Rorschach

Inner radius (of torus)

$$ \begin{align*} R &= 15e3\ m\ (30km\ long) \\ V &= 2e8\ m^3 \\ \\ Now, V &= (2\pi R) \times (\pi r^2)\ (toroidal) \\ \therefore r &= \frac{210^8}{2 \times \pi^2 \times 15 \times 10^3}\\ \bold{r }&\approx\ \bold{7 \times 10^3\ m} \end{align} $$

Orbital velocity

$$ \begin{align*} h\ (\text{perigee above atm}) &= 1.5e6\ m\\ r_{BB} &= 1.6e7\ m\\ M_{BB} &= 1.6e28\ kg\\ \\ V_o& = \sqrt{\frac{GM}{r+h}}\\ &= \sqrt{\frac{6.6 \times 10^{-11} \times 1.6 \times 10^{28}}{1.6 \times 10^7 \times 1.5 \times 10^{6}}} \\ \therefore \bold{V_o} &\approx \bold{200\ m/s} \end{align*} $$