= Solution
For a <circular orbit>, $v^2/r=GM/r^2$. The relation for <circular orbital speed from surface gravity> eliminates the planet's <gravitational parameter>: $GM=gR^2$. The orbital radius is $r=R+h=7.4\times10^6$ m, so
$$
\boxed{v=\sqrt{\frac{gR^2}{R+h}}=\sqrt{\frac{10(6.4\times10^6)^2}{7.4\times10^6}}\simeq7.44\times10^3\,\mathrm{m\,s^{-1}}.}
$$
This is about \b[7.4 km/s], under the circular-orbit and spherical-Earth approximation. Using surface $g$ unchanged at orbital height would overestimate the <centripetal acceleration>; its role here is to determine $GM$.
Back to article page
