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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]

The aerodynamic load factor $n=L/W$ compares lift with gravitational weight. The stated convention takes $L$ along the chosen positive lift direction, so $n$ is a signed force ratio, not simply the magnitude of total acceleration divided by $g$. A steady level condition can have $n=1$ with zero inertial acceleration.^[ref]

A focused example is a wings-level pull-up at the bottom of a vertical curved path. Neglect other vertical forces and take upward normal acceleration as positive. The force balance givesFor stipulated true airspeed $V=50m/s$, curvature radius $R=500\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the upward centripetal acceleration is $5m/{\mathrm{s}}^2$ and $n\simeq 1.510$. With $m=600\mathrm{kg}$, lift is $8886\mathrm{N}$, while the net upward force is $3000\mathrm{N}$. Neither is the other.

In the steady horizontal-turn model, the condition instead is $n=1/\cos \varphi$; bank angle derives that balance. It does not apply unchanged to the vertical pull-up. At fixed air mass density, wing area and usable maximum lift coefficient, the relation $L=\rho V^{2}S{C}_L/2$ gives limiting speed proportional to $\sqrt{n}$ for positive $n$. For this example, the factor is about $1.229$ relative to the corresponding one-load condition.

A force balance measures aerodynamic load; a trajectory measurement estimates inertial acceleration. Comparing them requires the gravitational weight term and correct directions. A reported load factor also needs its force-axis convention: thrust, side forces and coupled motion can matter to what an instrument actually records. The definition alone establishes neither structural capability nor available lift before aerodynamic stall.

The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

Suppose a body of mass $M_0$ has centre of mass ${\mathbf{R}}_0$, and material of mass $M_c<M_0$ with centre of mass ${\mathbf{R}}_c$ is removed. Additivity of the first mass moment gives the remaining centre of massWhen the original centre of mass is at the origin, the remaining centre of mass points opposite the removed material's first mass moment. For a constrained rigid body rotating about a fixed axis, this displaced centre of mass undergoes centripetal acceleration even if the applied axial torque is zero.

A mechanism in which rotating weights move outward as angular speed increases, allowing their displacement to indicate or regulate a machine's speed. The steady configuration follows the balance of actual applied forces and centripetal acceleration, or equivalently of applied forces and centrifugal force in a rotating reference frame. Geometry and the restoring mechanism determine the speed-displacement relation.

For a velocity field $\mathbf{u}=U(r)e_{\varphi }$ in spherical coordinates, the convective acceleration is $(\mathbf{u}\cdot \nabla )\mathbf{u}=-U(r{)}^2(e_r+\cot \theta e_{\theta })/r$, away from the polar axis. Each streamline is a circle of cylindrical radius $r\sin \theta$, so this is centripetal acceleration of magnitude $U^2/(r\sin \theta )$ directed toward the polar axis. The terms containing $U^{\prime }(r)$ cancel because a particle on such a streamline stays at fixed $r$ and $\theta$.

A centripetal force is the inward resultant force needed for circular motion. Newton's second law and centripetal acceleration give magnitude $m{v}^2/r$ for a particle of mass $m$, speed $v$, and path radius $r$. It is a role played by an actual interaction, such as tension or an electric force, rather than an additional interaction to add to a free-body diagram. A purely radial centripetal force is perpendicular to the velocity, so it does no work and leaves the speed unchanged.

A circular orbit maintains constant distance from its centre of attraction. Its velocity changes direction even when its speed stays constant. The inward centripetal acceleration has magnitude $v^2/r$, so the radial force must satisfy $F_r=-m{v}^2/r$. For a Newtonian gravitational field with standard gravitational parameter $\mu$, this gives $v=\sqrt{\mu /r}$.

For a test body on a circular Kepler orbit about a point mass $M$, the centripetal acceleration $v_c^2/r$ equals $GM/r^2$, so $v_c=\sqrt{GM/r}$. For a spherical extended source use the enclosed mass $M(<r)$ instead. The formula is not valid for an arbitrary nonspherical gravitational potential.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

A path has constant speed when the magnitude of its velocity does not vary with time. Its acceleration may still be nonzero: uniform circular motion has turning velocity and inward centripetal acceleration. Constant velocity is stronger, requiring unchanged direction as well as unchanged speed.

Take an ion of mass $m$ and electric charge $q\ne 0$, with velocity perpendicular to a uniform magnetic field of magnitude $B>0$. In the classical mechanics limit, the Lorentz force provides the centripetal acceleration:Here $r$ is the local radius of curvature and ${\omega }_c$ is the cyclotron frequency, expressed as an angular frequency. A half-turn takesIncreasing speed increases the path length in exactly the same proportion. Drive the gap at angular frequency ${\omega }_{\mathrm{RF}}={\omega }_c$, or ordinary frequency $f_{\mathrm{RF}}=|q|B/(2\pi m)$. In each half-turn the gap electric field reverses once, just as the particle reverses its direction of passage through the gap. The particle therefore receives two accelerating kicks per complete turn, using one oscillation per complete turn. This is the fundamental resonance; transit time through the gap and departures from a uniform magnetic field have been neglected.

For a circular orbit, gravity supplies centripetal acceleration, giving $v_c^2/r=\mu /r^2$. ThusA body in a circular orbit is bound despite its continuous free fall. At the same $r$, escape needs twice its kinetic energy per unit mass. An instantaneous forward tangential velocity increment from the circular orbit requires $\Delta v=(\sqrt{2}-1)v_c$, rather than the full escape speed from rest. In the synthetic example this is exactly $2(\sqrt{2}-1)m/s\approx 0.828427m/s$.^[ref]

Escape speed is a coasting threshold, not a required continuously maintained speed or a measure of rocket propellant. A powered vehicle can start at rest and acquire energy gradually. The rocket equation relates an ideal velocity increment to rocket exhaust speed and rocket mass ratio; rocket thrust, drag, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative velocity and the inertial frame used in the calculation.

Reaching another body is a finite-destination spacecraft trajectory problem. In the synthetic radial example, a launch at $2m/s$ reaches $r=4\mathrm{m}$ while remaining bound. A spacecraft can similarly encounter the Moon on a bound geocentric transfer; actual planning must include the Moon's gravity, target motion, arrival velocity, and any propulsion needed for capture. Escaping Earth alone does not imply escaping the Sun, because the relevant reference frame and attracting mass have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.

One motivating laboratory demonstration compares two balls initially at the same height. A switch releases one vertically when the other leaves a horizontal track; arranging their paths to meet gives a collision as the readout of equal downward displacement over equal time. Equal initial height and synchronized release control the comparison. This tests the local independence of horizontal motion and free fall, under nearly constant gravitational acceleration; it does not measure an inverse-square law.^[ref]

A complementary astronomical comparison uses the Moon's distance and orbital period to infer inward centripetal acceleration, $a=4{\pi }^{2}r/T^2$, and compares it with surface gravitational acceleration scaled by $(R/r{)}^2$. The approximately sixtyfold distance ratio gives the approximately $1/3600$ prediction. The inference links terrestrial free fall to a lunar orbit. Distance calibration matters; the circular orbit approximation and the Sun's tidal force limit this elementary comparison.^[ref]

A fixed-wing aircraft obtains most of its aerodynamic support from aircraft wings that do not continuously rotate around a shaft to generate lift. The word fixed distinguishes the lifting arrangement; it does not require every surface to be immovable or perfectly rigid. Both powered airplanes and an unpowered glider can use fixed wings.

Thrust and aerodynamic support are distinct. Relative motion through air produces lift and drag; an engine can supply the thrust needed to balance drag in steady level flight. The components must be resolved as vectors, since their directions change with the motion and attitude.^[ref]

For an original ideal turning model, suppose a point-mass aircraft maintains horizontal circular flight at constant airspeed $V$, with negligible vertical thrust and no net vertical acceleration. If its lift vector is inclined through a bank angle $\varphi$, thenThe horizontal component supplies centripetal acceleration. Eliminating $L$ gives $L/W=1/\cos \varphi$ and $R=V^2/(g\tan \varphi )$. With $V=40m/s$, $\varphi =30{}^{\circ }$ and $g=9.81m/{\mathrm{s}}^2$, the radius is about $282.5\mathrm{m}$ and the required lift is about $1.1547W$. The wing remains fixed to the body while its force direction changes in space.

This is a Newtonian force model, not a prediction that an arbitrary airplane can produce the required lift. The attainable lift coefficient, wing loading, propulsion and aerodynamic moments determine whether a particular steady condition can exist. In a steady unpowered descent, the balances instead lead to the glide angle relation.

Free fall is motion under gravity alone: there is no supporting normal force, engine thrust or appreciable drag force. In the Newtonian mechanics model, a test particle of mass $m$ obeys $m\ddot{\mathbf{r}}=m\mathbf{g}(\mathbf{r})$, hence $\ddot{\mathbf{r}}=\mathbf{g}(\mathbf{r})$. The mass cancels. Bodies released at the same position with the same velocity follow the same path in this model, even if their gravitational weights differ. ^[ref]

A dropped stone is one example; a coasting satellite is another. Free fall does not require motion straight down or decreasing distance to the attracting body. An ideal circular orbit has constant radius while its velocity continually turns under centripetal acceleration. The circular orbit as continuous free fall connects the familiar drop with orbital motion.

Near a surface, treating the gravitational acceleration as constant gives a useful first prediction. Over larger heights, the inverse-square law changes the acceleration during the fall. Radial free fall from rest uses conservation of mechanical energy to account for that change exactly, within the specified Newtonian gravitational field. The contrast matters: the simplest formula is an approximation to a physical model, rather than a law with the same numerical $g$ everywhere.

History also makes the idealization tangible. Galileo's Two New Sciences (1638), Third Day, Theorem II, derives distance proportional to the square of time for uniformly accelerated motion from rest; its first corollary gives successive distances in the ratios $1,3,5,7,\dots$ over equal time intervals. This is a kinematic statement, predating the gravitational inverse-square law. ^[ref]

On Apollo 15's final lunar surface excursion, David Scott released a hammer and a feather together. The near-vacuum removed the large difference in drag force familiar on Earth. NASA's report describes their arrival together within the accuracy of the release; this demonstration illustrates the free fall prediction without establishing infinite experimental precision. ^[ref]

An orbit around an isolated source constrains its standard gravitational parameter $\mu =GM$ rather than independently measuring the source's mass and the gravitational constant. In the test particle approximation, a circular orbit of radius $a$ and orbital period $T$ satisfiesThe left equation equates the required centripetal acceleration to the gravitational acceleration. More generally, a Kepler orbit depends on the same $\mu$. In the exact relative two-body problem, replace $GM$ by $G(M+m)$.

For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted test particle trajectory unchanged. This is the gravitational mass degeneracy: more observations of the same orbit can refine $GM$ without separating its factors. A laboratory source whose mass is calibrated independently supplies the missing information.

Consequently an astronomical mass expressed in kilograms as $M=\mu /G$ inherits measurement uncertainty from $G$, even if $\mu$ is known much more precisely. To first order,When the relative standard uncertainty of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source masses inferred using the same $G$ is $M_1/M_2={\mu }_1/{\mu }_2$, so this common factor cancels. That distinction explains how excellent orbit determination can coexist with less precise absolute astronomical masses.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are