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For a spherical body of radius $R$, mass $M$ and mean density $\bar{\rho }=3M/(4\pi R^3)$, a low circular orbit has period $P\simeq 2\pi \sqrt{R^3/(GM)}$, by equating the gravitational acceleration from Newton's law of universal gravitation to the centripetal acceleration. Substitution proves the displayed relation. Thus similar low-orbit periods indicate similar mean densities, even for bodies with very different masses and radii. At altitude $h$, the right side is multiplied by $(1+h/R{)}^3$; the simple density inference requires $h\ll R$ and neglects rotation, nonsphericity and other gravitating bodies.

Mass spectrometry separates and detects ions according to mass-to-charge ratio. It can distinguish isotopes which belong to the same chemical element but have different atomic masses. The instrument measures ions: interpreting the result as a neutral atomic mass requires accounting for electron removal and its ionization energy. For a singly positive ion in its ground state, $m_{+}=m_{\mathrm{a}}-m_e+I/c^2$, where $m_e$ is the electron rest mass, $I$ is the ionization energy, and $c$ is the speed of light.

For a simple illustrative analyzer, positive ions of mass $m$ and electric charge $q=ze$ gain kinetic energy $qV=m{v}^2/2$ across an electric potential difference $V$ before entering a transverse uniform magnetic field $B$. The Lorentz force supplies centripetal acceleration, soAt fixed $V$ and $B$, the measured bending radius $r$ separates mass-to-charge ratios, not masses independently of electric charge. This ideal model assumes negligible initial kinetic energy and speeds small compared with the speed of light; the historical focusing mass spectrograph used the geometry described below.

The experimental achievement is to turn invisible differences between individual ions into separated, calibrated positions. Aston velocity focusing explains how an early mass spectrograph sharpened its images; the velocity-selected magnetic mass analyzer provides an independent model whose equations can be checked directly. The magnesium isotope mass spectrum shows why position, width, and intensity carry different information.

At a fixed point beside a river, the Eulerian coordinate description measures $\mathbf{u}(\mathbf{x},t)$. Its local acceleration ${\partial }_t\mathbf{u}$ measures change at that fixed point. Following a floating parcel gives a Lagrangian trajectory $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives its physical material acceleration:The extra term is the convective acceleration. For a river that narrows steadily, an observer at one fixed position measures a constant velocity, yet a parcel speeds up as it reaches the narrowing. A parcel following a steady bend also experiences a centripetal acceleration even when its speed is constant. An Eulerian calculation of the physical acceleration must include the same convective acceleration; the two descriptions do not predict different forces.

The Newtonian achievement combined a dynamical framework, a universal gravitational interaction and a distinctive standard of physical explanation. It was problematic partly because its mathematical successes exceeded what it supplied as a physical mechanism. Calling it a system should not conceal the difference between Isaac Newton's own writings, their later development and simplified versions promoted by his followers.

The Philosophiae Naturalis Principia Mathematica by Isaac Newton began with definitions and the Newton's laws of motion. Inertia distinguished persistence in uniform rectilinear motion from changes requiring impressed force. Newton's second law related such changes to impressed force; Newton's third law required reciprocal interactions. The account also distinguished absolute and relative quantities, raising questions about the meaning of true motion. In modern notation, a central part of the programme is expressed byHere momentum, force, masses and separation have their modern notation, and $G$ is the gravitational constant; the display is not a quotation of Isaac Newton's symbols. Combining Newton's laws of motion with Newton's law of universal gravitation made it possible to explain a wide range of terrestrial and celestial phenomena together.^[ref]

The quantitative content matters. For an ideal circular orbit of a small body around a much larger mass $M$, centripetal acceleration gives, in modern notation,This restricted illustration recovers the period–radius relation in Kepler's third law. It is not a reconstruction of the full original proof for an elliptical orbit. More generally the same interaction could account for the Moon's departure from a straight path, the paths of comets and mutual perturbations. Universal attraction made celestial irregularities potential consequences of a common law rather than reasons to assign a separate nature to each class of object.

The associated Newtonian programme extended beyond gravity. In optics, experiments on refraction supported the heterogeneous composition of white light: a prism could separate constituents with different refrangibilities rather than manufacture colours from a previously homogeneous ray. This challenged accounts that treated colour primarily as a modification introduced by the medium. It belongs to the broader Newtonian programme, although the 1704 Opticks by Isaac Newton must not be conflated with the content or publication date of the 1687 Philosophiae Naturalis Principia Mathematica by Isaac Newton.^[ref]

The greatest conceptual difficulty concerned action at a distance. For a defender of a contact mechanism, the mutual attraction of bodies through empty space appeared to reintroduce an occult quality instead of explaining an interaction. Cartesian vortices offered at least a material picture of how bodies might be carried around, whereas an inverse-square force law did not specify what transmitted the influence. Isaac Newton could point to the difficulties of reconciling such vortices with planetary periods and the varied paths of comets, but defeating that particular mechanism did not itself furnish an alternative mechanical cause.^[ref]

It is particularly important that Isaac Newton did not simply identify unmediated attraction as an essential property of matter. In his letter to Richard Bentley of 25 February 1692/3, he rejected that attribution and left open whether the relevant agent was material or immaterial. In the General Scholium by Isaac Newton, he distinguished establishing how gravity acts from assigning a cause not warranted by the phenomena.^[ref][ref]

Other difficulties followed from the standard of demonstration itself. The demanding geometry made the work hard to master; approximations and mutual interactions meant that apparent agreement had to be earned through calculation, not assumed from a slogan about universal attraction. Absolute motion and divine government also raised philosophical questions that a successful orbital calculation could not settle. Initial resistance thus included reasonable disputes about mechanism, evidence and explanatory adequacy. The system's later authority came from its capacity to sustain successful investigation despite those unresolved foundations, not from having answered every foundational question at publication.

Isaac Newton made the greater contribution if the criterion is the scope and explanatory integration of mathematical natural philosophy; Galileo Galilei has the stronger claim if the criterion is the formative transformation of how terrestrial motion could be investigated. The first criterion gives a defensible overall verdict for Isaac Newton, but a serious comparison must show what each changed rather than award credit according to who worked later.

Galileo Galilei's contribution was not simply to attach numbers to familiar events. In the Third Day of Two New Sciences by Galileo Galilei, he defines uniformly accelerated motion by equal additions of speed in equal intervals of time, then derives consequences. Starting from rest, the modern notation is $v=at$; the mean speed over the interval is $at/2$, soHence distances in successive equal intervals have the ratios $1:3:5:7:\dots$. The novelty lies in making a precisely specified kind of motion the subject of mathematical demonstration, with observed falls assessed against its consequences. This is not the same as deriving the underlying physical cause of gravity.^[ref]

The reported apparatus shows how such an idealization could be investigated. A smooth bronze ball descended in a straight grooved wooden channel, raised at one end. Water discharged through a small opening into a vessel during the descent; its collected mass served as a measure of elapsed time. Repeating runs over different portions of the channel compared the predicted squared dependence of time on distance. The inclined plane slowed the event, while repetition and weighing made a temporal comparison possible without a modern stopwatch. The published report is evidence of Galileo Galilei's experimental presentation, not a surviving raw-data archive from which every reported precision can now be reconstructed.^[ref]

A necessary physical qualification is often missed. A ball in rolling without slipping is not a point particle sliding without friction. For a solid sphere on an ideal inclined plane, modern rigid body dynamics gives $a=5g\sin \alpha /7$, rather than $g\sin \alpha$. Both are constant accelerations, so both give $s\propto t^2$. The trial can therefore test the temporal form without directly establishing the free-fall coefficient. This distinction makes the experiment more intelligible: the evidence supports a specified relation under approximations, rather than every detail of a modern idealization.

The Fourth Day extends this approach to projectile motion. Combine uniform horizontal velocity $u$ with constant downward acceleration $g$, taking downward displacement positive. Then $x=ut$ and $y=g{t}^2/2$. Eliminating time givesa parabola. Actual projectiles experience air resistance, but the ideal curve organizes the effects that need separate treatment. Galileo Galilei thus gave a powerful example of how mathematical abstraction and physical observation could cooperate.^[ref]

Isaac Newton went further by connecting kinds of motion through a general account of force. Newton's second law links the change of momentum to the impressed force; for constant mass, its modern differential expression is $\mathbf{F}=m\mathbf{a}$. Newton's third law makes interaction reciprocal, enabling a consistent analysis of interacting bodies. The Philosophiae Naturalis Principia Mathematica by Isaac Newton develops consequences of central forces and then asks which laws fit astronomical phenomena. Mathematical deduction now connects measurable orbits to an underlying force law, rather than just describing each trajectory separately.^[ref]

A simple modern reconstruction shows the increase in explanatory reach. For a circular orbit of radius $r$ and orbital period $T$ about a much more massive central body, centripetal acceleration is $4{\pi }^{2}r/T^2$. Kepler's third law, $T^2\propto r^3$, therefore implies an acceleration proportional to $r^{-2}$. This circular argument is illustrative; Isaac Newton's treatment of elliptical orbits is more general. With Newton's law of universal gravitation,the two-body orbital period is tied to the semimajor axis $a$ and both masses. The same theory connects the Moon's orbital acceleration with terrestrial gravity, and supports further investigations of tides, comets and departures from a simple two-body orbit.^[ref]

This achievement depended on more than two individuals. Johannes Kepler's relations supplied crucial astronomical constraints, and both authors inherited geometry, observational records and technical practices. Nor should the Philosophiae Naturalis Principia Mathematica by Isaac Newton be described simply as the first application of symbolic calculus: its published arguments are largely geometrical. What distinguishes Isaac Newton is the demonstrative network uniting terrestrial and celestial problems, together with explicit idealizations whose failures generated further calculable questions.

On breadth, quantitative unification and the inference from motion to force, Isaac Newton's contribution was greater. Galileo Galilei's work supplied an exemplary way to mathematize terrestrial motion and test its idealized relations. Giving the former the overall verdict should therefore acknowledge the latter's distinct foundational achievement, not portray the Philosophiae Naturalis Principia Mathematica by Isaac Newton as a creation without predecessors.

With the proton treated as fixed and electron mass $m$, the classical energy and angular momentum areFor a circular orbit, $L=mrv$ and Coulomb's law determines its centripetal acceleration:Eliminating $v$ gives $r=4\pi {ϵ}_0{n}^2{\hslash }^2/(m{e}^2)$. Thus the Bohr radius and orbit radii areThe kinetic energy is half the magnitude of the attractive potential energy. HenceThis is the circular-orbit Bohr model; including proton recoil replaces $m$ by the reduced mass.

Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.

The area element in plane polar coordinates is $rdrd\theta$. Thus the axial moment of inertia isFor the removed material, the angular width is $2\pi /13$, andIts moment of inertia is therefore $I_{\mathrm{cut}}=\pi {\rho }_0{a}^5/160$, leavingThe torque equation $I\dot{\Omega }=\tau$ gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is

To find the centre of mass after removing material, let ${\mathbf{e}}_1$ point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along ${\mathbf{e}}_1$ isbecause the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\mathrm{cut}}{\mathbf{e}}_1$. Dividing by the given mass $k{\rho }_0{a}^3$ gives the body-fixed centre of massIt lies away from the missing sector. In the inertial frame, let $t_{\ast }$ be the moment the applied torque stops and let $\psi ={\psi }_{\ast }+\Omega (t-t_{\ast })$ give the orientation of that sector's bisector. Then, with $d=11a\sin (\pi /13)/(96k)$,The centre of mass has inward centripetal acceleration of magnitude ${\Omega }^{2}d$. The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is $k{\rho }_0{a}^3\ddot{\mathbf{R}}$; a zero applied axial torque does not imply a zero resultant force.

On this circle the radial coordinate is constant, $r=2$, so $\dot{r}=\ddot{r}=0$. The velocity is a tangent vector to the circle in the anticlockwise direction, represented by $i{e}^{i\theta }$ when $\dot{\theta }>0$. Substituting into part (b),The inward component is the centripetal acceleration; the transverse component allows the speed to change. Anticlockwise motion does not imply uniform speed or $\ddot{\theta }=0$. The preceding figure includes the required tangent arrow.

For point masses $m_1,m_2$ separated by distance $r$, Newton's law of universal gravitation gives an attractive central force of magnitudeThe vector force on body 2 is $-G{m}_1{m}_2({\mathbf{r}}_2-{\mathbf{r}}_1)/|{\mathbf{r}}_2-{\mathbf{r}}_1{|}^3$. For extended bodies one adds their elemental interactions. The shell theorem explains why a spherically symmetric body acts externally like its total mass placed at its centre.

The law has experimental and astronomical support, rather than a deductive proof from definitions. Kepler's laws provide one test. For a circular orbit of radius $r$ and period $T$, the required centripetal acceleration is $4{\pi }^{2}r/T^2$. With $T^2\propto r^3$, it varies as $r^{-2}$. The full elliptical-orbit analysis, with the law of areas identifying a central interaction, gives the same inverse-square dependence and predicts the orbit's form.

The Earth–Moon comparison connects celestial motion to terrestrial falling. Using illustrative mean values $R_{\oplus }\simeq 6.37\times 10^6\mathrm{m}$, $r_M\simeq 3.84\times 10^8\mathrm{m}$ and $g_0\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$, the inverse-square prediction isThe circular estimate $4{\pi }^2{r}_M/T_M^2$, with $T_M\simeq 27.3$ days, is about $2.72\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-2}$. Their agreement is the relevant check; the numbers are approximate mean-orbit values, not exact observations of a circular Moon.

The Cavendish experiment adds a laboratory test. Small suspended masses twist a fine fibre when larger source masses are moved nearby. Comparing the torque with the suspension's calibrated restoring response measures the weak attraction between laboratory bodies. Reversing source positions helps separate the gravitational signal from a persistent instrumental offset. Planetary perturbations and additional laboratory geometries provide further tests of the common law. Its excellent accuracy in ordinary weak-field conditions does not make it an exact theory in every regime.

The magnitude of the gravitational force remains approximately $mg$ during a laboratory swing. Its tangential component changes with position, however. For simple pendulum length $\ell$ and angle $\theta$ measured from the downward vertical,This restoring component is zero at the lowest point and largest in magnitude at the turning points. It always accelerates the bob towards the lowest point. The radial gravitational component is $mg\cos \theta$; the string tension $S$ must additionally supply the centripetal acceleration, so $S-mg\cos \theta =m{v}^2/\ell$. Thus it is the radial resultant, not gravity alone, that bends the trajectory. The bob has maximum speed at the bottom even though its tangential acceleration is then zero.

For a small angular amplitude, the small-angle approximation gives simple harmonic motion with oscillation period $T=2\pi \sqrt{\ell /g}$. Measure $\ell$ from pivot to the bob's centre of mass, time many complete swings, divide the elapsed time by their number, and calculateRepeat with several lengths and plot $T^2$ against $\ell$; the ideal slope is $4{\pi }^2/g$. The first-order relative uncertainty obeys $\delta g/g\simeq \delta \ell /\ell -2\delta T/T$, with absolute uncertainty contributions combined according to their dependence. Keep the amplitude small: the finite-amplitude period is $T=T_0(1+{\theta }_0^2/16+\cdots )$, with ${\theta }_0$ in radians. Using the longer finite-amplitude period in the small-amplitude formula would underestimate $g$. Pivot friction, a finite bob and support motion are further experimental corrections.^[ref]

A geostationary orbit has the same angular velocity as the Earth. Let $M_E=4\pi \rho R^3/3$ and $r_g=\eta R$. Balancing the required centripetal acceleration with the exterior gravitational field givesThe orbital plane must be equatorial and the motion prograde for the satellite to remain over one point. The uniform density assumption supplies $M_E$ in terms of the specified $\rho$ and $R$.

Treat the carriage as translating along the circular track, neglecting wheel rotational kinetic energy and its finite size. At the top, gravity and the inward normal force supply the centripetal acceleration:For unilateral contact, $N\ge 0$, so the limiting speed is $v_{\mathrm{top}}^2=gr$. Conservation of energy from rest at height $h$ then givesThis top contact condition is the most restrictive: at angle $\varphi$ measured from the bottom, $N/(mg)=2h/r-2+3\cos \varphi$, whose minimum occurs at the top. A mechanically captive carriage able to sustain an outward rail reaction has a different constraint; the result here uses the usual requirement of positive normal contact.

Neglecting the planets' masses compared with the Sun's mass, Newton's law of universal gravitation supplies the centripetal acceleration of each circular orbit:Thus Kepler's third law, calibrated by the Earth's orbital period of one year and orbital radius of one astronomical unit, givesBoth orbits are taken to have the same direction. Their relative angular speed is $2\pi (1/T_J-1/T_S)$, so the Sun-centred interval between successive conjunctions is the synodic periodUsing the supplied rounded orbital radii explains the small difference from the commonly quoted approximately twenty-year interval.

From the Earth, the two lines of sight are displaced differently because Jupiter and Saturn are at different distances. The Earth's annual motion along its orbit therefore shifts the dates of apparent conjunctions by amounts of order months. The broad recurrence remains about twenty years, but successive apparent conjunctions need not be equally spaced: apparent retrograde motion can produce a group of three conjunctions separated by months during one encounter. The Earth's daily rotation adds a much smaller change of viewpoint. This is a geometrical comparison in the stipulated coplanar model; daylight and the horizon can prevent a particular conjunction from being visible at a chosen surface location.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

Let $N$ be circumferential tension. A small arc subtending $d\theta$ has inward resultant $Nd\theta$ and mass $\rho Ard\theta$. Its centripetal acceleration gives $N=\rho A{\omega }^2{r}^2$. Circumferential stress is $N/A$ and the strain is $\Delta (2\pi r)/(2\pi r)=\Delta r/r$. Thus the centrifugal strain of a thin elastic hoop isUse the unstrained radius and cross-section at this order. The approximation requires $\rho {\omega }^2{r}^2/E\ll 1$, a slender hoop and linear elastic response; the area cancels.

A stationary person relative to the rim has centripetal acceleration $a{\omega }^2=g$. If $u$ is a signed walking speed relative to the rim, the inertial tangential speed is $a\omega +u$, so the inward normal reaction satisfiesConsequently the support force per mass, and hence apparent weight, is $4g$ for $u=+a\omega$ and zero for $u=-a\omega$. In the latter case the walker has zero inertial speed and needs no centripetal support; ordinary weight-bearing contact cannot be assumed to persist without a restraint. For artificial gravity in a rotating ring, the centrifugal force and the Coriolis force alone are insufficient: the walker also has relative-path centripetal acceleration $u^2/a$. Omitting it would give the wrong apparent weights at these large walking speeds.

A horizontal pressure gradient force accelerates air from higher toward lower pressure. In a rotating frame the Coriolis force deflects the flow. Away from strong friction, at large spatial scales and small Rossby number, a slowly changing flow can reach geostrophic balance:where $f=2\Omega \sin \phi$ is the Coriolis parameter. The geostrophic wind runs along atmospheric isobars, with speed $|{\nabla }_{h}p|/(\rho |f|)$. Closely spaced atmospheric isobars therefore indicate stronger geostrophic wind when density and latitude are comparable. In the northern hemisphere lower pressure lies to the left of this wind, producing counterclockwise circulation around a low; the sense reverses in the southern hemisphere.

A curved flow also has centripetal acceleration. For cyclonic circular flow in the northern hemisphere, the gradient wind relation isNear the ground, friction reduces wind speed and weakens its Coriolis contribution, so air crosses atmospheric isobars toward lower pressure. The simple geostrophic formula fails near the equator, where $f$ becomes small, and where rapid changes or tight curvature make neglected acceleration important. Vertically, large-scale motion is often approximated by hydrostatic balance, $dp/dz=-\rho g$.

Bernoulli's principle can relate speed changes to pressure changes along an appropriate inviscid streamline, but its simple dynamic-pressure relation is not a replacement for the rotating force balance governing a broad atmospheric pressure map.

The radial force is $F_r=-dU/dr=-3A{r}^2$. A nontrivial circular orbit therefore requires $A>0$, so that the force is inward. Balancing its magnitude against the required centripetal acceleration givesThusThis is the circular-orbit energy ratio in a power-law potential, $2K=r{U}^{\prime }(r)=sU$ for $U=A{r}^s$, with $s=3$. It uses the zero of potential energy specified by $U=A{r}^3$; adding a constant would change the ratio without changing the motion. For $A<0$, the force is outward and no such free circular orbit exists.

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.