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The source expression $E=\frac{1}{2}\mu v^2-GMm/r$ is total mechanical energy in the centre-of-mass frame, not kinetic energy alone. Take $v_0$ as the relative speed at a large initial separation where the mutual potential is negligible. At closest approach, the radial relative velocity is zero, so conservation of angular momentum and conservation of energy giveUsing $Mm/\mu =M+m$ eliminates $v_c$:An impact occurs when $r_c\le R$, where $R=r_M+r_m$. Since the right side increases with positive $r_c$, the collision threshold isThe mutual escape velocity is the relative speed at contact required to reach infinite separation with zero residual speed. Setting total mechanical energy to zero givesThis includes target recoil. If the given initial speed were instead at a finite non-negligible separation, its potential energy would have to be retained; $v_0$ would then not equal the asymptotic speed used in the focusing formula.

In the classical smooth setting, a deformation in continuum mechanics is a sufficiently regular, injective map $y:\Omega \to {\mathbb{R}}^3$ with $\det Dy>0$. Injectivity excludes material interpenetration, and a positive determinant preserves orientation and local volume. We use $C^1$ regularity for these kinematics and additional differentiability where stress-divergence identities are applied.

Cauchy's stress theorem states that the contact force on an oriented surface element at a current point depends on the point and its unit normal, rather than on the shape of the chosen subbody. Thus the contact force on the boundary of a current subbody $v$ has the form ${\int }_{\partial v}t(z,n)da$. Static force balance, for a body force $b$ per unit current volume, isfor every subbody. A shrinking pillbox gives $t(z,-n)=-t(z,n)$; a shrinking tetrahedron then gives linear dependence on the normal, $t(z,n)=S(z)n$. Applying the divergence theorem and using arbitrary subbodies givesIf body force is specified per unit mass instead, replace $b$ by $\rho b$. Symmetry of $S$ follows additionally from conservation of angular momentum in the absence of body couples; it is not a consequence of force balance alone.

Place two equal point masses of $1\mathrm{kg}$ each on opposite sides of a fixed origin, at equal radius $r$, and let them share a signed angular velocity $\omega =\dot{\theta }$ in a plane. Their total mass is $M=2\mathrm{kg}$ and their moment of inertia about the perpendicular axis is $I=M{r}^2$. Radial pulling forces have zero torque about that origin, soTake an initial radius $r_i=1\mathrm{m}$ and angular velocity ${\omega }_i=1rad/s$, and a final radius $r_f=0.5\mathrm{m}$. ThenandThese are endpoint states with ${\dot{r}}_i={\dot{r}}_f=0$. During the pull, the full kinetic energy also contains radial motion: $K=M{\dot{r}}^2/2+K_{\mathrm{rot}}$. Assume the pulling apparatus supplies all net work to the two point masses; the work-energy theorem then fixes its net work at $+3\mathrm{J}$, including any radial acceleration and braking. Each mass spirals inward during the pull, allowing its inward force to do positive work. Conservation of angular momentum therefore predicts the final angular velocity without requiring kinetic energy conservation. For an apparatus included in a larger system, its stored energy supplies the increase consistently with conservation of energy. OpenStax uses the same mechanism to explain a skater pulling her arms inward. ^[ref]

For conservative binary mass transfer and $R_L\propto a(M_d/M{)}^{1/3}$,The conservation of angular momentum gives $d\log a/d\log M_d=-2(1-q)$. Since a mass loss has $d\log M_d<0$, the donor moves inside its lobe when ${\zeta }_{\mathrm{ad}}>{\zeta }_L$. An ideal fully convective donor with ${\zeta }_{\mathrm{ad}}=-1/3$ is therefore stable for $q<2/3$ and unstable for $q>2/3$ under these assumptions.

Let $\theta$ be the scattering angle between incoming and outgoing velocities, so forward scattering is $\theta =0$ and exact reversal is $\theta =\pi$. Define the length $a=\kappa /(2E)$. ThenFor $b>0$, this gives $0<\theta <\pi$; $b=0$ is the head-on limiting case $\theta =\pi$. Large deflections select small impact parameters, rather than requiring every incident alpha particle to reverse.

A short derivation uses conservation of angular momentum and conservation of energy. Choose the polar angle $\varphi =0$ at closest approach, and write $u=1/r$. The Binet equation for the repulsive Coulomb potential givesThe last equality follows by substituting this orbit into $E=m{\dot{r}}^2/2+L^2/(2m{r}^2)+\kappa /r$. The hyperbola reaches infinity at $\varphi =\pm {\varphi }_{\infty }$, with $\cos {\varphi }_{\infty }=1/ϵ$. Its scattering angle is $\theta =\pi -2{\varphi }_{\infty }$, henceThis proves the relation without assuming that the deflection is small. The symbol $ϵ$ here is a dimensionless orbital parameter, distinct from ${\epsilon }_0$.

For inward accretion rate $\dot{M}>0$, the steady conservation of angular momentum equation is $\mathcal{G}=\dot{M}(h-h_{\mathrm{in}})$ when the inner torque vanishes. The viscous torque in an accretion disk is $\mathcal{G}=2\pi qh\bar{\nu }\Sigma$, where $h=r^2\Omega$ and $q$ is the orbital shear parameter. ThusThe familiar Keplerian accretion disk follows when $q=3/2$ and $h\propto r^{1/2}$. The general form also applies to the non-Keplerian Paczyński-Wiita circular orbit rotation law.

For a positive inner magnetic torque $T_m$, define $\lambda =T_m/(\dot{M}{h}_{\mathrm{in}})$. Integrating steady conservation of angular momentum gives $\mathcal{G}=T_m+\dot{M}(h-h_{\mathrm{in}})$. With Keplerian rotation, this determines the surface density of a disk in the displayed formula.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

For steady injection into a Keplerian disk between two zero-torque boundaries, conservation of mass gives $\dot{M}={\dot{M}}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}$. Conservation of angular momentum gives $\dot{M}{h}_0={\dot{M}}_{\mathrm{in}}{h}_{\mathrm{in}}+{\dot{M}}_{\mathrm{out}}{h}_{\mathrm{out}}$, where $h(R)=\sqrt{GMR}$. Eliminating the outward mass flux gives the displayed inward fraction. This fraction does not depend on the viscosity law: the two boundary fluxes must carry away the injected mass and angular momentum. Equivalently, continuity of the viscous torque in an accretion disk joins $\nu \Sigma \sqrt{R}={\dot{M}}_{\mathrm{in}}(\sqrt{R}-\sqrt{R_{\mathrm{in}}})/(3\pi )$ inside the source to ${\dot{M}}_{\mathrm{out}}(\sqrt{R_{\mathrm{out}}}-\sqrt{R})/(3\pi )$ outside.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

In an axisymmetric thin disc with no vertical mass or torque flux, integrate mass and conservation of angular momentum over height. With fixed height-independent $l=r^2\Omega$, the equations are ${\partial }_t\Sigma +r^{-1}{\partial }_r(r\Sigma v_r)=0$ and ${\partial }_t(\Sigma l)+r^{-1}{\partial }_r(r\Sigma v_{r}l-r^3\bar{\nu }\Sigma {\Omega }^{\prime })=0$. Their difference gives $r\Sigma v_r{l}^{\prime }=(r^3\bar{\nu }\Sigma {\Omega }^{\prime }{)}^{\prime }$. The density-weighted viscosity of a disk supplies the integrated stress.

Subtract local specific angular momentum times mass conservation from conservation of angular momentum including wind losses S,T. With outward radial mass flux $\mathcal{F}$ and outward viscous torque in an accretion disk $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$, this gives $\mathcal{F}{h}^{\prime }+{\mathcal{G}}^{\prime }=2\pi r(Sh-T)$. For Keplerian rotation and $T=qSh$, substitution in mass conservation yields the displayed equation. A wind with q=1 removes local angular momentum but supplies no additional drift term; q>1 can drive accretion even with zero viscosity.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.