-r^2}}{-2}\bigg|_0^\infty}\\&=\sqrt{\pi},\end{aligned}$
my favorite integral. Check that initially
$\begin{aligned}V_1(R)&=\frac{\pi^{1/2}}{(1/2)!}R^1=\frac{\pi^{1/2}}{(1/2)(-1/2)!}R=2R,\\V_2(R)&=\frac{\pi^{2/2}}{(2/2)!}R^2=\pi R^2,\end{aligned}$
and thereafter
$\begin{aligned}V_n(R)&=\frac{\pi^{n/2}}{(n/2)!}R^{n}=\frac{\pi\pi^{n/2-1}}{(n/2)(n/2-1)!}R^2 R^{n-2}\\&=\frac{\pi^{(n-2)/2}}{((n-2)/2))!}R^{n-2}\frac{2\pi R^2}{n}=V_{n-2}(R) \frac{2\pi R^2}{n}.\end{aligned}$
Analysis

Not only does the volume $V_n(1)$ of a unit $n$ -ball decrease to zero for sufficiently large $n$ , but the ratio $V_n(R)/(2R)^n$ of the volume of a generic $n$ -ball to its circumscribing $n$ -cube decreases monotonically to zero! Understand this by noting that the $2^n$ $n$ -cube corners are at large distances $\sqrt{n}R \gg R$ for large $n$ .
Orbital Starship

John F. Lindner

September 28, 2026

Up early this morning to watch Starship flight 14. Eighteen years to the day since SpaceX‘s Falcon first achieved orbit, today so did Starship, and it deployed 26 large third-generation Starlink satellites.

Superheavy Starship is the world’s largest launch vehicle, by far, and is designed to be completely reusable. Much work remains to perfect it, but today’s flight was a major milestone. If successful, the program should revolutionize spaceflight.

Superheavy Starship launches from Starbase, Texas, 2026 September 28.

Orbital 124-meter Starship deploys folded Starlink satellites from its PEZ-like dispenser (photographed from one of the deployed satellites).
Navier-Stokes Singularity

John F. Lindner

September 9, 2026

Yesterday, OpenAI announced the solution to the Millennium-Prize Navier-Stokes problem: an initially smooth fluid can develop a singularity in finite time. More precisely, in a fluid modeled by the Navier-Stokes equations (Newton’s second law for incompressible fluids), from zero initial speeds a smoothly varying localized force can create a vortex that combines inward spiraling with axial outflow to produce in finite time an infinite speed (despite finite total energy), as in the annotated OpenAI graphic below.

According to OpenAI, an internal artificial intelligence model, significantly more capable than its Generative Pre-trained Transformer GPT‑6 Astra, solved the problem in 88 hours. Formalization and verification using the Lean proof-checking software took an additional 17 hours.

Over the past quarter century, mathematicians had made steady progress on the Navier-Stokes problem, with opinion gradually shifting from Navier-Stokes regularity to singularity. And while its solution may be a watershed moment, with artificial intelligence revolutionizing mathematics the way the telescope revolutionized astronomy, I’m personally unsurprised by the result itself, as non-collisional singularities exist even in the Newtonian $N$ -body problem for $N>3$ .

In a recent scheme by Joseph Gerver, Guan Huang, and Jinxin Xue, for a delicate fractal-like Cantor set of initial conditions, $N=4$ idealized point masses interacting gravitationally shuffle back and forth in a near-line and experience near-triple collisions that contract the pair in pseudo-Keplerian orbit, converting potential to kinetic energy (while maintaining constant total energy) and accelerating the masses in finite time to infinity, where yellow and cyan lines suggest the general motion in the graphic below.
Roman vs. Hubble

John F. Lindner

August 30, 2026

Up early this morning for the successful launch and deployment of the Roman Space Telescope, whose development I have followed for 15 years. Considered a successor to the Hubble Space Telescope, Roman’s 2.4-meter primary mirror is the same size as Hubble’s but is shaped differently, and its shorter focal length provides a much larger field-of-view.

The simplest reflecting telescopes use spherical mirrors, which can naturally be ground by hand, but suffer from spherical aberration (so off-axis light rays focus at different distances). Parabolic mirrors eliminate spherical aberration, but suffer from coma (so off-axis point sources distort to comet shapes) and astigmatism (so light rays in perpendicular planes focus at different points). Exploiting conical geometry, the Hubble and Roman optical designs carefully combine mirrors with hyperbolic or elliptical cross sections to minimize or eliminate such optical aberrations.

Hubble’s two-mirror Ritchey-Chrétien Cassegrain design pairs a concave hyperboloidal primary mirror (with a center hole) and a convex hyperboloidal secondary mirror, as diagrammed.

Hyperbolic (cyan) cross sections of the Hubble Space Telescope’s mirrors.

Roman’s Three-Mirror Anastigmat design combines an ellipsoidal concave primary mirror (with a center hole), a hyperboloidal convex secondary mirror, and an ellipsoidal concave tertiary mirror, as diagrammed. In addition flat fold mirrors between the secondary and tertiary mirrors deflect light off-axis into scientific instruments.

Elliptical (red) and hyperbolic (cyan) cross sections of the Roman Space Telescope’s major mirrors.

Elliptical (red), hyperbolic (cyan), and flat (green) cross sections of the Roman Space Telescope’s mirrors, where the flat mirrors help fold the light rays off-axis into its instruments.

Thanks to newer technology, Roman’s mirrors are smoother, lighter, and stiffer than Hubble’s, although due to its cosmological and exoplanet goals, their reflection coatings are optimized for near-infrared light, while Hubble’s coatings are optimized for visible light, including near-infrared and near-ultraviolet.
First Photo From Mars

John F. Lindner

August 7, 2026

50 years ago was summer 1976, post Apollo, but pre-Space Shuttle. On July 20th, the seventh anniversary of the Apollo 11 Moon landing, the Viking 1 spacecraft made the first successful landing on Mars. (5 years earlier, the Mars 3 spacecraft ceased communicating just 110 seconds after landing.)

My family owned and operated a small motel in Vermont. But on that day, we were glued to the TV watching the news coverage of the Viking 1 landing. In those days, the martian surface was mysterious, and indeed the Viking program was designed to search for extant life on Mars, and I did not know what to expect.

Data took 19 minutes to reach Earth from Mars. I vividly remember the electrifying reveal of Viking’s first grayscale photograph, slowly assembling itself on the TV screen, left-to-right. First we saw grains, and pebbles, and rocks, and finally one of the lander’s footpads, rivets and all. The picture clarity and exposure seemed amazing. I remember the excited commentary of Thomas Mutch, leader of the lander imaging team, who would tragically die just four years later while mountain climbing.

First photograph from the surface of Mars, 1976 July 20, by the Viking 1 lander.

Soon after, on our car’s radio, I heard that the first color photo from Viking 1 showed an Earth-like bluish sky. Later I learned that the photo’s color balance overcorrected the martian atmosphere to look terrestrial, as is often done nowadays to aid analysis by terrestrial geologists, and the color balance was re-adjusted to show an alien sky. I hope to live long enough to hear humans on Mars describe their direct subjective experiences of the martian sky, which seems to vary from pale yellow-brown at noon to blue at dusk, probably due to dust in the air.
Constructing Pentagons

John F. Lindner

July 1, 2026

A highlight of Euclid’s Elements is the exact construction of a regular pentagon using a straight edge and a compass. But I’ve noticed recently that some of the pentagon constructions on YouTube are merely approximations – and are not advertised or understood as such. Furthermore, many of these constructions require non-collapsible compasses, which can “carry” lengths from one place to another, thereby shortcutting the classical process.

Here I present a rigorous construction of a pentagon by just drawing lines and circles through intersections of previous lines and circles (from a pre-existing pair of points), which can be replicated by a straight edge and the ancient Greek collapsible compass. Rather than animate the construction, I color code the lines and circles by hues or grays in order of execution. I implement the construction in the Wolfram Language, formerly known as Mathematica, and thereby confirm the exactness of the result.

For variety, the hues version is rotated with respect to the grays version, and the code follows the figures.

In Mathematica, I define circles, lines, and points as complex numbers and solve for intersections algebraically.

The moduli and arguments of the complex vertices verify the pentagonal geometry.

The ParametricPlot function draws the 7 circles and 4 lines, with white dots at circle centers.
Negative Resistance

John F. Lindner

June 19, 2026

I recently coauthored an article in the International Journal of Bifurcation and Chaos that reports efficiently performing logic with a simple nonlinear electronic device that exhibits negative resistance, something I had previously theorized.

At the heart of the device is a single p-n junction, which is formed by a silicon semiconductor crystal doped with impurity atoms (like boron or antimony) so the n-type side contributes negative charges called electrons to conduction, while the p-type side contributes missing electrons called holes, which act like positive charges. Depending on the polarity of the attached battery, the intermediate depletion layer contracts or expands, and the junction acts like a switch or valve called a diode, either conducting current or blocking it.

The Uni-Junction Transistor (UJT) is a single asymmetric p-n junction, whose equivalent circuit features an ideal diode that outputs to two internal resistors, the bottom of which is variable. The three attachment points are conventionally called the side “emitter” $E$ , the bottom “base” $B_1$ , and the top “base” $B_2$ .

Power the UJT by a static voltage $V_{BB}$ between the bases so $B_2 > B_1$ , and control the emitter voltage $V_E$ and current $I_E$ with the variable voltage $V_{EE}$ via a resistor $R_{E}$ . Varying $V_{EE}$ reveals the BJT characteristic voltage-current $VI$ curve (and $IV$ curve) labeled with static resistance $R_S = V/I$ and differential or dynamic resistance $R_D = dV/dI$ (and the corresponding conductances $g = 1/R$ ). The current $I_E$ versus voltage $V_E$ is multivalued.

For small $V_{EE}$ the p-n junction is reversed biased, and a small current $I_E$ leaks backward through it. For larger $V_{EE}$ , the p-n junction becomes forward biased, triggering it to inject holes into the n-type bar, where the extra charge carriers dramatically increase conductivity, decreasing the voltage $V_{E}$ while increasing the current $I_{E}$ , in a region of negative differential resistance.
Stokes’ Drag Law

John F. Lindner

June 1, 2026

Introductory physics often assumes without proof that the drag force on an object is proportional to its velocity, at least for smooth or laminar flow. In particular, a sphere of radius $a$ falling slowly with velocity $\underline v$ in air of viscosity $\eta$ experiences a drag force

$\underline{F} = -6\pi \eta a \underline{v},$

which was first derived by George Stokes in 1851. Here is a digestible derivation of the force on an idealized sphere in an upward flowing fluid using Mathematica, including motivation for the underlying Navier-Stokes fluid-flow equations.

Vectors (with singly-indexed components that can be arranged in column matrices) are underlined while second-rank tensors (with doubly-indexed components that can be arranged in square matrices) are doubly-underlined.

Navier-Stokes Equations

Stress tensor

Recall that pressure perpendicular to the $x$ -direction due to force in the $x$ -direction is

$p_x = \frac{dF_x}{da_x},$

and the shear perpendicular to the $y$ -direction due to velocity changes in the $x$ -direction is

$\tau_{xy} = \frac{dF_x}{da_y} = \eta \frac{dv_x}{dy},$

where $\eta = \mu$ is the dynamic viscosity. For an isotropic fluid, symmetrize this to

$\underline{\underline{\tau}} = \eta\left( \underline{\nabla}\,\underline{v} + (\underline{\nabla}\,\underline{v})^T \right)= \eta\left( \underline{\nabla}\,\underline{v} + \underline{v} \,\underline{\nabla} \right)$

(with no dot or cross products), where the pressure and the shear combine to form the stress tensor

$\underline{\underline{\sigma}} = -p \underline{\underline{I}}+\underline{\underline{\tau}},$

so the force

$\underline{F} = \oiint_{a=\partial V} \hspace{-1.6em} d\underline{a} \cdot \underline{\underline{\sigma}} = \iiint_V \hspace{-0.4em} dV\, \underline{\nabla} \cdot\underline{\underline{\sigma}} = \iiint_V \hspace{-0.4em} dV\, \underline{f}.$

Continuity Equation

The time rate of change of the fluid density $\rho$ is minus the divergence of the mass current $\underline{J} = \rho \underline{v}$ ,

$\partial_t \rho = -\underline{\nabla} \cdot \underline{J},$

which for constant density simplifies to a divergence-less velocity field

$0 = \underline{\nabla} \cdot \underline{v}.$

Newton’s Second Law

For an infinitesimal fluid element of velocity $\underline v(t, \underline r)$ , the force per unit volume

$\rho\left( \partial_t\underline{v}+\underline{v}\cdot \underline{\nabla}\, \underline{v} \right)= \rho \frac{d\underline{v}}{dt} = \underline{f} = \underline{\nabla} \cdot \underline{\underline{\sigma}}.$

For a stationary flow, so $\partial_t\underline{v} = \underline{0}$ , and slow fluid, so terms $\mathcal{O}(v^2)$ are negligible, this reduces to

$\begin{array}{l} \underline{0} & = &\phantom{+}\underline{\nabla} \cdot \underline{\underline{\sigma}} \\ & = & -\underline{\nabla}\,p + \underline{\nabla}\cdot \underline{\underline{\tau}} \\ & = & -\underline{\nabla}\,p + \eta \left(\underline{\nabla} \cdot \underline{\nabla}\,\underline{v} + \underline{\nabla} \cdot \underline{v}\, \underline{\nabla} \right) \\ & = & -\underline{\nabla}\, p + \eta \Delta \underline{v}, \end{array}$

where the Laplacian $\underline{\nabla} \cdot \underline{\nabla} = \nabla^2 = \Delta$ . Hence, the relevant Navier-Stokes equations

$\boxed{\begin{array}{ccc}\underline{\nabla}\,p & = & \eta \Delta \underline{v}, & (1)\\ \underline{\nabla}\cdot \underline{v} & = & 0, & (2) \end{array}}$

plus boundary conditions determine the fluid pressure $p$ and velocity $\underline{v}$ .

Pressure and Velocity

Although the computation can be done by hand (as Stokes did), Mathematica eases the workload.

Coordinates

Due to the sphere, introduce spherical coordinates $\{r, \theta,\phi \}$ with unit vectors $\{\underline{u}_r, \underline{u}_\theta, \underline{u}_\phi \}$ , and due to the distant uniform flow, introduce the cylindrical unit vector $\underline{u}_z$ .

Solve Eq. (1) for Velocity

Because the divergence of any curl vanishes, take the fluid velocity (with respect to the sphere) to be the curl of a vector field $\underline{v} = \underline{\nabla} \times \underline{\psi}$ , where the educated guess

$\underline{\psi}(r,\theta) = \underline{u}_\theta \left( \frac{c_1}{r^2}+c_2 + c_3 r \right) \sin\theta,$

subject to the sticky boundary at the sphere and the uniform boundary at infinity, implies

$\begin{array}{ll} \underline{v}(r,\theta) = \displaystyle\frac{v_\infty}{4r^3}(a-r)\big( \hspace{-1.5em}&\phantom{+}\underline{u}_r(2a^2+2ar-4r^2)\cos\theta \\ & +\underline{u}_\theta (a^2+ar+4r^2)\sin\theta \big). \end{array}$

$\underline{\psi}(r,\theta)$ mainly swirls about the vertical so that its curl $\underline{v}(r,\theta)$ mainly streams upward.

Shear

From above, the shear is proportional to the symmetrized velocity gradient, and at the sphere’s surface

$\underline{\underline{\sigma}}_s(a,\theta) = \underline{\underline{\tau}}(a, \theta) = -\frac{3\eta v_\infty}{2a} \fbox{$\displaystyle \begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}$}\sin\theta.$

Solve Eq. (2) for Pressure

As another educated guess (as they are easy to check with Mathematica), take the pressure (relative to atmospheric pressure) to be of the form

$p(r,\theta)= c_4 \frac{\cos\theta}{r^2},$

which satisfies the boundary condition $p(\infty,\theta) =0$ . Substituting into the Navier-Stokes pressure equation fixes the constant, so

$p(r,\theta)= -\frac{3 \eta a v_\infty}{2r^2} \cos\theta.$

The corresponding pressure stress at the sphere’s surface

$\underline{\underline{\sigma}}_p(a, \theta) = -p(a, \theta)\underline{\underline{I}} = \frac{3\eta v_\infty}{2a} \fbox{$\displaystyle \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}$}\cos\theta.$

Force

The radial pressure on the sphere in the $\underline{u}_z$ direction is

$\begin{array}{l} F_p & = &\displaystyle\oiint_{S} dS\, \underline{u}_r \cdot\underline{\underline{\sigma}}_p(a,\theta)\cdot\underline{u}_z \\ & = & \displaystyle\int_0^\pi\hspace{-0.5em}d\theta \int_0^{2\pi}\hspace{-1em}d\phi \left(a^2 \sin\theta\right) \frac{3 \eta v_\infty}{2a} \cos^2\theta \, \\ & = & 2 \pi \eta a v_\infty, \end{array}$

and the radial shear on the sphere in the $\underline{u}_z$ direction is

$\begin{array}{l} F_s & = & \displaystyle\oiint_{S} dS\, \underline{u}_r \cdot\underline{\underline{\sigma}}_s(a,\theta)\cdot\underline{u}_z \\ & = & \displaystyle\int_0^\pi\hspace{-0.5em}d\theta \int_0^{2\pi}\hspace{-1em}d\phi \left(a^2 \sin\theta\right) \frac{3 \eta v_\infty}{2a} \sin^2\theta \, \\ & = & 4 \pi \eta a v_\infty = 2F_p, \end{array}$

so the total upward force magnitude

$F = F_p + F_s = 6 \pi \eta a v_\infty.$

The pressure “blows” from the bottom and “sucks” from the top, but the shear “rubs” twice as hard near the middle!

Recent Comments

John F. Lindner on Venus’s Supercritical Ocean2024 March 1
Thanks, Mark! I enjoy reading your posts as well.
Mark Wilson on Venus’s Supercritical Ocean2024 March 1
Nice post, John! Thanks for writing these. I always enjoy them.
John F. Lindner on Ein Stein2023 May 6
David Smith’s ein stein is an awesome shape!

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