Codex @codex  0

Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

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]

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Along the holonomic constraint $z=x^2/(4a)$, we have $\dot{z}=x\dot{x}/(2a)$ and hence speed squared $(1+x^2/(4a^2)){\dot{x}}^2$. The smooth wire's normal reaction is perpendicular to the permitted velocity and does no work. Conservation of mechanical energy, with release from rest at height $h$, givesThe turning points are $x=\pm x_0$ with $x_0=2\sqrt{ah}$. One full oscillation has twice the travel time between them:Set $x=x_{0}u$ and $\beta =h/a$. After cancelling the scale factors,so the bead on a parabolic wire hasFor $\beta \ll 1$, expand the numerator uniformly on $[-1,1]$:The integrable endpoint weight gives ${\int }_{-1}^1(1-u^2{)}^{-1/2}du=\pi$ and ${\int }_{-1}^1{u}^2(1-u^2{)}^{-1/2}du=\pi /2$, for example by $u=\sin \theta$. Therefore the requested small-oscillation period isThe leading frequency is $\sqrt{g/(2a)}$, and the first finite-amplitude correction increases the period. At $h=0$ the ring is stationary; the expression is the limiting period of nonzero small oscillations.

By Newton's second law, the central force obeysDifferentiating the cross product givessince the acceleration is parallel to $\mathbf{r}$. Hence $\mathbf{n}=\mathbf{n}(0)$ is constant. Since $\mathbf{r}\cdot \mathbf{n}=0$, the entire motion lies in the fixed plane through the origin with nonzero normal vector $\mathbf{n}(0)$.

Choose polar coordinates in that plane, orienting $\theta$ so that its positive normal vector points along $\mathbf{n}(0)$. The velocity isThereforeMultiplication by $m$ identifies this with conservation of angular momentum; geometrically, the swept area per unit time is $h/2$.

Taking the dot product of Newton's second law with the velocity givesThusThis expresses conservation of mechanical energy, the sum of kinetic energy and potential energy.

Finally, for $u=1/r$ and a prime denoting a derivative with respect to $\theta$, the chain rule givesSubstitution into the conserved energy yields

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.

Let $\mathbf{N}$ be the normal reaction of the stationary smooth surface of revolution. Its normal lies in the span of ${\mathbf{e}}_r,{\mathbf{e}}_z$, so it has no ${\mathbf{e}}_{\theta }$ component. The transverse component of Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{N}-mg{\mathbf{e}}_z$, is thereforewhich gives $\boxed{h=r^2\dot{\theta }=\text{constant}}$.

The velocity is tangent to the stationary surface, so $\mathbf{N}\cdot \dot{\mathbf{r}}=0$. Take the dot product of Newton's second law with $\dot{\mathbf{r}}$:Using $z=f(r)$ proves conservation of mechanical energy per unit mass:The ideal constraint does no work. These equations describe motion while the particle remains constrained to the surface.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

Interpret $h$ as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Before impact, rolling without slipping gives $\omega =u/r$, so conservation of mechanical energy givesUse the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If $v$ and $V$ are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution giveDuring the subsequent simple pendulum swing, conservation of mechanical energy givesThe cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then $m+I/r^2=3m/2$. Writing $J$ for the impulse delivered to the bob gives $(3m/2)(v-u)=-J$, $2mV=J$ and $V-v=u$. Hence $V=6u/7$ and $\cos {\theta }_{\max }=25/49$, giving $59.32^{\circ }$. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

The original diagram has a horizontal bottle neck, so weight does not enter the axial force balance. Gauge pressure acts on the cork's end area and supplies outward force $F_g=\pi a^2{P}_g$. The radial contact force is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs whenThe assumption of equal static and kinetic friction coefficients removes a separate force drop when slipping starts.

Initially the pressure is marginal at $x=0$, so $\pi a^2{P}_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure force stays constant. The net force is therefore $F(x)=\pi a^2{P}_{g}x/l$. The cork's mass is $m_c=\rho \pi a^{2}l$, giving the pressure-driven cork ejection equationFor $x(0)=0$ and $\dot{x}(0)=v_0>0$, solving the constant-coefficient equation givesThe equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced force grows.

Multiplying the equation of motion by $\dot{x}$ and integrating gives ${\dot{x}}^2=v_0^2+b^2{x}^2$. The contact ends at $x=l$, soEquivalently, the net work is ${\int }_0^{l}F(x)dx=\pi a^2{P}_{g}l/2$. Exactly zero initial velocity would leave the ideal marginal configuration stationary. The exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges as $v_0\to 0^{+}$, even though the limiting exit speed is finite.

The density is $500\mathrm{kg}{\mathrm{m}}^{-3}$, so $v_{\mathrm{exit}}\simeq \sqrt{400000/500}=28.3\mathrm{m}{\mathrm{s}}^{-1}$. For an ideal upward launch at this speed, conservation of mechanical energy gives height above the exitThis neglects air resistance and the small gravitational work during extraction compared with the launch energy. Actual gas pressure need not stay constant, so it is an idealized estimate.

Adding a stopper that doubles the moving mass, while preserving the same pressured area and contact friction, leaves the extraction work unchanged. The squared exit speed and the ballistic height are therefore halved:

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

The tunnel is taken thin enough that removal of its material does not significantly disturb spherical symmetry. For $a\le r\le 2a$, the enclosed mass isBy the spherical shell theorem, the inward gravitational acceleration has magnitudeThe released gravitational potential energy is the positive work done by gravity while the mass falls from radius $2a$ to $a$:By conservation of mechanical energy,Inside the cavity, the spherical shell theorem gives zero gravitational field, so the velocity stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance $2a$ apart. ThereforeThe quantity $G\rho$ has units ${\mathrm{s}}^{-2}$, so this is a time. There are no numerical values for $a$ or $\rho$ in the paper. The transit time is independent of $a$ and the falling mass $m$. The cavity's gravitational potential is constant, not necessarily zero; only its gradient vanishes. The calculation is a hollow-sphere gravitational transit.

The force is conservative, with $\mathbf{F}=-\nabla U$ and potential energy zero at the origin:Released from rest at $(c,c)$, the trajectory has $x=y=c\cos \omega t$, givingThe right panel of the figure shows this nonnegative oscillation: it starts at a maximum $k{c}^2$, reaches zero at $t=\pi /(2\omega )$, and repeats after $\pi /\omega$. Its frequency is twice the coordinate frequency. Conservation of mechanical energy gives $K(t)=k{c}^2{\sin }^2\omega t$, so the missing potential energy is kinetic energy, not an energy loss.

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

Take $+x$ along the upper ship's initial velocity, and take $+y$ from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial velocities $v_1\hat{\mathbf{x}}$ and $-v_2\hat{\mathbf{x}}$. Their equal masses put each at distance $l/2$ from the centre of mass, soBy momentum conservation, the centre of mass keeps moving with constant velocityIn that frame the ships have opposite velocities of magnitude $u=(v_1+v_2)/2$. Their initial angular momentum about the centre of mass is nonzero even though their trajectories were straight:The tether exerts central internal forces, so conservation of angular momentum givesEach ship follows uniform circular motion of radius $l/2$ around the translating centre of mass, clockwise viewed from $+z$. More explicitly, put the initial centre at the origin and defineThe kinetic-energy decomposition about the centre of mass verifies conservation of mechanical energy directly:At closest approach the relative radial velocity is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its tension is perpendicular to each ship's velocity relative to the centre of mass and does no net work on the pair at fixed $l$.