A star 2.5 times the mass of the sun and collapsed to a size of 42 km

1.	A star 2.5 times the mass of the sun and collapsed to a size of 42 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 × 1030 kg).
Answer» 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 (F_G) > Centrifugal force (F_c) Then the object on the equator remains stuck on the star. Mass of Star, M = 2.5 (M_Sun) = 2.5 × 2 × 10³⁰ kg = 5 × 10³⁰ kg Radius of Star = 12 km = 1.2 × 10⁴ m ⇒ F_G = \(\frac {GMm}{R^2}\) = \(\frac {6.67 \times 10^{-11} \times 5 \times 10^{30}}{(1.2 \times 10^4)^2}\) = \(\frac {6.67 \times 5}{1.44}\) × 10¹¹ × m = 2.316 × 10¹² × m N F_c = mr ω² Where, r = radius of Star = 12 km ω = angular speed = 2 × π × (1.2) ⇒ F_c = m × 1.2 × 10⁴× (2.4π)² = 6.82 × 10⁵ × m N Since F_c < F_G the object is stuck to Star’s surface.

A star 2.5 times the mass of the sun and collapsed to a size of 42 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 × 1030 kg).

Answer»

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 (F_G) > Centrifugal force (F_c)

Then the object on the equator remains stuck on the star.

Mass of Star, M = 2.5 (M_Sun)

= 2.5 × 2 × 10³⁰ kg

= 5 × 10³⁰ kg

Radius of Star = 12 km = 1.2 × 10⁴ m

⇒ F_G = \(\frac {GMm}{R^2}\) = \(\frac {6.67 \times 10^{-11} \times 5 \times 10^{30}}{(1.2 \times 10^4)^2}\)

= \(\frac {6.67 \times 5}{1.44}\) × 10¹¹ × m

= 2.316 × 10¹² × m N

F_c = mr ω²

Where, r = radius of Star = 12 km

ω = angular speed = 2 × π × (1.2)

⇒ F_c = m × 1.2 × 10⁴× (2.4π)²

= 6.82 × 10⁵ × m N

Since F_c < F_G the object is stuck to Star’s surface.