Answer
$2.5\times 10^{-3}\ rad/s$
Work Step by Step
Given:
Radius of circular turn $ r = 3220 km= 3220 \times 10^{3} m$
Speed of spaceship $v = 29000\ km/h =29000\times \frac{1000m}{3600 s }=8055.55\ m/s$
Therefore, the angular speed is;
$\omega = \frac{v}{r}$
$\omega = \frac{8055.55\ m/s}{3220 \times 10^{3} m}$
$\omega = 2.5\times 10^{-3}\ rad/s$
