For the prograde captured orbit, the departing half has instantaneous velocity tangent to the original orbit, in the original direction, with speedIt exceeds the local escape speed: since $V\ge \sqrt{2}{u}_c$, $u_{\mathrm{depart}}\ge (2\sqrt{2}-1)u_c>\sqrt{2}{u}_c$. This is its speed immediately after the explosion, not its later speed at infinity.

If a counterrotating circular captured orbit is allowed, its signed velocity is $-u_c$, the outgoing speed is instead $2V+u_c$, and the required generated energy is $m(V+u_c{)}^2/2$. The usual answer takes the prograde orbit; specifying that choice removes the directional ambiguity.

Put $V=v_{\max }$ and let $u_c=\sqrt{GM/r_0}$ be the circular orbit speed. Take the captured half to move in the original tangential direction, the minimum-energy choice. Conservation of momentum during the short explosion gives the other half speed $2V-u_c$. The gravitational potential energy is unchanged at the instant of separation, so the necessary generated energy is the increase in kinetic energy:The captured half has zero radial velocity and exactly the required tangential speed, so it can enter the stable Newtonian circular orbit at this radius. Internal self-binding and any mass lost to radiation are neglected.

The lines of centres from either lower cylinder to the upper cylinder make $60^{\circ }$ with the horizontal. Let $N_L,N_R$ be the upward normal forces on the top cylinder from the left and right cylinders, and let $Q$ be the lower cylinders' mutual normal force. While all three contacts remain closed, the triangular configuration translates horizontally with common acceleration $a$; frictionless normals exert no torques about the cylinder centres.

For the top cylinder, Newton's second law givesThus $N_L=m(g/\sqrt{3}+a)$ and $N_R=m(g/\sqrt{3}-a)$. The right lower cylinder has horizontal equation $N_R/2+Q=ma$, soA contact may push but cannot pull: $N_L,N_R,Q\ge 0$. The limiting inequalities are thereforeAt the lower bound, $Q=0$ and the lower pair are just losing compression; at the upper bound, $N_R=0$ and the upper-right contact is just losing compression. All three pairwise normal forces are strictly positive only inside the open interval. The ground normal forces are positive throughout this range. The applied horizontal force on the left cylinder is $F=3ma$, as follows from the whole system's horizontal momentum balance.

The two positive powers below require $\beta >0$; the negative power requires $\beta <0$, so that the force is attractive. To leading order, $r\simeq r_0[1+ϵ\cos (\sqrt{k+2}\varphi +\delta )]$. The azimuthal advance from one radial maximum to the next isFor $k=2$, ${\omega }_r/\Omega =2$: there are two radial oscillations per revolution. The isotropic harmonic oscillator has orbits that are exact ellipses centred on the force centre, and the perturbed circle is such an ellipse.

For $k=7$, ${\omega }_r/\Omega =3$: the leading perturbation has three radial lobes per revolution, with consecutive radial maxima separated by $2\pi /3$. Its small-amplitude sketch is a rounded three-lobed orbit.

For $k=-7/4$, ${\omega }_r/\Omega =1/2$: the radius completes one oscillation while the body makes two revolutions. The leading curve closes after $\Delta \varphi =4\pi$, with a double-wound, generally self-crossing shape; the nearest and furthest points occur in the same spatial direction one revolution apart.

The rational ratios for $k=7$ and $k=-7/4$ describe closure of the linearized near-circular curves. They do not imply that every finite-amplitude orbit closes exactly: higher-order changes in the radial-to-orbital frequency ratio generally produce apsidal precession. Bertrand's theorem identifies the harmonic potential $k=2$ and the Kepler potential $k=-1$ as the exceptional attractive central potentials with all nearby bounded orbits closed.

Write $r=r_0+\eta$ and keep $L$ fixed. The small radial oscillations in a power-law potential have squared angular frequencyThe circular orbital angular frequency has magnitude $\Omega =|L|/(m{r}_0^2)$, so ${\Omega }^2=(\beta k/m)r_0^{k-2}$. HenceA restoring radial oscillation exists for $k>-2$, subject to the attractive-force condition in part (b). For $k<-2$ the radial stationary point is unstable; $k=-2$ is the degenerate boundary. This agrees with circular-orbit stability for a power-law central force, since the actual force is proportional to $-r^{k-1}$.

A circular orbit of radius $r_0>0$ is a stationary point of the effective potential:For $L\ne 0$, $\beta k>0$ and $k\ne -2$, this givesThe condition $\beta k>0$ makes the radial force attractive. At $k=-2$, the equation instead requires the fine tuning $L^2/m=-2\beta$; then the centrifugal and potential terms cancel at every radius and there is no isolated restoring minimum. If that equality fails, there is no positive-radius circular orbit for this special power.

A central force confines nonradial motion to a plane. In plane polar coordinates, the Lagrangian for $V(r)=\beta r^k$ isThe azimuth $\varphi$ is a cyclic coordinate, so $L=m{r}^2\dot{\varphi }$ is conserved angular momentum. The Lagrangian has no explicit time dependence, giving conserved energyAt fixed $L$, the radial equation of motion is $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$. The effective potential packages the transverse kinetic energy as a centrifugal term alongside the actual potential energy. It gives a one-dimensional radial problem without introducing an extra physical interaction.

Direct fall towards a point Sun. If the Sun is treated as a point target, the projectile must lose its heliocentric angular momentum. The cheapest change is retrograde, cancelling Earth's tangential orbital velocity. It then starts its radial fall with zero heliocentric speed, so $w=v_E$ andThis is the usual radial-fall estimate. Merely launching towards the Sun while retaining Earth's tangential velocity does not remove the angular momentum barrier.

Minimum for impact on a finite Sun. A literal collision with the solar surface does not require zero angular momentum. The limiting solar impact by a retrograde launch instead has apocentre $a$ at Earth's orbit and pericentre $R_{\odot }$ at the solar surface. The vis-viva equation with semi-major axis $(a+R_{\odot })/2$ gives the heliocentric departure speedIt remains prograde about the Sun, but the Earth-relative departure is retrograde with $w=v_E-v_a$. Therefore the surface launch speed for grazing the finite solar surface isHere $a\simeq 149.598\times 10^6\mathrm{km}$ and $R_{\odot }\simeq 695700\mathrm{km}$.^[ref][ref] Slightly reducing the departure speed below $v_a$ makes the trajectory enter the solar surface rather than just graze it. The $31.81$ and $29.15\mathrm{km}{\mathrm{s}}^{-1}$ values correspond to the explicitly different point-target and finite-radius idealizations, not to two ways of adding escape speeds. These are direct-flight estimates; planetary gravity assists and Earth's rotation are outside the model.

To escape the Solar system in the single-Sun approximation, the heliocentric specific orbital energy must reach zero. At Earth's orbital radius, the required heliocentric escape speed is $\sqrt{2}{v}_E$. The smallest change from Earth's initial velocity is prograde, parallel to its motion:Combining the terrestrial and solar energy requirements givesThe asymptotic heliocentric trajectory is a parabolic Kepler orbit; larger launch speeds can give positive solar orbital energy.

The snow remains aboard, so the moving mass grows as $M+\sigma t$. Incoming snow contributes no horizontal momentum, and there is no horizontal external force. ThereforeThe combined sled–snow system follows a variable-mass system balance; setting $(M+\sigma t)\dot{v}=0$ would omit the incoming mass term.

Snow now leaves with zero horizontal velocity in the ice frame, the same as its incoming horizontal velocity. The net horizontal momentum flux is zero:Any forward momentum given to the snow on capture is returned to the sled while sweeping it off. “Perpendicular” specifies a different outgoing velocity here than in case A; changing the reference frame without transforming that velocity would miss this difference.

Snow leaves perpendicular to the motion in the rider's frame, so its outgoing horizontal velocity in the ice frame is $v$. During $dt$, incoming snow brings zero horizontal momentum and outgoing snow carries $\sigma vdt$. ThusAlthough the sled does not retain snow, it must continually supply the snow's departing horizontal momentum.

The photon propagates along $-x$ in the tunnel frame and has energy $E_{\gamma }=hc/{\lambda }_0$ and momentum $p_x=-E_{\gamma }/c$. A Lorentz transformation into the approaching runner's frame givesSince $E_{\gamma }^{\prime }=hc/{\lambda }^{\prime }$, the approaching-source relativistic Doppler effect isIt is a blueshift. With $q={\lambda }^{\prime }/{\lambda }_0=500.0/589.1$, solving $q^2=(1-\beta )/(1+\beta )$ givesThis shift depends on the relative velocity along the light ray, not on the tunnel length or the location of detection.

Use the tunnel frame $S$, with the entrance at $x=0$, the far end at $x=L$, and $t=0$ at both runner entry and photon emission. The runner has $x_R=vt$, while the approaching photon has $x_{\gamma }=L-ct$. At detection,For $0<v<c$, $0<f<1/2$: the photon meets the runner before the runner reaches the tunnel midpoint.