Define the energy function $\mathcal{E}={\sum }_i{\dot{q}}_i{p}_i-L$. Along solutions of the Euler-Lagrange equations,ConsequentlyFor an ordinary isolated mechanical system with $L=T-U$, a time-independent quadratic kinetic energy and a velocity-independent potential energy, ${\sum }_i{\dot{q}}_i{p}_i=2T$, so $\mathcal{E}=T+U$. For a more general Lagrangian, the conserved quantity is the displayed energy function; identifying it with $T+U$ requires these assumptions.

A cyclic coordinate $q_j$ is absent explicitly from the Lagrangian: $\partial L/\partial q_j=0$, even though ${\dot{q}}_j$ may occur. The Euler-Lagrange equation then givesThus a continuous coordinate-translation symmetry gives a conserved conjugate momentum. The conserved quantity need not be ordinary linear momentum: an absent azimuthal angle can give conserved angular momentum. The derivative is taken holding the other coordinates and velocities fixed.

Use a year of $365.25$ days, so $t=9.46728\times 10^8\mathrm{s}$, and $k=\rho A=10^{-11}\mathrm{kg}{\mathrm{m}}^{-1}$. With $M_p=722\mathrm{kg}$ and $v_{\infty }\simeq 1.752\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$, e givesThe corresponding cloud thickness is $x_f=(M_f-M_p)/k\simeq 1.50\times 10^{13}\mathrm{m}$. Checking $M_f{v}_f=M_p{v}_{\infty }$ verifies conservation of momentum; using the initial speed for all thirty years would overestimate the collected mass.

For constant-area capture of stationary dust, let $x=0$ at cloud entry, $M(0)=M_p$, $v(0)=v_{\infty }$, and define $k=\rho A$ with fixed collecting area $A$. Every swept volume $Adx$ adds stationary mass, soThere is no external longitudinal force, and the incoming dust has zero momentum in the cloud frame. The variable-mass system therefore satisfies $d(Mv)/dt=0$, rather than $M\dot{v}=0$. Using $Mv=M_p{v}_{\infty }$ givesHenceThe kinetic energy decreases during sticking: $dK/dt=-\frac{1}{2}\dot{M}{v}^2$. The lost energy becomes internal energy. Radiation, gravity, sputtering and changes in area are excluded by the ideal complete-collection model. The zero-density limit recovers $M=M_p$, $v=v_{\infty }$ and $x=v_{\infty }t$.

Setting $1/r\to 0$ in b gives the asymptotic speedThe radicand is positive, so the ideal probe has positive total mechanical energy and escapes the Sun. The later dust-cloud model takes this limiting speed as its initial speed.

Retain the source's idealized solar radial trajectory. Use ${\mu }_{\odot }\simeq 1.32712\times 10^{11}\mathrm{k}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, $r_i=1.5\times 10^9\mathrm{km}$ and $v_i=22\mathrm{km}{\mathrm{s}}^{-1}$. At $r=1.8\times 10^{10}\mathrm{km}$,The speed has decreased because the probe gains gravitational potential energy while moving away from the Sun. This is the supplied simplified model, rather than a reconstruction of Voyager 1's actual three-dimensional trajectory or planetary encounters.

The Lagrangian has no explicit time dependence, so its energy function is constant. With total speed $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$, division by $M_p$ givesConsequently, between the initial point $(r_i,v_i)$ and any other point,This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; angular momentum is then zero.

Let $\mu =G{M}_{\ast }$ and use plane polar coordinates in the orbital plane. The Newtonian gravitational potential energy is $U=-\mu M_p/r$, soA central force keeps the orbit in a plane. The star is treated as stationary, appropriate when $M_p\ll M_{\ast }$; otherwise the two-body reduction uses the reduced mass and the relative separation.

The capsule continues inertially at $v_c/c=350/353$ while the ship reduces its station-frame speed to $0.7c$. Transform the ship's new velocity into the capsule frame:Thus its speed is $0.952778c$, and the negative sign means it recedes behind the capsule along the chosen axis. The same transformation compares the subsequent local inertial velocities; the details of the intervening deceleration are not needed.

The velocity-addition formula combines the ship's station-frame velocity and the capsule's ship-frame velocity:For the distance convention in i, station-frame arrival takes $D/v_c=(353/350)\mathrm{day}=1.008571\mathrm{day}$. Transforming its two endpoint events back to the ship frame gives ${\gamma }_s[D/v_c-0.85D/c]$, agreeing with i. The capsule remains below the speed of light in either frame.

The berth moves at $0.85c$ in the ship's initial rest frame, so it too is subject to length contraction:The ship remains $100\mathrm{m}$ long in this frame. The two length comparisons use different simultaneous endpoint events; this is the same relativity of simultaneity that resolves the ladder paradox. Physical docking also requires deceleration, whose later effects are outside these initial inertial measurements.

For the approaching ship, ${\beta }_s=0.85$ and ${\gamma }_s=(1-0.85^2{)}^{-1/2}$. Its station-frame length contraction gives $L_s=100/{\gamma }_s\mathrm{m}$. Adding the two station-frame clearances givesThis is the berth's proper length, since the berth is at rest in the station. It is a simultaneous station-frame allocation, not a ship-frame guarantee of a safe docking maneuver.

For $\theta =30^{\circ }$, the result in b gives $1/3=2E_0/(E+E_0)$, hence $E=5E_0$. Acceleration from rest supplies only the change in kinetic energy:The accelerating electric potential difference is a magnitude; its sign depends on which electrode is taken as the reference, since the electron has negative electric charge.

Let $p$ be the incident momentum and $q_1,q_2$ the outgoing magnitudes in the laboratory frame. Transverse conservation of momentum gives $(q_1-q_2)\sin \theta =0$. For the nondegenerate symmetric scattering, $q_1=q_2=q$, and the two outgoing energies are equal, $ϵ=(E+E_0)/2$. ThenCombining these mass shell relations yieldsHere $E$ includes the incident electron's rest energy. As $E\downarrow E_0$ the limiting angle is $45^{\circ }$, consistent with the nonrelativistic equal-mass elastic collision; at large $E/E_0$ the outgoing tracks become forward directed.

For a frame moving at velocity $V\hat{\mathbf{x}}$, put $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The Lorentz transformation of the energy-momentum four-vector isExpansion givesThe transverse components are unchanged, hence $\boxed{E^{\prime 2}-c^2|{\mathbf{p}}^{\prime }{|}^2=E^2-c^2|\mathbf{p}{|}^2}$. For a particle of rest mass $m$ this invariant is $m^2{c}^4$; the same algebra also applies to the total four-momentum of a system.

For the shorter uniform rod, the parallel axis theorem gives $I=m(a^2/12+x^2)$. Its equivalent simple pendulum length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the periods therefore requiresA pivot on the shorter rod requires $0<x\le a/2$, leavingEither side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.

Let $\theta =0$ place the centre of mass vertically below a horizontal pivot axis. The restoring torque and rotational kinetic energy are $-mgh\sin \theta$ and $I{\dot{\theta }}^2/2$, soFor small angles, $\sin \theta =\theta +O({\theta }^3)$, and the equation reduces to simple harmonic motion. Thus the physical pendulum hasThis assumes negligible friction and a horizontal axis, with $h>0$. For a tilted axis the projected gravitational restoring torque changes; an axis through the centre of mass has no gravitational restoring torque.

Choose parallel axes along $z$, one through the centre of mass and one displaced transversely by $\mathbf{a}$. Write the perpendicular position as ${\mathbf{r}}_{\perp }={\mathbf{r}}_{\perp }^{\prime }+\mathbf{a}$. Direct expansion of the moment of inertia givesThe integral in the cross term is zero by the definition of the centre of mass. Therefore the parallel axis theorem isThe axes must be parallel and the reference axis must pass through the centre of mass; the formula is not generally valid between arbitrary parallel axes using either axis as the reference.