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The speed at which the rupture front advances along a fault plane. It is distinct from the local particle slip velocity and the propagating seismic wave speeds. Length equals $V_{r}T$ only for the stated approximately constant-speed propagation geometry.

Rutherford scattering is elastic scattering by a Coulomb potential. In the repulsive classical case, a positively charged projectile bends around a positively charged atomic nucleus. A close approach can turn it backwards without a contact collision. The calculation connects an invisible impact parameter with a measurable outgoing angle, then with a differential scattering cross-section and a detector count.

Assume a projectile of mass $m$ and electric charge $ze$, a stationary point atomic nucleus of electric charge $Ze$, and positive $z,Z$. Here $e$ is the elementary charge. Neglect nuclear recoil, electron screening effect, energy loss, and the nuclear force. With vacuum permittivity ${\epsilon }_0$, defineHere $E$ and $v$ are the incident kinetic energy and speed far from the atomic nucleus, $b\ge 0$ is the impact parameter, and $L$ is the magnitude of angular momentum. The nonrelativistic limit requires $v\ll c$, with $c$ the speed of light. The fixed-target approximation requires the target mass to be much greater than $m$. In SI, $\kappa$ has units $\mathrm{J}\mathrm{m}$, $E$ has units $\mathrm{J}$, and $b$ has units $\mathrm{m}$.

A screw motion combines a rotation in three dimensions about a fixed line with a translation parallel to that line. With unit axis direction $e$, axial speed $v$, and rotation matrix $R_t$ fixing $e$, each map ${\Phi }_t(x)=R_{t}x+tve$ is a proper Euclidean motion of Euclidean three-space, and together these maps form a one-parameter subgroup. Each off-axis group orbit is a circular helix when both rotational speed and axial speed are nonzero. A pure rotation or translation is a limiting case.

Seafloor spreading creates oceanic crust at a spreading ridge and carries it away with the adjoining plates. With symmetric steady spreading, an age interval maps to distance by the half-spreading rate, not the full relative opening speed. A time sequence of field polarities can therefore become paired marine magnetic stripes.

For the sign convention $u_t-6u{u}_x+u_{xxx}=0$ and any $\kappa >0$, the displayed negative solitary wave travels at speed $4{\kappa }^2$. To verify it, put $\xi =x-ct$ and write $u=U(\xi )$. The equation becomes $(-cU-3U^2+U^{\prime \prime }{)}^{\prime }=0$. A decaying profile has zero integration constant, so $U^{\prime \prime }=cU+3U^2$. Direct differentiation of the displayed profile gives $U^{\prime \prime }=4{\kappa }^{2}U+3U^2$, proving the result. Its depth is $2{\kappa }^2$, so narrower and deeper pulses travel faster. Changing the sign of the nonlinear term reverses the amplitude sign. This wave belongs to the soliton family of the Korteweg-De Vries equation.

For a P ray incident from a layer of speed ${\alpha }_1$ onto a faster layer ${\alpha }_2$, the transmitted ray is tangential when $\sin i_c={\alpha }_1/{\alpha }_2$. Angles are measured from the interface normal.

The adiabatic bulk modulus divided by density, $\Phi =K_S/\rho ={\alpha }^2-4{\beta }^2/3$ for an isotropic solid. It has units of squared speed and is a key input to the Adams-Williamson density relation.

The propagation speed of a seismic disturbance, depending on elastic properties, density, composition, temperature and microstructure.

Three identical like point charges, initially at rest at an equilateral triangle's vertices, remain equilateral as they move away from the fixed centroid. Each resultant force is radial by symmetry. If the centroid radius is $r$, its magnitude is $k_e{q}^2/(\sqrt{3}{r}^2)$. In a nonrelativistic model neglecting radiation, energy conservation gives the final speed $v=\sqrt{2k_e{q}^2/(ma)}$ for initial side $a$ and charge mass $m$.

The propagation speed of a shear elastic wave. In an ideal isotropic elastic medium it is $V_s=\sqrt{\mu /\rho }$, where $\mu$ is shear modulus and $\rho$ mass density. Seismic anisotropy, frequency-dependent relaxation and melt can complicate the inference from an observed shear-wave velocity to temperature or composition.

For a flat layer of thickness $h$ and shear speed ${\beta }_1$ above a faster half-space of speed ${\beta }_2$, unfolding $m$ down-and-up traversals gives the displayed reflected-ray travel time. Its slope is the horizontal ray slowness. At $x=2mh\tan i_c$, with $\sin i_c={\beta }_1/{\beta }_2$, the corresponding head-wave branch $t=x/{\beta }_2+2mh\sqrt{{\beta }_1^{-2}-{\beta }_2^{-2}}$ is tangent to this curve.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

For ideal clocks in an inertial frame separated by fixed distance $L$, transporting a clock at small speed $u$ produces lag $\delta =(L/u)(1-\sqrt{1-u^2/c^2})\sim Lu/(2c^2)$. Hence the limit agrees with Einstein synchronisation, assuming the clock hypothesis. A relabelling $t_{\kappa }=t+\kappa x/c$ retains an endpoint offset $\kappa L/c$, so the result does not remove the choice of distant-time convention.

A slow neutron has low kinetic energy relative to the fast neutrons commonly emitted in nuclear reactions. The term covers an energy range, rather than a single speed. A neutron moderator transfers kinetic energy to surrounding matter through collisions. Slowing a neutron can increase its chance of causing particular neutron capture reactions, even though each collision has reduced its kinetic energy. The Fermi slow-neutron activation experiment made this surprising distinction experimentally visible.

For a mass $m$ passing at relative speed $u$ and impact parameter $b$, integration of $Gmb/(b^2+u^2{t}^2{)}^{3/2}$ gives the transverse kick $2Gm/(bu)$. Weak deflection requires $b\gg Gm/u^2$, up to the equal-mass order-one factor. This scale supplies a small-impact cutoff for the weak-scattering diffusion approximation, not a physical exclusion of stronger encounters.

At speed $v\ll c$, an endpoint separation $L$ measured simultaneously in its own rest frame becomes $L\sqrt{1-v^2/c^2}$ in the moving frame, shorter by approximately $L{v}^2/(2c^2)$. The stable evaluation $L{\beta }^2/[1+\sqrt{1-{\beta }^2}]$ avoids subtracting nearly equal floating-point numbers.

In the circular-Earth-orbit, unpowered patched conic approximation, the least Earth-relative excess speed for direct solar-surface contact lowers the heliocentric tangential speed to that of an ellipse with apocentre $a$ and pericentre $R_{\odot }$. The vis-viva equation gives $v_a=v_E\sqrt{2R_{\odot }/(a+R_{\odot })}$, so the retrograde excess magnitude is $v_E-v_a$ and the terrestrial surface launch speed is the title expression. Cancelling all heliocentric angular momentum instead produces radial fall to a point Sun and the larger estimate $\sqrt{v_{e,E}^2+v_E^2}$. Rotation, planetary gravity assists and atmospheric drag are outside this direct-flight model.

A nonrelativistic particle accepted only at a fixed flight time through an ideal uniform-field tube has fixed parallel speed $v_{\parallel }=l/t_0$. On-axis entry and wall clearance bound the independent perpendicular speed by $|q|Ba/(2m)$, provided a full gyration occurs before exit. Adding parallel and perpendicular kinetic energies gives the displayed ideal energy window, where $K_0=m{v}_{\parallel }^2/2$. The upper endpoint is a grazing limit; strict wall exclusion makes it a supremum.