A star 2.5 times the mass of the sun and collapsed to a size of 12 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).

Asked by Pragya Singh | 1 year ago |  149

##### Solution :-

Any matter/ object will remain stuck to the surface if the outward centrifugal force is lesser than the inward gravitational pull.

Gravitational force, fG

$$\dfrac{GM\;m}{R^{2}}$$= ​ [Neglecting negative sign]

Here,
M = Mass of the star = 2.5× 2 × 1030 = 5 × 1031 kg
m = Mass of the object
R = Radius of the star = 12 km = 1.2 ×104 m
Therefore, fG

$$\dfrac{(6.67\times 10^{-11})\times (5\times 10^{31})\;m}{(1.2\times 10^{4})^{2}}$$

$$= 1.334\times 10^{13}m\;N$$

Now, Centrifugal force, fc= m r ω2
Here, ω = Angular speed = 2πν
ν = Angular frequency = 10 rev s–1
fc = m R (2πν)2

fc = m × (1.2 x 104) × 4 × (3.14)2 × (1.2)2

= (6.82 ×105m) N

As fG > fC, the object will remain stuck to the surface of black hole.

Answered by Sudhanshu | 1 year ago

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