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Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

For a nearly radial Kepler orbit, $q\simeq h^2/(2GM)$, so modest changes in specific angular momentum can strongly change pericentre. A planetary torque acting for a reference orbital time gives $\delta h\sim (m_p/M)\sqrt{GM{a}_p}$. Coherent kicks change $q$ by order itself after $N\sim (M/m_p)\sqrt{q/a_p}$ orbital-scale encounters, whereas uncorrelated kicks require an order-$N^2$ random-walk count. Both estimates are conditional on encounter geometry: a close encounter with a smaller impact parameter can give a much larger kick.

Suppose a body of mass $M_0$ has centre of mass ${\mathbf{R}}_0$, and material of mass $M_c<M_0$ with centre of mass ${\mathbf{R}}_c$ is removed. Additivity of the first mass moment gives the remaining centre of massWhen the original centre of mass is at the origin, the remaining centre of mass points opposite the removed material's first mass moment. For a constrained rigid body rotating about a fixed axis, this displaced centre of mass undergoes centripetal acceleration even if the applied axial torque is zero.

For a fixed mass on a Keplerian circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}$ has derivative $M_s{r}_0\Omega /2$. The torque on the satellite is minus the torque $\Gamma$ on the disk. This gives the displayed drift law. In the density-slope satellite torque model, ${\dot{r}}_0=-(8\chi /9)\beta q({\Sigma }_0{r}_0^2/M_{\ast })r_0\Omega (H/r_0{)}^{-2}$.

A local odd mass density slope $\Sigma (x)-\Sigma (-x)=2{\Sigma }_0\beta x/r_0$ changes the cancelling one-sided torques into this net torque. A local outer cutoff $L$ adds the factor $1-(H/L{)}^2$. Positive slope transfers net angular momentum from satellite to disk. An infinite linear mass density law is only a formal local extension, since it cannot remain nonnegative on both sides.

At fixed outer boundary velocity, extending the inclusion velocity rigidly through its interior makes the particle flow an admissible competitor for the particle-free Minimum-dissipation theorem for Stokes flow. Its interior rate-of-strain tensor vanishes. For a linear background, the increase in viscous dissipation is force times relative translation, torque times relative rotation, plus the contraction of the particle stresslet tensor with background strain.

At a flat shear-free impermeable plane, a tangential torque is reflected with opposite sign because torque is an axial vector. A parallel axial rotlet dipole therefore has an opposite signed moment at the reflected source. Its velocity vanishes at the point directly beneath the image, but its velocity gradient can turn the swimmer.

Two integrations of the steady Keplerian viscous diffusion equation give this general transported quantity. $A$ fixes the mass accretion rate, while $B$ fixes the additive angular-momentum flux or inner torque. A zero-torque inner boundary condition gives $B=-A\sqrt{r_{\mathrm{in}}}$; a nonaccreting constant-torque disk has $A=0$.

The hydrodynamic torque on a body about a reference point is $T={\int }_{S}r\times (\sigma \cdot n)dS$, with $r$ measured from that point and $n$ pointing from the body into the fluid. The stress includes pressure and viscous contributions, and $\sigma \cdot n$ is the fluid traction on the body. Changing the reference point by $d$ changes the torque by $-d\times F$, where $F={\int }_S\sigma \cdot ndS$ is the fluid force. A torque-free body has zero resultant torque, including any external torque.

Use $l=\sqrt{G{M}_{\ast }r}$ and ${\Omega }^{\prime }=-(3/2)\Omega /r$ in the vertically averaged viscous disk equations. The radial drift is $v_r=-3(\Sigma \sqrt{r}{)}^{-1}{\partial }_r(\sqrt{r}\bar{\nu }\Sigma )$. Conservation of mass gives this nonlinear diffusion equation. It assumes the central rotation law remains fixed and excludes wind/surface torque terms.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

A steady Keplerian disk can transport angular momentum by viscous stress with zero mass flux. The radial drift vanishes, but the torque is constant and nonzero. An inner source and outer sink must supply its boundary angular-momentum exchange. This is distinct from a decretion flow, which transports mass outward.

Multiply encounter mass flux $\Sigma (x)Sxdx$ by specific angular-momentum gain $r_0\Delta v_y$. The net disk torque replaces $\Sigma (x)$ by $\Sigma (x)-\Sigma (-x)$, so a symmetric mass density gives zero net torque. For constant surface density, Keplerian shear and the full-flyby value $\chi =2$, the one-sided coefficient is $8/27$ in $q^2\Sigma r_0^4{\Omega }^2(r_0/H{)}^3$, matching the primary coplanar impulse normalization.