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Changing basis and acceleration. Attach orthonormal unit vectors ${\mathbf{e}}_i(t)$ to the rotating frame, with a common origin fixed in the inertial frame. An infinitesimal rotation $d\mathbf{\theta }=\mathbf{\omega }dt$ changes a vector fixed in those axes by $d\mathbf{A}=d\mathbf{\theta }\times \mathbf{A}$: its tip moves tangentially to the circle about the rotation axis. More generally, write $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$. Differentiating the components and the rotating basis separately proves the rotating-frame derivative formula,

Apply the formula first to position: ${\dot{\mathbf{r}}}_0=\mathbf{v}+\mathbf{\omega }\times \mathbf{r}$, where $\mathbf{v}=(d\mathbf{r}/dt{)}_S$. Apply it again to both terms. For constant $\mathbf{\omega }$,One cross term comes from differentiating $\mathbf{v}$ in the inertial frame and another from differentiating $\mathbf{r}$ in $\mathbf{\omega }\times \mathbf{r}$. The displayed source formula implicitly requires constant angular velocity and an unaccelerated common origin. For changing rotation rate there is also $\dot{\mathbf{\omega }}\times \mathbf{r}$; an accelerating origin supplies its own translational acceleration. Moving the rotation terms to the force side gives the apparent Coriolis force and centrifugal force with negative signs.

Southward motion on Earth. Use local east, north and up axes, with latitude $\lambda$ positive northward. Then $\mathbf{\Omega }=\Omega (0,\cos \lambda ,\sin \lambda )$ and $\mathbf{v}=(0,-v,0)$. The Coriolis acceleration in local Earth coordinates givesIt is westward in the northern hemisphere, eastward in the southern hemisphere and zero for exactly southward motion at the equator. There is no vertical Coriolis component for this particular velocity.

Correct steering. On the desired meridional track, drag has no eastward component because it is antiparallel to the velocity. If the thrust makes signed angle $\varphi$ east of south, its eastward component is $T\sin \varphi$. The cross-track force balance is thereforesoThe argument must have magnitude at most one for this simple thrust-only correction to be feasible. The signed angle changes sign at the equator: correction is east of south in the north and west of south in the south. Thus the phrase “east of south” cannot describe a positive correction throughout the entire flight.

Wrong-sign correction and the miss distance. A literally constant velocity vector cannot develop a cross-track displacement. The physically useful interpretation of the last task is constant speed, with an erroneous lateral steering correction. In the usual leading-track model, thrust and drag balance along the actual motion, and the applied transverse correction has the correct magnitude but the wrong sign. It therefore reinforces rather than cancels the Coriolis force. Starting at the equator, use the unperturbed southward track $\lambda \simeq -vt/R$, and let $x_E$ be the eastward departure from that track. To first order in the departure,There is a factor two relative to uncompensated Coriolis deflection because the wrong correction adds an equal lateral force. Integration gives the wrong-sign Coriolis steering error,For southern angular distance $\Lambda =16\pi /180=0.27925$, $R=6.4\times 10^6\mathrm{m}$ and $v=800/3.6=222.2\mathrm{m}{\mathrm{s}}^{-1}$, the nominal flight time after the equator isTaking $\Omega \simeq 2\pi /(86400\mathrm{s})$ givesThe small-latitude approximation $\sin (vt/R)\simeq vt/R$ gives ${\ddot{x}}_E\simeq 4\Omega v^{2}t/R$ and $x_E\simeq 2\Omega v^2{t}^3/(3R)$, or $1.95\times 10^5\mathrm{m}$, confirming the scale and direction.

This is an estimate under an explicit steering model, not an exact trajectory forced by the wording. For example, retaining spherical transport of the local axes while still linearizing in the cross-track motion gives, with $s=-\lambda$, $\dot{u}-(v/R)u\tan s=4\Omega v\sin s$. Consequently $u\cos s=2\Omega R{\sin }^{2}s$ and the longitude departure is $\Delta \ell =2\Omega R(\tan \Lambda -\Lambda )/v$. The distance along the destination parallel is then $R\cos \Lambda \Delta \ell \simeq 1.93\times 10^5\mathrm{m}$. Corrections from the changed track and a more detailed control law are not fixed by the problem. If “keeps the velocity constant” instead means the full vector, the pilot would need an additional compensating force and there would be no miss; if thrust is kept at a specified geographical bearing despite drift, its along-track component and lateral drag must also be modelled.

Let a coordinate frame rotate with angular velocity $\mathbf{\Omega }$ relative to an inertial frame. For any vector $\mathbf{u}$,Applying this twice to position, with coincident nonaccelerating origins, givesNewton's second law written using rotating coordinates therefore contains the Coriolis forceThe other inertial corrections are the centrifugal force and, for changing rotation rate, the Euler force. An accelerating origin introduces an additional translational fictitious force.

The Coriolis force depends on velocity and is perpendicular to it, so ${\mathbf{F}}_C\cdot {\mathbf{v}}_{\mathrm{rotating}}=0$: the Coriolis force does no work in the rotating frame. It bends trajectories without directly changing their rotating-frame speed. It is absent for motion parallel to the rotation axis.

For locally horizontal motion on Earth at latitude $\lambda$, take $x$ east, $y$ north and $z$ up. Keeping the vertical component $\Omega \sin \lambda$ givesHorizontal motion is deflected right in the northern hemisphere and left in the southern hemisphere. In the ideal absence of other horizontal forces these equations give inertial circles with angular frequency $|f|$. The same velocity-dependent correction explains the precession of a Foucault pendulum, whose precession-rate magnitude is $\Omega |\sin \lambda |$. Terrestrial vertical motions also involve Earth's horizontal rotation component, so the reduced horizontal equations are not the full force law.

Let the rotating frame have origin acceleration ${\mathbf{A}}_0$ and angular velocity $\mathbf{\Omega }(t)$ relative to an inertial frame. For a vector referred to the moving axes, the transport rule isApply it twice to position. With rotating-frame velocity $\mathbf{v}$ and acceleration ${\mathbf{a}}_R$,Thus Newton's equation expressed in the rotating frame isThe added terms are translational, Coriolis, Euler and centrifugal fictitious forces. They reflect the frame's acceleration, rather than additional interactions with nearby objects.

The centrifugal force points outward from the rotation axis and, for constant $\mathbf{\Omega }$, has potential $-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. The Coriolis force is perpendicular to velocity, so does no work on the particle measured in that rotating frame; it changes direction rather than speed directly. The Euler force is present when the frame's rotation rate or direction changes.

On Earth, gravity and centrifugal acceleration combine into effective gravity. For slow horizontal motion at latitude $\lambda$, the vertical component of Earth's rotation gives Coriolis parameter $f=2\Omega \sin \lambda$: an initially northward motion deflects eastward in the northern hemisphere. This supplies a useful local explanation of rotating flows and Foucault precession. A bucket of rotating liquid illustrates the centrifugal potential: its steady free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$, giving $z={\Omega }^2{r}^2/(2g)+\mathrm{constant}$. These equations require a declared frame; adding a centrifugal force while retaining inertial-frame acceleration would count the same kinematic effect twice.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

Signed radial balance. Let $r$ increase outward from the vortex centre and $U$ be signed azimuthal velocity, positive anticlockwise. At $45^{\circ }$N, $f=2{\Omega }_{\oplus }\sin 45^{\circ }>0$. The positive speed is $V=|U|$, $f$ is the Coriolis parameter, the density is $\rho$, and $p_r=\partial p/\partial r$. In circular motion the inward centripetal acceleration is $-U^2/r$, while the outward radial acceleration from the Coriolis force is $fU$. The pressure gradient acceleration is $-p_r/\rho$. HenceThe term $U^2/r$ is an acceleration, not an additional physical force in the rotating-frame momentum equation. It can be moved to the force-balance side and depicted as outward centrifugal force in a frame following the circular parcel.

For a cyclone, $U=+V$ and $p_r/\rho =V^2/r+fV>0$: pressure increases outward and the centre is a low. Its inward pressure force balances the outward Coriolis and centrifugal terms. For a conventional anticyclone, $U=-V$ and $G=-p_r/\rho >0$; the balance isThe inward Coriolis force balances the outward pressure and centrifugal terms, and pressure has a maximum at the centre.

Anticyclonic branches and near-centre regularity. The anticyclonic gradient-wind branches follow from the quadratic balance. Solving givesThe minus root is the normal branch connected continuously to geostrophic flow for small forcing. The plus root is an anomalous high-speed branch and must not be discarded algebraically, even though it often fails the stability conditions relevant to large-scale anticyclones. Reality requiresA finite smooth central pressure therefore has zero radial gradient at the origin, with $-p_r/\rho$ no larger than order $r$. For $p=p_c-\rho C{r}^2/2+\cdots$, the limiting curvature obeys $C\le f^2/4$. A pressure maximum with a finite nonzero limiting outward pressure force cannot support the required circular anticyclonic balance arbitrarily close to its centre.

For positive $\mathrm{Ro}=V/(fr)$ and geostrophic magnitudes,The cyclonic gradient wind is subgeostrophic and the normal anticyclonic wind supergeostrophic. On the latter branch $0\le \mathrm{Ro}\le 1/2$. At a point on the eastern side of a cyclone, geostrophic and actual winds point north but the actual arrow is shorter; the ageostrophic difference points south. On the eastern side of a normal anticyclone both geostrophic and actual winds point south, with the actual arrow longer and its ageostrophic addition also south.

At lower latitude, the same $V$ and radius give a larger Rossby number and larger differences from geostrophic wind. The contour topology remains a central low or a conventional central high, but for fixed $V,r$ the required radial pressure gradients change with $f$: the cyclonic gradient is $\rho (V^2/r+fV)$ and the anticyclonic outward pressure force is $fV-V^2/r$. The latter conventional branch cannot be maintained once the reduced Coriolis contribution is too weak. Actual spatial gradients, rather than an arbitrary spacing of contour values, determine how tightly drawn contours should be.

The high-centre conclusion is conditional on the conventional anticyclonic branch. Gradient balance alone also allows a sufficiently rapid clockwise circulation about a low: if $V>fr$, then $p_r/\rho =V^2/r-fV>0$. For example, $V=2fr$ gives $p_r/\rho =2f^{2}r>0$. Such a configuration is outside the usual geostrophic anticyclone regime. Tornado-scale vortices have very large Rossby number, so the dominant balance is nearly cyclostrophic, $p_r/\rho \simeq V^2/r>0$. Their centres are low pressure irrespective of the sign of rotation; low pressure supplies the inward acceleration needed to bend the fast flow.

Suppress primes on the angular velocity while retaining rotating components. The canonical momentum is $\partial L^{\prime }/\partial {\dot{\mathbf{r}}}^{\prime }=m({\dot{\mathbf{r}}}^{\prime }+\mathbf{\Omega }\times {\mathbf{r}}^{\prime })$. Varying $L^{\prime }$ with respect to position and applying the Euler-Lagrange equations givesThe terms are the physical force, the velocity-dependent Coriolis force, the outward centrifugal force, and the Euler force caused by changing angular velocity. The common origin is inertial; an accelerating origin would introduce another translational apparent force.

At latitude $\ell$, the Earth's angular velocity in east–north–up coordinates is $\mathbf{\Omega }=(0,\Omega \cos \ell ,\Omega \sin \ell )$. Assume a small horizontal patch, constant $\Omega$, negligible vertical velocity, and a vertical constraint balancing effective gravity. Absorb the steady centrifugal force into that effective gravity. With $f=2\Omega \sin \ell$ and physical horizontal force per unit mass $F_x^{\prime },F_y^{\prime }$, the Coriolis force givesLarge-scale atmospheric and oceanic flows deflect to their right in the northern hemisphere and left in the southern hemisphere; their balance with a horizontal pressure gradient produces geostrophic flow. A Foucault pendulum has an oscillation plane that precesses relative to the ground at angular rate $-\Omega \sin \ell$. Both effects vanish at the equator within this horizontal approximation.

Put $X={\omega }^2$ and $S={\omega }_1^2+{\omega }_2^2+4{\Omega }^2$. Its equation is $X^2-SX+{\omega }_1^2{\omega }_2^2=0$. If both effective stiffnesses are positive, the sum and product of roots are positive and $S^2-4{\omega }_1^2{\omega }_2^2\ge 0$, so both roots are positive. They can coincide in a degenerate zero-rotation isotropic case; “two” need not mean distinct.

If ${\omega }_1^2<0<{\omega }_2^2$, the product is negative, so one root is negative. Its imaginary frequency gives a growing exponential displacement: centrifugal softening has overwhelmed one elastic restoring direction, making the central configuration unstable. Coriolis coupling mixes the two coordinates but does not remove this negative-product instability.

For a rigid frame with constant angular velocity $\mathbf{\Omega }$, add the Coriolis force and centrifugal force to the inviscid magnetohydrodynamic momentum equation. A constant magnetic permeability and solenoidal magnetic-field constraint split the Lorentz force density into magnetic pressure and magnetic tension. This nonrelativistic conducting-fluid approximation neglects the displacement current, net electric-charge force and viscosity. Differential rotation belongs to the fluid's velocity field, not to a spatially varying rigid-frame rotation vector. A time-dependent frame requires the additional Euler force.

A rotating reference frame uses spatial axes whose orientation changes relative to an inertial frame. Coordinates in those axes can simplify an apparatus rotating at a known angular velocity $\mathbf{\Omega }$, but the equations of motion acquire inertial terms because the basis vectors themselves change.

For a shared fixed origin and constant $\mathbf{\Omega }$, a particle at rotating-frame position $\mathbf{r}$ and velocity $\mathbf{u}$ satisfiesThe additional terms are the Coriolis force and centrifugal force. A time-dependent $\mathbf{\Omega }$ adds an Euler term $-m\dot{\mathbf{\Omega }}\times \mathbf{r}$; a translating origin can add another term. The rotating-axes theorem derives these contributions by differentiating the position twice, rather than treating them as new interactions with another body.^[ref]