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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.

About a chosen origin $O$, a particle has angular momentum $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. A body has $\mathbf{L}={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$, or the corresponding integral for a continuous mass distribution. For a fixed origin in an inertial frame,Internal central pair forces cancel in the total torque. About a fixed origin, a central force conserves a particle's angular momentum; zero net force alone is insufficient if an external couple remains.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

Write $\mu =GM$ and let $v_c=\sqrt{\mu /r_0}$ be the original circular orbital speed. An impulse leaves the orbital angular momentum unchanged if it is radial: its angular impulse $\mathbf{r}\times \Delta \mathbf{p}$ is zero. The tangential velocity remains $v_c$, and an added radial component $w$ givesThus kinetic energy increases even though angular momentum does not. For a Radial kick from a circular Kepler orbit, the new specific orbital energy and squared specific angular momentum areThe orbital eccentricity is $e=\sqrt{1+2\epsilon j^2/{\mu }^2}=\sqrt{\beta -1}$, while the semi-latus rectum is $p=j^2/\mu =r_0$. HenceFor $\beta =2$ the trajectory is a parabolic Kepler orbit and has no finite furthest radius; its periapsis is $r_0/2$. For $\beta >2$ it is a hyperbolic orbit, with the same formula for $r_{\min }$ but unbounded radius:One must not use the negative denominator in the bound-orbit expression as a negative physical apocentre. On a departing unbound branch the conic's periapsis may already lie in the past; the future minimum from the kick event is then $r_0$. These are ideal conic orbits; a periapsis inside a real planet would cause impact before the ideal continuation.

A straight diameter at nonzero angle $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial torque, the total axial angular momentum of the Earth and body is conserved. The body's contribution to the moment of inertia decreases toward the centre, so the Earth spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.

For a body of mass $m$ constrained to follow a tunnel fixed in the Earth, an illustrative angular-momentum balance isHere $I_E$ is the Earth's axial moment of inertia, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial angular momentum with the rotating Earth. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi =0$ and is largest for an equatorial diameter.

Choose the smaller disc's initial sense as positive. For either uniform disk, the axial moment of inertia is $I=\frac{1}{2}m{r}^2$, so $I_1=I_2/2$. The signed initial angular velocities are ${\Omega }_1=2{\omega }_2$ and ${\Omega }_2=-{\omega }_2$, giving total axial angular momentum zero. The frictionless axle supplies no axial torque, and the pins' negligible mass supplies no extra stored kinetic energy. ThusThese equations allow the original state or its simultaneous reversal. A nontrivial elastic pin impact selects reversal of the relative contact velocity, rather than continued interpenetration. ThereforeThe requested magnitudes are unchanged: $|{\Omega }_1^{\prime }|={\omega }_1$ and $|{\Omega }_2^{\prime }|={\omega }_2$. The intersecting pin paths make the impact possible; no additional lever arm is needed once axial angular momentum and energy conservation have been imposed.

Conservation of angular momentum means that the total angular momentum of a system about a chosen fixed origin in an inertial frame stays constant. Its dynamical condition isInternal torques cancel for the complete isolated mechanical system. If only the component of external torque about an axis vanishes, only that angular momentum component is necessarily conserved. The skater's upright spin uses this axial statement: vertical weight and support provide negligible vertical-axis torque, and frictional torque is neglected. The skater need not conserve rotational kinetic energy while changing the internal configuration.

Using rotational kinetic energy $K=I{\omega }^2/2$ at the two steady endpoint configurations,At fixed angular momentum $J$, $K=J^2/(2I)$ increases when the moment of inertia decreases. The skater supplies work by pulling the arms inward; chemical energy in the muscles is converted into additional rotational kinetic energy, with possible extra dissipation in a real skater. This does not require an external axial torque.

For the ideal rotating point masses, an inward radial force supplies the required centripetal acceleration. During contraction both that inward force and the radial displacement point inward, so their scalar product gives positive work. At the initial and final steady configurations radial motion has ceased; any transient radial kinetic energy must also be included in a full intermediate-time balance. The quoted difference compares the rotational endpoint energies only.

The first term, $m{\dot{r}}^2/2$, is radial kinetic energy. The second, $L^2/(2m{r}^2)$, is tangential kinetic energy, since conservation of angular momentum gives $L=m{r}^2\dot{\phi }$. The final term, $-GMm/r$, is the gravitational potential energy, with zero at infinite separation. ThusThe second and third terms together form the radial effective potential for a Kepler orbit. At an apsis the radial velocity vanishes, so $E=U_{\mathrm{eff}}$; for a bound noncircular orbit the two accessible roots give the nearest and furthest radii.

Write the lost mass fraction as $f$, with $f=0.02$, and let $\mu =GM$. Initially the circular orbit has $v_0^2=\mu /R$, zero radial velocity, and specific angular momentum $h^2=R^2{v}_0^2=\mu R$. The stipulated rapid symmetric loss leaves the satellite's instantaneous position and velocity unchanged, but reduces the gravitational parameter to ${\mu }^{\prime }=(1-f)\mu$. There is no impulsive kick from contact with ejecta in this model.

The new specific orbital energy isand $h$ remains unchanged because the post-loss force is still central. At a radial turning point,Inserting the known constants and factoring yieldsFor $0<f<1/2$ the two radii are $R$ and $R/(1-2f)$. The initial radius is the pericentre: immediately after the loss, $\ddot{r}=h^2/R^3-{\mu }^{\prime }/R^2=f\mu /R^2>0$, so the satellite begins moving outward. ThereforeEquivalently, the new semi-major axis is $49R/48$ and the orbital eccentricity is $1/49$. These values also satisfy $r_{\min }=a(1-e)$ and $r_{\max }=a(1+e)$. For loss of half the original mass the ideal trajectory becomes parabolic, and larger loss makes it unbound. Slow adiabatic loss would give a different, nearly circular expansion; it is not the instantaneous limit stipulated here.

The figure retains the same angular momentum before and after the change. Once the expelled material is far away, a spherically symmetric outer shell exerts no interior gravitational force by the shell theorem; direct ejecta interaction and the shell's finite transit time are neglected as in the source idealization.

For the attractive central force $\mathbf{F}=-k\mathbf{r}$, with $k>0$, the potential energy is $U(r)=k{r}^2/2$ up to an additive constant. Conservation of angular momentum confines the motion to a plane; writing its magnitude as $J=m{r}^2\dot{\phi }$, the kinetic energy isAt fixed $J$, the radial energy equation becomes $E=m{\dot{r}}^2/2+U_{\mathrm{eff}}(r)$, whereThe effective potential combines the true potential energy with the angular-motion contribution. Its negative derivative gives the radial equation $m\ddot{r}=-U_{\mathrm{eff}}^{\prime }=J^2/(m{r}^3)-kr$; the first term is the centrifugal contribution arising from elimination of the angular coordinate.

A circular orbit is a fixed radial position at a stationary point:For $J\ne 0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=4k>0$, so this is a stable minimum. Small radial disturbances have angular frequency $2\sqrt{k/m}$. For $J=0$, there is no nonzero-radius circular orbit.

Allowing azimuthal motion makes the endpoint speed satisfy $v^2=l^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$. The light-rod spherical pendulum therefore hasThe azimuth $\phi$ is a cyclic coordinate, since the Lagrangian is independent of it. Its conjugate canonical momentum is the conserved vertical angular momentum:For a finite uniform rod, replace $m{l}^2$ by $I$ and $mgl$ by $A$ from i. The same azimuthal symmetry then gives $p_{\phi }=I{\sin }^2\theta \dot{\phi }$.

The Hamiltonian is the Legendre transform of this Lagrangian. With $p_{\theta }=I\dot{\theta }$,There is no explicit time dependence, so $dH/dt=\partial H/\partial t=0$. Here $H$ also equals the total mechanical energy: the constraint is stationary, the kinetic energy is quadratic in generalized velocities, and gravity is conservative. The apparent singularity of the momentum expression at $\sin \theta =0$ is a spherical-coordinate singularity; trajectories reaching a pole must be described consistently, and nonzero $J_z$ prevents reaching it at finite energy.

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.

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.

The azimuth $\varphi$ is a cyclic coordinate, since $\partial \mathcal{L}/\partial \varphi =0$. Its Euler-Lagrange equation gives conservation of the vertical component of angular momentum:This $h$ is the actual angular momentum, not specific angular momentum. The Lagrangian also has no explicit time dependence. The conserved energy isAlong solutions, $\dot{E}=-\partial \mathcal{L}/\partial t=0$. These are the conserved quantities associated with rotational symmetry about the vertical and time-translation symmetry, respectively.

Fixed-angular-momentum form. Eliminating $\dot{\varphi }$ using the conserved angular momentum $h$ givesEquivalently, the polar motion has effective potentialFor the steady conical pendulum at ${\theta }_0$, $\ddot{\theta }=0$, soThe sign of $h$ agrees with the sign of $\Omega$; substituting $h=m{l}^2{\sin }^2{\theta }_0\Omega$ verifies the constant-angle result in part (d).

Period of small polar oscillations. Put $\theta ={\theta }_0+\alpha (t)$, with $|\alpha |\ll 1$ and $h$ held fixed. This is the appropriate small nutation about a conical pendulum; keeping $\dot{\varphi }$ fixed instead would violate the angular momentum constraint during the oscillation. Ifthen $f({\theta }_0)=0$ andThe linearization is $\ddot{\alpha }+{\omega }_{\theta }^2\alpha =0$, givingSince $0<{\theta }_0<\pi /2$, the coefficient is positive and these small polar oscillations are stable. To first order, $\dot{\varphi }=\Omega (1-2\alpha \cot {\theta }_0)$, making the compensating change in azimuthal angular velocity explicit. The limit ${\omega }_{\theta }/|\Omega |\to 2$ as ${\theta }_0\to 0$ concerns radial modulation of a small nonzero circular orbit; the exactly planar $h=0$ pendulum is a different, degenerate limit of these polar coordinates.

For a sphere with uniform mass density, $M=(4\pi /3)\rho a^3$. Its moment of inertia of a uniform solid sphere is $I=2M{a}^2/5$, and rotation about a diameter gives angular momentum magnitude $L=I\omega$. The spheres have the same angular frequency $\omega$, soThe mass grows by a factor of two, and the extra squared lever arm supplies a further factor of four. Equal angular frequencies alone do not mean equal angular momenta.

Let $\mathbf{r}$ be the orbiting point mass's position relative to the attracting body, and let its velocity be $\mathbf{v}=\dot{\mathbf{r}}$. Its angular momentum isConservation of angular momentum keeps both the magnitude and direction of $\mathbf{L}$ fixed. Since $\mathbf{L}$ is perpendicular to $\mathbf{r}$, a nonradial orbit lies in a fixed plane. During a short interval $dt$, the radius vector sweeps a thin triangle of areaThus the areal velocity is constant:Equal time intervals give equal swept areas: this is Kepler's second law. In plane polar coordinates, the same statement is $L=m{r}^2\dot{\varphi }$ and $\dot{A}=r^2\dot{\varphi }/2$. The attracting body is at the relevant focus of the ellipse, not its geometric centre. The area argument needs conserved angular momentum, not specifically an ellipse; a central force provides this conservation because its torque $\mathbf{r}\times \mathbf{F}$ vanishes.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.

Put $\mu =GM$. Balancing Newton's law of universal gravitation with the required centripetal force gives $m{v}_0^2/r_0=\mu m/r_0^2$, henceThe circular Kepler orbit is stable against small radial perturbations: its effective radial potential is $U_{\mathrm{eff}}=J^2/(2m{r}^2)-\mu m/r$, and at the circular value $J^2=\mu m^2{r}_0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=\mu m/r_0^3>0$. Here $J$ is the conserved angular momentum.