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Let $v_K=\sqrt{\mu /a}$ be the speed of the circular Kepler orbit at radius $a$. The release velocity relative to the planetesimal isThe particle starts at the same position as its parent, so the change in specific orbital energy isUsing the velocity components from part (c),Since $C=-\mu /(2a)$ and $C^{\prime }=-\mu /(2a^{\prime })$, rearrangement gives

Solved by gpt-5.6-sol high.

For $T_p=5/2$ and $I=0$, the fixed-Tisserand parameter curve begins atIt falls smoothly toand then rises asymptotically back toward $e=1$ as $a/a_p\to \infty$. Since $T_p<3$, this coplanar locus never reaches a circular Kepler orbit.

Solved by gpt-5.6-sol high.

Uniform ${\rho }_0,p_0,$ and ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$ obey the unperturbed ideal magnetohydrodynamics equations whenIndeed $({\mathbf{v}}_0\cdot \nabla ){\mathbf{v}}_0=0$, while the Coriolis and tidal terms cancel:

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations areEliminating $B_x,B_y,v_x,v_y$ and writing the Alfvén speed as $v_A^2=B_0^2/(4\pi {\rho }_0)$ givesSince $d{\Omega }_0^2/d\log r=-2q{\Omega }_0^2$, a root has ${\omega }^2<0$ precisely when the constant term is negative:This is the magnetorotational instability criterion.

For a circular Kepler orbit, $q=3/2$. With $K=k^2{v}_A^2$, the unstable branch isMinimizing it givesand henceThe e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2{v}_A^2<3{\Omega }_0^2$, making the critical wavelength of the magnetorotational instabilityFor a thin isothermal disk, $H=c_{s,0}/{\Omega }_0$. Setting ${\lambda }_{\mathrm{crit}}=2H$ gives $v_A^2=3c_{s,0}^2/{\pi }^2$, and thereforeAbove this field strength the shortest unstable vertical MRI wavelength exceeds the full disk thickness, so no such vertical mode fits inside the disk. Magnetic tension then stabilizes this local mode; the result corresponds to a magnetic-to-gas pressure ratio $B^2/(8\pi p)=3/(2{\pi }^2)$.

Solved by gpt-5.6-sol high.