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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]

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.

Astrometry measures the positions and motions of celestial objects. Repeated position measurements can distinguish stellar parallax, caused by the observer's motion, from proper motion of the source across the sky.

An astrometry measurement needs a date and a reference frame, as well as an angle. In an equatorial coordinate system, right ascension and declination label two viewing directions. A catalogue's reference date states when its quoted position applies; comparing entries at different dates without accounting for proper motion can make the same star appear misplaced. The observational problem is to infer a consistent history from successive incoming light directions, while accounting for the observer's changing position and measurement error.

To see the separation clearly, consider just one angular coordinate $\theta$, with $t$ the numerical elapsed time in years. Assume a simplified circular orbit of the Earth, a fixed source Euclidean distance, constant proper motion in this coordinate and no other accelerations. Choose the direction and the starting date so that the annual stellar parallax projection is the cosine termHere ${\theta }_0$ is the coordinate without the annual stellar parallax offset at the reference date; the observed $\theta (0)$ is ${\theta }_0+p$. The proper motion coefficient $\mu$ is quoted in arcseconds per year, and $p$ is the annual stellar parallax in arcseconds. This is an illustrative choice of a full-amplitude projection. General sky directions have different projected amplitudes and phases in their two coordinates; their apparent paths need not be circles. Real astrometry uses the observer's calculated position rather than imposing this single cosine. ^[ref].

Set ${\theta }_0=0$, $\mu =100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$ and $p=50\mathrm{mas}$, where $1\mathrm{mas}$ is one thousandth of an arcsecond. The model gives $\theta (0)=50$, $\theta (0.5)=0$ and $\theta (1)=150\mathrm{mas}$. Same-season observations one year apart differ by $100\mathrm{mas}$ because the annual stellar parallax contribution repeats. Subtracting that fitted straight trend leaves an annual signal between $-50$ and $+50\mathrm{mas}$. The existing stellar parallax distance relation then gives $20$ parsecs.

Observing only at the same annual phase would recover the proper motion trend but leave $p$ indistinguishable from a constant offset. Sampling several seasons as well as several years separates the repeating signal from the accumulating displacement. The two signatures answer different questions using the same measurements.

For a circular orbit, fixed total mass, transfer from $M_2$ to $M_1$, negligible spin and $R_L\propto a(M_2/M{)}^{1/3}$, the orbital angular momentum $J=M_1{M}_2\sqrt{Ga/M}$ givesIf the donor follows a radius sequence with stellar radius response exponent $\zeta$, has no additional radius change at fixed mass, and maintains contact, equating its radius derivative to this expression proves the displayed contact equation. Negative external $\dot{J}$ drives donor mass loss when $\zeta +5/3-2q>0$. The appropriate equilibrium or adiabatic response and its timescale must be specified separately; the equation alone is not a universal dynamical-stability criterion.

For fixed component masses, total mass $M=M_1+M_2$ and reduced mass $\mu =M_1{M}_2/M$, in the weak-field slow-motion circular orbits, $E=-GM\mu /(2a)$ and $J=\mu \sqrt{GMa}$ obey $dE/dJ=\Omega$. The gravitational-wave energy and angular-momentum balance makes the secular loss tangent to this circular family. Substituting $v^2=GM/a$ into the instantaneous quadrupole luminosity of a Kepler binary and differentiating $E$ gives the displayed inspiral equation. Integration givesThe orbit is quasicircular, with small radial drift. The formal coalescence time is not a valid extension of the weak-field point-mass model into the final strong-field or stellar-contact phase.

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]

In a spherically symmetric gravitational potential $\Phi (r)$, a circular orbit has $r{\Omega }^2={\Phi }^{\prime }(r)$, specific orbital energy $\epsilon =\Phi +r^2{\Omega }^2/2$ and specific angular momentum $\mathbf{h}=r^2\Omega \mathbf{n}$, with $|\mathbf{n}|=1$. Differentiating gives ${\epsilon }^{\prime }=2r{\Omega }^2+r^2\Omega {\Omega }^{\prime }=\Omega (r^2\Omega {)}^{\prime }$. Changes in orbital orientation have $\mathbf{n}\cdot d\mathbf{n}=0$, so $\mathbf{\Omega }\cdot d\mathbf{h}=\Omega d(r^2\Omega )=d\epsilon$. For a particle of variable mass $M$ and total angular momentum $\mathbf{H}=M\mathbf{h}$, this implies $d(M\epsilon )=(\epsilon -\Omega h)dM+\mathbf{\Omega }\cdot d\mathbf{H}$. Moreover $d(\epsilon -\Omega h)/dr=-h{\Omega }^{\prime }$, which is positive when the orbital angular velocity decreases outward.

A circular orbit in a central potential $U(r)=A{r}^s$ requires inward force $U^{\prime }(r)=m{v}^2/r>0$. Thus $2K=r{U}^{\prime }(r)=sU$, giving $K/U=s/2$ for the specified zero of potential. The ratio depends on that choice of zero; adding a constant to $U$ leaves the orbit unchanged but changes $K/U$. This is the circular special case of the virial theorem.

For a fixed mass on a Keplerian circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}$ has derivative $M_s{r}_0\Omega /2$. The torque on the satellite is minus the torque $\Gamma$ on the disk. This gives the displayed drift law. In the density-slope satellite torque model, ${\dot{r}}_0=-(8\chi /9)\beta q({\Sigma }_0{r}_0^2/M_{\ast })r_0\Omega (H/r_0{)}^{-2}$.

For unit mass with specific angular momentum $l$ in $V(r)=-e^{-r}$, the effective potential is $-e^{-r}+l^2/(2r^2)$. A positive-radius circular orbit satisfies the displayed equation. Since $r^3{e}^{-r}$ has its unique maximum $27/e^3$ at $r=3$, exactly two radii occur for $0<l^2<27/e^3$. Their radial stability follows from $V_{\mathrm{eff}}^{\prime \prime }=e^{-r}(3/r-1)$: the inner orbit is stable at fixed angular momentum, the outer unstable. The case $l=0$ has no finite-radius circular orbit despite satisfying the upper bound.

For the radial acceleration $-k{r}^n$ with $k>0$, a circular orbit of radius $a$ has squared specific angular momentum $h^2=k{a}^{n+3}$. At fixed $h$, its radial small oscillations have squared frequency $(n+3)k{a}^{n-1}$, so the orbit is stable for $n>-3$. This is a special case of the effective potential stability criterion.

In a slowly changing spherical gravitational potential, angular momentum is exactly conserved and the radial action is an adiabatic invariant. An initially circular orbit remains circular in the adiabatic limit, so $J_r=0$ and $L^2=GrM(r)$ give the displayed shell relation. If dark-matter shells do not cross, their enclosed dark mass is also conserved. A density profile alone does not specify this orbital model: eccentric-orbit populations require their full action distribution.

For a distant coplanar circular orbit about a binary star, the averaged quadrupole potential of a circular binary givesAt fixed outer radius $x$, the quadrupole shortens the period relative to the monopole result. If the inner binary evolves slowly, the outer specific angular momentum stays constant in the axisymmetric averaged potential, but the outer energy can change. The outer radius must then be allowed to adjust when computing a period derivative.

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.

View a circular orbit along its perpendicular spin axis. Let $f$ be the orbital angle measured from a fixed direction, choosing an arbitrary zero for the circular limit of true anomaly, and let $\theta$ be the direction of an outward arrow painted on the body, measured from the same fixed direction. The direction from the body to its companion is $f+\pi$. Thus an inward-facing arrow satisfiesMore generally synchronous rotation keeps $\theta -f$ constant; its particular value selects which surface direction faces the companion. Over one orbital period $T$, both $f$ and the unwrapped $\theta$ increase by $2\pi$. This is the short geometric explanation for one turn per orbit. For a nonspinning body, $\dot{\theta }=0$ instead, and the companion direction sweeps through all surface longitudes. On an eccentric orbit, mean $1:1$ spin does not force $\theta -f$ to stay exactly constant: true anomaly advances nonuniformly, producing the longitudinal component of libration.

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 flattened power-law Newtonian gravitational potential has equipotential spheroids $R^2+{\lambda }^2{z}^2=\text{constant}$. For ${\Phi }_0<0$ and $0<\beta <1$, its midplane circular orbit angular frequency satisfies ${\Omega }^2=-\beta {\Phi }_0{R}_0^{\beta }{R}^{-\beta -2}$.