Q: A star 2.5 times the mass of the sun is reduced to a size of 12 km and rotates with a speed of 1.5 rps. Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the Sun = 2 × 10^{30} kg).

Sol: Acceleration due to gravity, g = GM/R^{2}

$\large g = \frac{6.67 \times 10^{-11} \times 2.5 \times 2 \times 10^{30}}{(12000)^2} $

= 2.3 × 10^{12} m/s^{2}

Centrifugal acceleration

$\large = r \omega^2 = r(2 \pi f)^2 $

$\large = 12000(2 \pi \times 1.5)^2 $

= 1.1 × 10^{6} m/s^{2}

Since, g > r ω ^{2},the body will remain stuck with the surface of star.