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The method of images replaces induced charges or currents on a boundary by fictitious sources outside the physical region, chosen to satisfy that boundary's conditions. For example, a point charge $q$ a height $d$ above a grounded conducting plane has an image $-q$ at the reflected point. The sum has zero potential on the plane, and the electrostatic uniqueness theorem identifies it with the physical solution in the upper half-space. Images do not describe a second physical charge below the metal.

For the superconducting plane, use the different boundary condition $B_z=0$ at its surface: magnetic flux cannot enter the ideal bulk. Choose the real ring moment $\mathbf{m}=m\hat{z}$, where $m=I\pi R^2$, at $z=h$. Its image has moment $-m\hat{z}$ at $z=-h$. This superconducting plane image dipole makes the normal fields cancel at $z=0$, while their tangential fields add. The image is a computational continuation; the physical field inside the plane is zero. The original sketch shows the dipole-scale field, appropriate to $h\gg R$; the fine field close to the ring wire is outside that approximation.

In the vacuum region the magnetic scalar potential isAt the surface $z=0$, it reduces to $-mh/[2\pi ({\rho }^2+h^2{)}^{3/2}]$. Its tangential derivative therefore givesBoth $B_z$ and $B_{\varphi }$ vanish there; the sign is inward radially for the chosen positive moment. The expression is the leading dipole result, not an exact finite-ring formula at arbitrary $R/h$.

For the force, calculate the external image field on the axis near the real dipole:Differentiate this field with the image location held fixed, then evaluate at $z=h$. The dipole levitation force above a superconducting plane isdirected upward. Differentiating a substituted separation $2h$ while moving both dipoles in the force derivative would incorrectly double the result.

To levitate ring mass $M$, balance this force against $Mg$:A light, thin wire and the smallest practical gap reduce current. At fixed gap and wire cross-section $s$, $M=2\pi Rs{\rho }_w$ gives $I_{\mathrm{lev}}\propto h^2\sqrt{s{\rho }_w}/R^{3/2}$, favouring a larger ring until the dipole-limit constraint on $R/h$ matters. There is no unique least current without geometric and material constraints. As a geometrically consistent order-of-magnitude example, take $R=1\mathrm{mm}$, $h=10\mathrm{mm}$, wire radius $50\mu \mathrm{m}$ and an assumed metallic-scale density ${\rho }_w=6.0\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Then $M=2.96\times 10^{-7}\mathrm{kg}$ and $I_{\mathrm{lev}}\simeq 2.8\times 10^2\mathrm{A}$. The large value reflects the weak $h^{-4}$ force in the stipulated far-field geometry. Uniformly shrinking this geometrical example reduces current as size to the power $3/2$, but increases required electric current density; arbitrarily thin wire is therefore not a free optimization.

This balance estimate is not a verified superconducting design. The actual wire must carry that current without quenching, its self-field counts toward the limit, the plane must remain in its assumed flux-excluding state, and leads or supports add load. Transport critical-current measurements specify temperature, applied field and the voltage criterion rather than a material-independent limiting current.^[ref]

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

For a central potential, angular momentum is conserved and the motion lies in a plane. With $L=m{r}^2\dot{\theta }$, radial energy conservation becomesAllowed radii have $\mathcal{E}\ge V_{\mathrm{eff}}$; equality gives radial turning points. A well with two finite turning points gives regular radially bound motion. Access to arbitrarily large radii gives an unbound branch, while access to the origin can lead to collision in a singular potential. Stationary points are circular-orbit candidates; a strict local minimum gives radial stability. For a potential tending to zero at infinity, energy relative to that asymptote is useful, but a singular inner potential and centrifugal barrier must still be examined.

Integrating the angular-momentum relation with the origin of angle at $t=0$ givesFor $V=-A/r+B/r^2$ at $L=L_0$, the effective potential is $-A/r+(L_0^2+2m{B}_0)/(2m{r}^2)$. It is exactly the Kepler effective potential with $L_{\mathrm{K}}=\sqrt{L_0^2+2m{B}_0}$, provided that expression is real. Equal radial energy and initial radial phase therefore give identical $r(t)$, although the angular motion differs. Writing $\psi$ for the Kepler angle, $\dot{\theta }/\dot{\psi }=L_0/L_{\mathrm{K}}=k$. With matched periapsis origins,Here $r_0=L_{\mathrm{K}}^2/(mA)$ and the eccentricity is fixed by radial energy. For $B_0>0$, $k<1$ and successive periapses are separated by $2\pi k$, so there is retrograde apsidal motion relative to a closed Kepler ellipse. The radial equivalence is the relevant content of the Newton's theorem of revolving orbits.

For the pure inverse-cube central force set $A=0$. The free-particle reference orbit at $B=0$ is a straight line, $r=d/\cos (\psi -{\psi }_0)$, not a finite-$A$ conic with its old $r_0$ held fixed. To cover attractive as well as repulsive cases completely, putwhere primes denote angle derivatives and $L\ne 0$. The inverse-cube central orbit classification gives the following physically positive-radius branches.

For $\lambda >0$, write $u=C\cos [\sqrt{\lambda }(\theta -{\theta }_0)]$ with $C>0$, so $r=C^{-1}\sec [\sqrt{\lambda }(\theta -{\theta }_0)]$. These are scattering orbits with positive energy and an inner turning radius $1/C$; the angular interval between the two infinite-radius ends is $\pi /\sqrt{\lambda }$. Repulsion ($B>0$) gives an interval smaller than $\pi$; weak attraction ($-L^2/(2m)<B<0$) gives a larger interval and can cause winding. The $B=0$ limit is the straight line.

For $\lambda =0$, $u=C+D(\theta -{\theta }_0)$. If $D\ne 0$, reciprocal-linear spirals connect an infinite-radius end to collision at infinite accumulated angle. If $D=0$ and $C>0$, every radius admits a circular solution at zero energy: attraction exactly balances centrifugal acceleration. Each is a marginal circular orbit under an inverse-cube force, with no radial restoring force.

For $\lambda <0$, put $\beta =\sqrt{-\lambda }$ and $u=C_{+}{e}^{\beta \theta }+C_{-}{e}^{-\beta \theta }$. Up to angle shifts and reversal, the three energy classes arewith positive-radius domains understood. The negative-energy branch is radially bounded but reaches the singular centre on either side of its outer turning point; it is not a regular closed orbit. Collision requires additional physics to continue through $r=0$. Zero angular momentum gives radial trajectories and is excluded from the angular parametrization above.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

In the thin-gap Couette flow approximation, the tangential velocity varies linearly through the gap. At radius $r$, its difference between the two discs is $r\omega$, so a Newtonian fluid exerts shear stressAn annulus has area $2\pi rdr$, and its tangential force acts at lever arm $r$. Integrating its torque givesThe driving torque is in the direction of rotation; the liquid's resisting torque has the opposite sign. This parallel-disc viscometer result assumes laminar shear with no slip, negligible edge effects and negligible secondary circulation. Its fourth-power radius dependence comes from the combined radial dependence of speed, annular area and lever arm.

An electric dipole consists of equal and opposite electric charges separated by a displacement $\mathbf{d}$. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field areThere is no monopole term because the net electric charge is zero. The axial electric field points along $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$, while the equatorial electric field points oppositely with half that magnitude.

A uniform applied electric field exerts zero net force on an ideal fixed electric dipole, but it exerts torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$. Its orientation-dependent potential energy is $U=-\mathbf{p}\cdot \mathbf{E}$, so alignment with the electric field is stable. A nonuniform applied electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment.

A permanent electric dipole moment exists without an applied electric field. An induced electric dipole moment instead develops in response to the local electric field; in an isotropic linear model $\mathbf{p}=\alpha \mathbf{E}$. Creating that induced moment requires accounting for the work of polarization, so its mechanical potential energy in a specified applied electric field is $-\alpha E^2/2$, rather than $-\alpha E^2$. In matter, the dipole moment per unit volume is the polarization density $\mathbf{P}$.

Coupled oscillators have more than one collective pattern of motion. For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and linked to each other by stiffness $\kappa$, small displacements obeyHere equal linear damping $b$ is included to make finite steady-state responses at resonance. The normal mode coordinates $q_{+}=(x_1+x_2)/\sqrt{2}$ and $q_{-}=(x_1-x_2)/\sqrt{2}$ diagonalize these equations. Their natural angular frequencies areThe symmetric normal mode leaves the coupling spring unextended; the antisymmetric normal mode stretches it and therefore has the higher frequency. With $F_{\pm }=(F_1\pm F_2)/\sqrt{2}$, their complex steady-state responses areThus the drive's spatial pattern matters as well as its frequency. Equal in-phase forces drive only the symmetric normal mode; opposite forces drive only the antisymmetric one; driving a single mass generally excites both. Finite damping broadens the resonance peaks and produces a phase lag. In the ideal undamped system, a resonant drive with nonzero projection onto a normal mode produces secular growth rather than a bounded resonant steady state.

The displacement of a particular mass adds contributions from both normal modes. They can cancel at an antiresonance. For $F_2=0$ and no damping, $x_1$ vanishes at ${\Omega }^2=(k+\kappa )/m$, while the partner still moves. This differs from the two nearby resonances, where a mode responds strongly; a small measured response need not mean that no motion or stored energy exists elsewhere.

A Newtonian fluid exerts viscous shear stress proportional to the rate of strain. For an incompressible fluid of constant dynamic viscosity $\eta$,The local momentum balance is the Navier-Stokes equation:The left side is density times material acceleration. A steady velocity field can still have convective acceleration when a moving parcel changes speed or direction; setting $\partial \mathbf{u}/\partial t=0$ does not eliminate it.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscous effects. At low Reynolds number, the fluid's own inertia can often be neglected, giving Stokes flow. This does not say that every immersed body has zero acceleration: its mass and the net pressure, viscous and body forces still enter its own equation of motion.

For a small rigid sphere moving through a slowly varying viscous background, Stokes drag is $-6\pi \eta a(\mathbf{v}-\mathbf{u})$. In still fluid, a simple settling model includes weight, buoyancy and drag:Ignoring fluid added mass and unsteady history forces, this gives relaxation time $m/(6\pi \eta a)$ and terminal velocityAt the terminal velocity the net force and acceleration vanish, while viscous drag and gravity remain individually nonzero. The dissipated power $6\pi \eta a{v}_t^2$ is supplied by the loss of gravitational potential energy relative to buoyant support. The creeping-flow and quasisteady-drag assumptions must be checked before applying this expression at large speed or immediately after a rapid start.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

Images and uniqueness. The method of images replaces a conducting boundary by fictitious charges outside the physical field region. Choose them so that the resulting electric potential satisfies the required conductor potentials. The Uniqueness of the Dirichlet problem then guarantees the same physical exterior field. For example, a point charge $q$ at height $h$ above a grounded plane has an image $-q$ at height $-h$: the plane is an equipotential at zero, and the charge is attracted toward it by $q^2/(16\pi {\epsilon }_0{h}^2)$. The image is a computational device; the physical conductor carries a continuous induced surface charge density.

Potential of a line. Surround an infinite line charge $\lambda$ by a coaxial Gaussian cylinder of radius $r$ and length $L$. Gauss's law gives $E_r(2\pi rL)=\lambda L/{\epsilon }_0$, hence $E_r=\lambda /(2\pi {\epsilon }_{0}r)$. Integrating $E_r=-dV/dr$ givesWriting an additive constant instead is equivalent. The reference length makes the logarithm dimensionless; an isolated infinite line cannot be assigned a finite zero potential at infinity.

Locating the two lines. Put the grounded plane at $x=0$ and the physical cylinder at $(x,y)=(d,0)$, with axis along $z$ and radius $a<d$. Set $c=\sqrt{d^2-a^2}$, and place line charges $+\lambda$ and $-\lambda$ at $(c,0)$ and $(-c,0)$. If $R_{+}$ and $R_{-}$ are their distances from an observation point, the cylinder-plane image charges giveOn $x=0$, $R_{-}=R_{+}$, so $V=0$. To check the cylinder, define $q=(d+c)/a$. The level set $R_{-}=q{R}_{+}$ is a circle: completing the square givesSince $d^2-c^2=a^2$, its centre is $d$ and radius is $a$. The positive line lies inside the cylinder; the negative line lies behind the plane. There is consequently no fictitious charge in the exterior physical region. A Gaussian contour around the cylinder encloses $+\lambda$ only, so the required physical charge per unit length is also correct. These boundary conditions and the far-field behaviour determine the physical solution by Uniqueness of the Dirichlet problem.

Cylinder voltage. The constant distance ratio on the cylinder immediately givesIts capacitance per unit length is $C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$. In the thin-cable limit, $\mathrm{arcosh}(d/a)\simeq \ln (2d/a)$.

Cable attraction. Because the cable is idealized as infinite, the meaningful answer is force per unit length. For $a=0.020\mathrm{m}$, centre height $d\simeq 10\mathrm{m}$ and $V_c=1.1\times 10^4\mathrm{V}$,The electrostatic attraction per unit length of a cylinder above a plane isFor example, this follows without any ambiguity about moving image positions from $F_d=\frac{1}{2}{V}_c^{2}d{C}^{\prime }/dd$ at fixed voltage, including the work of the voltage source. Its sign is negative because capacitance decreases with height. A cable length $L$ experiences approximately $fL$ downward when end effects are negligible. This is the electrostatic contribution only.

Field law and centre value. The Biot-Savart law for a steady thin wire isHere $d{\mathbf{\ell }}^{\prime }$ follows the conventional electric current and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the right-hand rule. Then

On-axis integration. At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing ${\mathbf{r}}^{\prime }=(a\cos \phi ,a\sin \phi ,0)$ gives axial cross-product element $a^{2}d\phi$, so the on-axis magnetic field of a circular current loop isIt points the same way on the two halves of the axis; its strength is even in $z$.

Directions near the axial point. At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete magnetic field lines return around the outside of the loop and form closed curves.

Radial component from zero divergence. Smooth axial symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$ and $B_{\rho }=O(\rho )$. By Gauss's law for magnetism,Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the near-axis radial magnetic field, $B_{\rho }=-\rho B_0^{\prime }(z)/2+O({\rho }^3)$. Differentiating the axial field therefore givesFor $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.

Force on the second loop. Put the second loop at $z=d$ and radius $\rho =a$. For $d\gg a$, the radial field due to the first loop is $B_{\rho }\simeq 3{\mu }_{0}I{a}^3/(4d^4)$. The element $d\mathbf{\ell }=ad\phi \hat{\mathbf{\phi }}$ experiences Lorentz forceThe radial contributions cancel as vectors around the loop. The axial contributions add, giving the force between distant coaxial current loops,The sign follows $\hat{\mathbf{\phi }}\times \hat{\mathbf{\rho }}=-\hat{\mathbf{z}}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll 1$.

Force direction. The second loop is pulled back toward the first: the force is attractive for the same current sense. Reversing one current reverses the force. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.

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.

Radial energy and effective potential. For a central force with potential energy $U(r)$, angular momentum conservation fixes $J=m{r}^2\dot{\varphi }$ and confines motion to a plane. The kinetic energy separates into radial and angular parts, givingThe second term is the centrifugal barrier. It converts the radial motion to a one-dimensional energy problem: allowed radii have $E\ge U_{\mathrm{eff}}$, turning points have equality, and a circular orbit occurs at a stationary point of $U_{\mathrm{eff}}$. Stability of that radial orbit requires a local minimum.

The first-order orbit equation. Write $A=GMm$. Since $f=-dU/dr=-A/r^2-A\gamma /r^3$, choose zero potential at infinity to getAlso $\dot{r}=(dr/d\varphi )J/(m{r}^2)$. Substitute this into the energy equation and multiply by $2m{r}^4/J^2$:Here $\gamma$ has units of length. A regular bound orbit with a centrifugal barrier requires $J^2>mA\gamma$ as well as the appropriate negative energy.

Solving in inverse radius. Let $u=1/r$ and defineDividing the first-order equation by $r^4$ givesDifferentiating, or directly using the Binet equation, gives $u^{\prime \prime }+qu=b$. The continuation through turning points follows from this smooth second-order equation. Its noncircular bound solution, with angular origin at a pericentre, isSubstitution into the first-order equation fixes the amplitude through this $e$. Therefore the requested representation holds withsince ${\alpha }_1/r-{\alpha }_2=\cos ({\alpha }_3\varphi )$. A general angular origin replaces $\varphi$ by $\varphi -{\varphi }_0$. This is the orbit equation for combined inverse-square and inverse-cube attraction. The circular limit $e=0$ is regular in $u=b/q$ but is not represented by finite ${\alpha }_1,{\alpha }_2$ in the source's normalized cosine form.

Precession and the inferred length. The last expression immediately givesSuccessive pericentres occur after angular advance $2\pi /{\alpha }_3$, so the apsidal precession per orbit isFor ${\alpha }_3$ near one, $1-{\alpha }_3^2\simeq \Delta \varpi /\pi$. Inferring an inverse-cube perturbation from apsidal precession consequently gives $\gamma \simeq [J^2/(mA)]\Delta \varpi /\pi$. For the stipulated nearly circular approximation, $J^2/(mA)\simeq r_0$, and Kepler's third law givesusing $P=88(86400)\mathrm{s}$, $M=2.0\times 10^{30}\mathrm{kg}$ and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. A century contains about $36525/88=415.1$ orbits. Convert the observed advance to radians per orbit before using it:HenceThe positive sign strengthens the attraction and gives prograde precession. The fractional perturbation $\gamma /r_0\simeq 1.6\times 10^{-7}$ validates the small-deviation expansion. This is the parameter fitting the proposed phenomenological force law under the specified small-eccentricity approximation, not a derivation of that law as the physical explanation of Mercury's anomaly.

Normal-mode idea. Near a stable equilibrium, expand the potential energy to quadratic order and the kinetic energy as a quadratic form in generalized velocities. The Euler-Lagrange equations then take the linear form $\mathsf{M}\ddot{\mathbf{x}}+\mathsf{K}\mathbf{x}=0$. A normal mode has fixed displacement ratios, $\mathbf{x}=\mathbf{q}\cos (\omega t+\phi )$, with $\mathsf{K}\mathbf{q}={\omega }^2\mathsf{M}\mathbf{q}$. Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. Superposition permits a general motion to be assembled from the modes. A free symmetry direction may instead be a zero mode, with constant or linearly changing coordinate.

Lagrangian and bead equations. Take each $x_i$ as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are $x_2-x_1$, $x_3-x_2$ and $x_1-x_3$. Equal equilibrium spring tensions give no net tangential force; their linear energy terms sum to zero. Dropping an irrelevant constant, the Lagrangian isFor example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three Euler-Lagrange equations therefore areThe wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the normal modes of three beads connected by springs on a circular hoop.