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Choose upward velocity positive during ascent. Gravity and linear drag both point downward, giving the linear differential equationAt the highest point $v=0$, soThe vertical ascent under linear drag time is smaller than $v_0/g$ for $\alpha >0$, because $\log (1+x)<x$. Its zero-drag limit is $v_0/g$, as required by constant-gravity projectile motion.

Write $\omega$ for the imposed constant angular velocity. In inertial polar coordinates, the bead has $v^2={\dot{r}}^2+{\omega }^2{r}^2$. Its constrained Lagrangian isThe reduced Lagrangian has no explicit time dependence, so its Lagrangian energy function is conserved:This is the Jacobi integral $J=E-\omega L_z$, since the bead's actual inertial energy and axial angular momentum areThe motor maintaining the stick's angular velocity supplies torque and does work. Indeed $\dot{E}=2m{\omega }^{2}r\dot{r}=\omega {\dot{L}}_z$, generally nonzero. A time-independent reduced Lagrangian therefore need not imply conservation of the bead's inertial kinetic energy when its imposed constraint is moving.

Treat the ladder as a uniform slender rigid body. While both contacts remain imposed, its centre of mass is $(l\sin \theta ,l\cos \theta )$, with speed $l|\dot{\theta }|$. Its central moment of inertia is $m(2l{)}^2/12=m{l}^2/3$. Adding translational and rotational kinetic energy givesTaking the floor as zero gravitational potential energy, the Lagrangian, conjugate momentum and Euler-Lagrange equation areThe positive sign is appropriate: gravity increases the angle from the vertical. These are the constrained equations of a frictionless sliding ladder. Real wall contact is unilateral: a wall cannot pull the ladder. Once the computed normal force would become negative, the two-contact model ceases to apply; the question explicitly assumes that contact persists.

In the pion rest frame, momentum conservation makes the muon and neutrino momenta opposite, with common magnitude $p$. For a massless neutrino, $E_{\nu }=pc$; energy conservation and the relativistic energy-momentum relation giveSubstitute $pc=m_{\pi }{c}^2-E_{\mu }$ into the second equation. The $E_{\mu }^2$ terms cancel, leavingFor completeness, the complementary relativistic two-body decay energy is $E_{\nu }=(m_{\pi }^2-m_{\mu }^2)c^2/(2m_{\pi })$. The result is a total energy, including the muon rest energy, rather than its kinetic energy.

For planets with negligible mass compared with the Sun, Kepler's third law gives $T^2\propto a^3$, where $a$ is the semi-major axis. Relative to Earth,Thus the requested orbital-distance estimates are about 0.39 and 5.2 times Earth's. Strictly, the period fixes the semi-major axis, not a time-averaged radius of an arbitrary eccentric orbit; identifying the two as a mean distance is the usual elementary orbital-size convention.

Einstein's postulates state that the laws of physics have the same form in every inertial frame, and that light in vacuum has the same speed of light $c$ in every inertial frame, independently of the motion of its source. For a primed frame moving with velocity $v{\mathbf{e}}_x$ relative to an unprimed frame, with coincident origins at $t=t^{\prime }=0$, the Lorentz transformations areThe following worldlines use Earth-frame time in years and distance in light years, so numerically $c=1$. The instantaneous turnaround is the idealized limit of a short acceleration stage; the elapsed proper time is calculated along each inertial segment.