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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.

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.

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.

If $S^{\prime }$ moves at $+v$ relative to $S$, the Lorentz transformations areA rod at rest in $S$ has length $L$. To measure it in $S^{\prime }$, select endpoint events simultaneous there: $\Delta t^{\prime }=0$, so $\Delta t=vL/c^2$ and $\Delta x^{\prime }=\gamma (L-v\Delta t)=L/\gamma$. This is length contraction.

Now let a clock at rest in $S^{\prime }$ cross the two rod endpoints. These are different events, with $\Delta x=L$, $\Delta t=L/v$ and $\Delta x^{\prime }=0$. Their transformed time interval isThusThe moving clock measures the proper time $\Delta t^{\prime }$. The rod's contracted length divided by its speed in that clock's frame gives exactly the same crossing time; time dilation and length contraction are consistent consequences of the same transformation. Measuring a length and measuring a traversal interval require different simultaneity conditions.

Put $\mu =GM$. At the turning point of a radial launch the speed vanishes, so conservation of energy givesThis result is exact within the exterior Newtonian gravitational potential. The condition $r_0\gg R$ means $v_0$ is just below the escape velocity: $1-v_0^2/v_{\mathrm{esc}}^2\ll 1$. Replacing gravity by a constant would be inappropriate in this regime.

For a tangential launch, conserved specific angular momentum is $j=Rv$. At the outer turning point the radial velocity vanishes but the tangential speed is $j/r_0$. ThusUsing a eliminates the gravitational potential difference and givesThe expression printed in the source is the leading correction for $r_0\gg R$, rather than an exact identity. Equivalently, $v^2=2\mu r_0/[R(r_0+R)]$, so the launch speed exceeds the circular value and the surface is the periapsis of the nonintersecting ideal orbit.

At a fixed launch radius the escape threshold is zero total mechanical energy. It depends on the initial speed alone:A direction aimed into the planet can of course cause impact before escape. If the planet rotates, the ground-relative launch velocity must be added to $\mathbf{\Omega }\times \mathbf{R}$ to obtain the inertial velocity. The condition becomes $|\mathbf{u}+\mathbf{\Omega }\times \mathbf{R}{|}^2\ge 2GM/R$: prograde launch needs less ground-relative speed than retrograde launch. For a spherical planet the threshold expressed as inertial speed at the same radius remains independent of direction.

As $m_1/m_2\to 0$, a gives $v_1\to 3u$. The upper ball then rises through height $v_1^2/(2g)$ above the near-ground mutual encounter:The rise above its original release level is therefore $8H$ in this idealization. Although its speed triples, the light ball's fraction of the total energy tends to zero, $f(r)\sim 9r$. This distinguishes maximum rebound height from the maximum energy-transfer fraction in b. With finite radii, $9H$ is the leading approximation in drop height relative to the collision height.

The external work equals the change in relativistic kinetic energy, $W=m_e{c}^2({\gamma }_f-{\gamma }_i)$. With $m_e{c}^2=511\mathrm{keV}$,The rest energy cancels between endpoints. At these speeds, the nonrelativistic estimate $\frac{1}{2}{m}_e{c}^2(0.11^2-0.10^2)=0.53655\mathrm{keV}$ is close but slightly smaller.

Applying the same work-energy theorem at the higher pair of speeds givesThe identical increment $0.01c$ now needs about two thousand times as much work. The growth follows from the rapidly increasing Lorentz factor near the speed of light.

Treat the carriage as translating along the circular track, neglecting wheel rotational kinetic energy and its finite size. At the top, gravity and the inward normal force supply the centripetal acceleration:For unilateral contact, $N\ge 0$, so the limiting speed is $v_{\mathrm{top}}^2=gr$. Conservation of energy from rest at height $h$ then givesThis top contact condition is the most restrictive: at angle $\varphi$ measured from the bottom, $N/(mg)=2h/r-2+3\cos \varphi$, whose minimum occurs at the top. A mechanically captive carriage able to sustain an outward rail reaction has a different constraint; the result here uses the usual requirement of positive normal contact.

For a fixed-speed orbital launch-direction error, the initial radius and speed are correct, so the specific orbital energy is still $\epsilon =-GM/(2R)$ and the new semi-major axis is still $a=R$. Resolve the launch velocity into radial and tangential components, taking $\alpha$ as the in-plane deviation from the intended tangent:For a nondegenerate orbit $\cos \alpha \ne 0$, the orbital eccentricity follows from $j^2=GMa(1-e^2)$:The intended circular orbit has radius $R$ everywhere, so the largest radial distance between the orbit curves occurs at either apsis:This is the geometric separation of the two orbit curves, not the distance between two satellites compared at equal elapsed times. The latter also depends on orbital phase. A purely radial launch is a degenerate zero-angular-momentum trajectory through the central body and requires separate treatment of the collision.

Use nonrelativistic mechanics and let $\mathbf{V}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/(m_1+m_2)$ be the constant centre-of-mass velocity. In the centre-of-mass frame, conservation of momentum gives $m_1{\mathbf{u}}_1+m_2{\mathbf{u}}_2=0$ before and after the elastic collision. Thus ${\mathbf{u}}_2=-(m_1/m_2){\mathbf{u}}_1$, and the kinetic energy isThe decomposition $T_{\mathrm{lab}}=T_{\ast }+\frac{1}{2}(m_1+m_2)V^2$ and constant centre-of-mass velocity show that the elastic collision preserves $T_{\ast }$, so the positive constant multiplying $u_1^2$ forces that speed to remain unchanged. The same momentum relation fixes the second speed:Only their directions can change; the two velocities remain antiparallel in this frame. This derivation does not require a head-on collision.

For equal masses, let the incoming laboratory frame velocity be $\mathbf{v}$ and the outgoing velocities be $\mathbf{a},\mathbf{b}$. Conservation of momentum and kinetic energy giveSquaring the first equation and subtracting the second yields $2\mathbf{a}\cdot \mathbf{b}=0$. HenceThe stated oblique collision ensures both outgoing speeds are nonzero. In the degenerate head-on case one particle can stop, and a direction cannot be assigned to its zero velocity.

Use the inverse Lorentz transformation, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$:Taking differentials along the object's trajectory givesThis is the collinear relativistic velocity addition law. It concerns the velocity of one object measured in two frames. By contrast the rate of change of the distance between two objects in one laboratory frame can be as large asattained by oppositely directed light signals; for massive objects it is approached but not attained. This is not an object's speed in the other's rest frame and cannot transmit a signal faster than $c$.

Take the light to travel along positive x, while the water moves with signed velocity $-v$. In the water's rest frame its light phase velocity is $c/n$. The relativistic velocity addition law from a gives the laboratory phase velocityExpanding to first order in $v/c$, with $nv/c\ll 1$, yieldsEquivalently $w\simeq c/n-v(1-1/n^2)$: the counterflow reduces the forward light speed. This is the nondispersive Fresnel drag coefficient. The refractive index is taken as constant over the relevant frequencies; dispersion and its frequency-shift corrections are outside this approximation.

Choose positive x along the rocket's motion. If its instantaneous total mass is $\mu$ and it ejects a positive amount $d\eta =-d\mu$, the exhaust velocity in the original inertial frame is $v-u_0$ to first order. Conservation of momentum over that increment givesIntegrating the rocket equation from $\mu =m+M$ to $\mu =m$ givesThe constant exhaust speed is relative to the rocket, not the initial frame. Kinetic energy of rocket plus exhaust is not conserved because stored fuel energy is released. The calculation is Newtonian and requires resulting speeds small compared with the speed of light.

Neglecting exhaust flight time, no momentum is stored in transit andSince the velocities have opposite signs, equal nonzero speeds imply equality of the instantaneous masses. For $m_B=2m_A$ and $M=5m_A$, this occurs at $6m_A-q=2m_A+q$, or $q=2m_A$. ThereforeThe common-speed event occurs after $40\%$ of A's original fuel has been transferred, rather than when either rocket has its final mass. At this event the rearward exhaust still overtakes B: $v_A-u_0=-0.5945u_0<v_B=-0.4055u_0$. Thus the interception restriction in b does not invalidate this event even though extending complete ballistic capture over the much later burn would need separate checking.

Put $m_A=M=m$ and $m_B=2m$. At reversal the transferred mass is $q_r=0.45m$, leaving A with $1.55m$ and, after capture of that exhaust, B with $2.45m$. The first burn givesAfter reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this velocity. Applying the rocket equation with reversed thrust to A givesBoth now move toward negative x, but A moves faster in that direction. The asymptotic closing speed isThis conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing speed closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual momentum after collecting that fixed amount.

Idealize the pole and stick as having negligible transverse thickness. Put the pole at the origin, the stick initially along the $y$ axis, and the struck lower end at $y=-l/2$, as in the source diagram. A short horizontal impulse $J$ along $+x$ gives the centre of mass a constant speed $V=J/m$. The angular impulse about the centre is $Jl/2$, giving constant counterclockwise angular velocityThe pole exerts no further force while the stick moves clear of it. Let $\psi =\Omega t$. The trailing upper end has coordinatesThe pole can lie on the translating straight stick again only when the stick becomes horizontal: for other orientations, its intersection with $y=0$ is the moving midpoint $x=Vt>0$. At the first horizontal orientation, $\psi =\pi /2$, the trailing end is atConsequently the grazing clearance of an impulsively struck rod occurs atFor $\beta >1/(2\pi )$ the whole stick has passed the pole by its first horizontal orientation, and the continually advancing centre makes all later passes safe. Equality is the limiting grazing contact of the endpoint; strict avoidance requires the strict inequality. If $\beta <1/(2\pi )$, the pole lies within the stick at that first horizontal pass, so contact occurs before the freely moving trailing end can pass the pole. A uniform thin stick has $\beta =1/12$, below the threshold: sufficient mass near the ends is needed to increase its moment of inertia and reduce the rotation relative to translation.