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The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant

Solved by gpt-5.6-sol high.

Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

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The kick changes the tangential velocity by $\gamma v_K\cos (\varphi -f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new specific angular momentum isEvery Kepler orbit satisfies $h^{\prime 2}=\mu a^{\prime }(1-e^{\prime 2})$. Combining this identity with the value of $a/a^{\prime }$ found in part (d) giveswhere $a/a^{\prime }$ is the explicit function of $e,f,\gamma ,\varphi$ in part (d).

Solved by gpt-5.6-sol high.

Here $a_p$ is the planet's semi-major axis, while $a,e,I$ are the particle's semi-major axis, orbital eccentricity, and orbital inclination relative to the planet's plane. The formula assumes the circular restricted three-body problem: the planet-to-star mass ratio is small, the particle has negligible mass, and its motion is approximately heliocentric and Keplerian away from brief encounters. The planet's own orbit is circular.

The Jacobi constant is exactly conserved in that ideal rotating problem. Expressing it in heliocentric orbital elements away from the planet gives the approximately conserved Tisserand parameterThe first term measures the particle's normalized binding energy. The second is twice the component of its specific angular momentum normal to the planet's plane, normalized by $\sqrt{G{M}_{\ast }{a}_p}$.

Solved by gpt-5.6-sol high.

Define the surface density of a disk and outward radial mass flux byVertical integration of mass conservation eliminates the surface term because $\rho u_z\to 0$, and axisymmetry eliminates the azimuthal derivative. Hence

For $u_{\varphi }\simeq r\Omega (r)$, the specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the viscous torque in an accretion diskSubtracting $h$ times the mass equation from the integrated angular-momentum equation givesFor an axisymmetric circular flow, ${\Pi }_{r\varphi }=\mu rd\Omega /dr$. Ifthen

Combining the two conservation laws givesThe first term is the local rate of change of angular momentum per radial interval, $\mathcal{F}h$ is outward advective angular-momentum flux, and $\mathcal{G}$ is outward stress-carried angular-momentum flux.

Solved by gpt-5.6-sol high.

At the inner edge, the imposed torque isThereforeThe angular-momentum conservation law says that $\mathcal{F}h+\mathcal{G}$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\mathrm{out}}$ carries the injected angular momentum together with the angular momentum $\dot{M}{h}_{\mathrm{in}}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more specific angular momentum.

Solved by gpt-5.6-sol high.