Solution
= Solution
Taking the <Earth>'s sidereal rotation period as $86164\,\mathrm s$, $\omega=2\pi/(86164\,\mathrm s)$, and $G=6.6743\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ gives
$$
\boxed{\eta\simeq6.61.}
$$
The <orbital radius> is therefore about $6.61R$, or an altitude of about $5.61R$. Replacing the sidereal day by $86400\,\mathrm s$ changes this estimate by less than $0.2\%$.