The equilibrium orbit is circular with radius a₀, so that $h_0=m a_0^2 \omega_0$ and, since it is co-rotating with the black hole, $\omega_0=\omega_{\rm Kerr}$ . With respect to the moment of inertia of the black hole $m_1 r_{\rm (1)g}^2 R_1^2=J_{\rm bh}/\omega_{\rm Kerr}$ , the moment of inertia of the asteroid can be neglected. According to Ashtekar & Krishnan (2004), $J_{\rm bh}/\omega_{\rm Kerr}=M_{\rm bh}R_{\rm h}^2$ , where $R_{\rm h}$ is the horizon radius of the black hole, which becomes $2~r_{\rm g}$ in the limit of small $a_{\rm Kerr}$ .

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The exchange $\alpha \rightarrow \alpha^\prime$ changes the unstable equilibrium state $\tilde r_-$ to the attractor $\tilde r_{\rm p}=1$ and vice versa with respect to the tracks shown in Hut (1982).

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... $\alpha^\prime=9720$ )

This is an unreasonably large number with respect to our problem, but sufficiently small and convenient to show the topology of the solution space.

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...Hut (1981)

For the sake of simplicity we neglect the contribution of work done by the tangential component of tidal force, which depends on the difference between the orbital and spin period.

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