Since the Star is rotating, there is a force pushing it outward (centrifugal force) and a force pulling it inwards, (gravitational force).
If gravitational force (FG) > Centrifugal force (Fc)
Then the object on the equator remains stuck on the star.
Mass of Star, M = 2.5 (MSun)
= 2.5 × 2 × 1030 kg
= 5 × 1030 kg
Radius of Star = 12 km = 1.2 × 104 m
⇒ FG = \(\frac {GMm}{R^2}\) = \(\frac {6.67 \times 10^{-11} \times 5 \times 10^{30}}{(1.2 \times 10^4)^2}\)
= \(\frac {6.67 \times 5}{1.44}\) × 1011 × m
= 2.316 × 1012 × m N
Fc = mr ω2
Where, r = radius of Star = 12 km
ω = angular speed = 2 × π × (1.2)
⇒ Fc = m × 1.2 × 104 × (2.4π)2
= 6.82 × 105 × m N
Since Fc < FG the object is stuck to Star’s surface.