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Acceleration is the time derivative of velocity, equivalently the second derivative of position:It is a vector with SI unit $m/{\mathrm{s}}^2$. A change in direction produces acceleration even if speed stays constant. For example, a car rounding a bend at steady speed needs an inward resultant force by Newton's second law. Thus acceleration does not mean only speeding up. When $v>0$, the dot product gives $dv/dt=\mathbf{v}\cdot \mathbf{a}/v$: a perpendicular acceleration changes direction without changing speed at that instant.

Take counterclockwise circular motion about the origin with radius $R=2\mathrm{m}$ and constant angular speed $\omega =0.5rad/s$. The position, velocity and acceleration areThe speed is $R\omega =1m/s$ throughout, while the inward centripetal acceleration has magnitude $R{\omega }^2=v^2/R=0.5m/{\mathrm{s}}^2$. At the four cardinal points, reached at $t=0,\pi ,2\pi ,3\pi \mathrm{s}$, the velocities in $m/s$ are $(0,1),(-1,0),(0,-1),(1,0)$; the accelerations in $m/{\mathrm{s}}^2$ are $(-0.5,0),(0,-0.5),(0.5,0),(0,0.5)$. The vectors turn together, remaining perpendicular. ^[ref]

The proof outline is just two applications of the chain rule to $\mathbf{r}(t)$; ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ then fixes both magnitudes. The usefulness is geometric: turning requires a force even when a speed reading remains steady, and doubling speed at fixed radius quadruples the required centripetal acceleration. The same mechanism appears in a circular orbit.

Historically, Newton's Principia, Book I, Section II, Proposition IV, relates the inward forces of uniform circular motion to the squared arcs traversed in equal times, divided by their radii. In modern notation, its dependence on speed and radius is $v^2/R$; Newton presents a geometric argument rather than the coordinate derivatives above. ^[ref]

For a spherical drop of constant mass density $\rho$, radius $a>0$ and downward velocity $v>0$, suppose its volume grows at rate $c\pi a^{2}v$ by collecting stationary mist and its upward quadratic drag is $k\rho \pi a^2{v}^2$, with $c,k>0$. The drop mass is $4\pi \rho a^3/3$. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional $c$ term alongside drag.

For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

A constant added to a function changes its reference level without changing its derivative:Geometrically, it is a vertical translation of the graph. For $F(x)=x^3/3$, changing $C$ moves every plotted value by the same amount while preserving every tangent slope. This is the geometric reason that antiderivatives form a family.

When $F$ represents a physical quantity, the physical dimension of $C$ must match that of $F$. When position is obtained from velocity, $C$ is a position offset. An unknown starting position remains unknown even when the rate of change is known exactly.

When describing the family of antiderivatives on an interval, the arbitrary additive constant is the constant of integration. On a disconnected open domain, the corresponding freedom is a locally constant function, as in antiderivatives on disconnected domains, rather than necessarily one constant on the whole domain.

For a smooth incompressible direct trajectory with total velocity field $\mathbf{U}$ and active scalar $\Theta$, the negative-constraint Lagrangian function in constrained optimization convention givesThe transpose term is the formal adjoint of $\delta \mathbf{u}\cdot \nabla \mathbf{U}$; the two couplings transpose scalar transport by advection and buoyancy. For terminal cost $\int \Theta (T{)}^2$, the terminal data are ${\mathbf{u}}^{†}(T)=0$, ${\theta }^{†}(T)=2\Theta (T)$. They are integrated backward along the stored direct trajectory. No-slip boundary conditions and Dirichlet boundary conditions for the adjoint velocity and homogeneous Neumann boundary conditions for the adjoint scalar remove the spatial boundary terms.

An airfoil (British spelling: aerofoil) is the cross-sectional profile of a lifting surface, such as a wing or blade. Its chord is the straight segment from leading edge to trailing edge, with length $c$; its mean camber line runs midway between its upper and lower surfaces, and its thickness measures their separation. Camber is the departure of that mean line from the chord. The angle of attack describes the profile's orientation relative to the incoming flow. These independent choices let a designer alter the distribution of pressure, lift and drag, rather than simply enlarging the surface. ^[ref]

The distinction between an airfoil and a finite wing matters. A two-dimensional section model assumes no variation along span and predicts lift per unit span $L^{\prime }$. A wing of span $b$ has a spanwise distribution $L^{\prime }(z)$ and total lift $L={\int }_{-b/2}^{b/2}{L}^{\prime }(z)dz$. Its tip flow and trailing vortices generate a downward induced velocity, called downwash, which changes the local flow seen by the sections. Thus one cannot generally multiply the isolated two-dimensional $L^{\prime }$ by $b$. The wing aspect ratio $b^2/S$ measures its span relative to its reference area and helps characterize these three-dimensional effects. ^[ref]

For a thin section with small camber slopes and small angle of attack, thin airfoil theory predictswhere both angles are in radians and ${\alpha }_{L=0}$ is the zero-lift angle of attack. The model assumes two-dimensional, steady flow, incompressible flow, inviscid flow, an attached outer flow, and smooth departure at a sharp trailing edge (the Kutta condition). This last condition selects the circulation used by the Kutta–Joukowski theorem; with counterclockwise circulation $\Gamma$ positive and ${\mathbf{U}}_{\infty }=U_{\infty }{\mathbf{e}}_x$, that theorem gives $L^{\prime }=-{\rho }_{\infty }{U}_{\infty }\Gamma$ for positive lift towards $+{\mathbf{e}}_y$. The beauty is that a detailed velocity field collapses to a simple force law, while the geometry enters through ${\alpha }_{L=0}$. ^{[ref] [ref]}

A symmetric airfoil has ${\alpha }_{L=0}=0$ in this model: at $\alpha =4{}^{\circ }=\pi /45$ the predicted lift coefficient is $c_l\simeq 0.439$, and reversing $\alpha$ reverses lift. A stipulated cambered section with ${\alpha }_{L=0}=-2{}^{\circ }$ instead predicts $c_l\simeq 0.219$ at zero angle of attack. These are ideal calculations, not measurements or guarantees for a particular profile. Air parcels on the upper and lower surfaces have no obligation to arrive together at the trailing edge; the pressure and velocity field must satisfy the flow problem. A symmetric lifting section already makes an unequal-path-length explanation inadequate. ^[ref]

Real airfoils have a viscous boundary layer, even when inviscid flow approximates the outer motion. At sufficiently demanding incidence, flow separation can become extensive and the lift coefficient departs from its approximately linear law, reaches a maximum and may fall: this is aerodynamic stall. Lift does not generally vanish at aerodynamic stall. The onset depends on profile, Reynolds number, surface condition and motion history; $2\pi (\alpha -{\alpha }_{L=0})$ predicts neither a universal stall angle nor post-stall behavior. The pressure/balance experiment in lift makes this limit observable by measuring changes in the surface pressure distribution alongside the total force. ^[ref]

An actual teaching experiment in Arbor Scientific's The Level Air Track uses one glider, an air track, two photogates, and a ruler. Turn on the blower and place the glider near the middle. Adjust the feet until it does not drift when released. Put the photogates at least half a track length apart and align them with the glider's flag. Measure the length $L$ that interrupts the beams. Give a gentle push, let go, and record each beam-blocking time $\Delta t_i$. The readout is the pair of speeds $s_i=L/\Delta t_i$. The worksheet asks how those speeds compare, which forces remain after the push, and which interaction the air track minimizes.^[ref]

The motivating contrast is between a brief push and the subsequent glide. The air cushion reduces contact friction; it does not remove the downward gravitational weight or the upward support force. Similar successive speeds after release are consistent with a small horizontal net external force, rather than with a continuing push. The experiment approximates the zero net external force model over a finite track and observation interval.

To strengthen the documented procedure, use independent measurement calibration: check the flag length and each timer against a known length and time reference; measuring $L$ sets the length scale but does not calibrate the timer. Repeat launches in both directions without contact during the timed passage, and check for drift at several track locations. These are additional controls, not reported steps or measured results in the worksheet. A residual inclined plane produces a downhill component of gravitational weight with the same laboratory direction on both runs; friction and drag force oppose the direction of travel. Reversing the launch can therefore help distinguish these effects.

A photogate directly measures a beam-blocking time; $L/\Delta t_i$ is an average speed during that passage, not an exact instantaneous velocity. Assign the direction separately. Equal readings at two gates do not prove zero acceleration everywhere between them: opposing changes could cancel. Measurement uncertainty from timer resolution, an incorrectly measured flag length, residual slope, airflow disturbances, and remaining friction or drag force limit the inference. No measured glider speeds are assumed here.

Ambipolar damping is attenuation of magnetic disturbances by friction between the magnetically coupled charged fluid and a neutral fluid. In a strongly collisional mixture, the fluids nearly move together, but their small relative velocity dissipates energy. With drag coefficient ${\rho }_i{\mu }_{in}={\rho }_n{\mu }_{ni}$, the frictional energy loss is $-{\rho }_i{\mu }_{in}\int |{\mathbf{u}}_i-{\mathbf{u}}_n{|}^{2}dV$. Neglecting other dissipative processes does not remove this collisional loss.

For slow rotation and low-frequency inertial motion, neglect the time derivative of the Eulerian mass density perturbation relative to the mass-weighted velocity divergence. This filters acoustic compressibility while retaining background mass density variation. A spherical background mass density is often sufficient at leading order in rotation.

The signed angle between a lifting surface's chord and its incoming relative flow. Positive incidence in the planar convention used in airfoil raises the leading edge relative to the trailing edge in a horizontal left-to-right flow. It is distinct from orientation relative to the ground; an induced velocity can change the local effective incidence.

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

Move the fixed origin from $O$ to $O^{\prime }=O+\mathbf{a}$, keeping the axes parallel and the inertial frame's velocity measurements unchanged. The new positions are ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so distributivity of the cross product givesHere $\mathbf{P}$ is total linear momentum. For the angular momentum example with $\mathbf{a}=(1,0,0)\mathrm{m}$, the new position is $(1,0,0)\mathrm{m}$ and $L_{O^{\prime },z}=6-3=3\mathrm{kg}{\mathrm{m}}^2/s$. If total momentum is zero, angular momentum is independent of this origin shift. A moving origin additionally contributes $-\dot{\mathbf{a}}\times \mathbf{P}$ when this expression is differentiated while retaining the original momenta; see angular momentum about the centre of mass for an important case where that term vanishes.

Uniform axial strain rate $E$ in an annular Newtonian fluid gives radial velocity $u_r=-Er/2+C/r$. For radii $R_i,R_e$, area parameter $A=R_e^2-R_i^2$, and constant gas pressures,The opposite signs of inner and outer curvature in the Young–Laplace equation give these expressions. The kinematic boundary condition implies $\dot{A}=-EA$ and

In antiplane perfect plasticity, velocity $w=v_3$ and shear stresses $a={\sigma }_{13},b={\sigma }_{23}$ depend only on $x_1,x_2$. Force balance is $a_{,1}+b_{,2}=0$, the fixed yield surface is $f(a,b)=k$, and the associated flow rule is $\nabla w=2\dot{\lambda }(f_a,f_b)$ with $\dot{\lambda }\ge 0$. Differentiating the yield equation in each coordinate and combining it with equilibrium shows that $a,b,w$ are constant along characteristic curves tangent to $(-f_b,f_a)$. For example $-f_b{a}_{,1}+f_a{a}_{,2}=f_b{b}_{,2}+f_a{a}_{,2}=0$. Nonzero regular yield gradient is needed for a characteristic direction.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.

A circular distant barycentric Kepler orbit around a circular star–planet binary need not have constant star-centred osculating orbital elements. Let $\mu$ be the planet's fraction of the binary mass, $\alpha =a_{\mathrm{pl}}/a\ll 1$, and $\epsilon =\mu /\sqrt{\alpha }\ll 1$. Subtracting the stellar reflex velocity produces radial and tangential perturbations $(-v_0\epsilon \sin \Delta ,v_0\epsilon \cos \Delta )$, where $\Delta$ is the planet's longitude minus the particle's longitude. The vis-viva equation gives the displayed semi-major axis variation. The eccentricity vector components follow from $e_r=r{v}_{\theta }^2/(G{M}_{\star })-1$ and $e_{\theta }=-r{v}_r{v}_{\theta }/(G{M}_{\star })$. The errors are $O(\mu +{\epsilon }^2)$; the spatial translation itself is only $O(\mu \alpha )$. These are changes of instantaneous orbital description, even when the assumed barycentric trajectory is fixed.