Gravitation

Question : 23 of 25

Marks: +1, -0

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 \times 10^{30}

kg).

Solution:

The object will remain stuck to the surface of star due to gravity, if the acceleration due to gravity is more than the centrifugal acceleration due to its rotation.

Acceleration due to gravity,

g = \frac{G M}{R^{2}}

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

= 2.3 \times 10^{12} \,\text{m}\,\text{s}^{-2}

Centrifugal acceleration

= r \omega^{2} = r (2 \pi v)^{2}

= 12000 (2 \pi \times 1.5)^{2}

= 1.1 \times 10^{6} \,\text{m}\,\text{s}^{-2}

Since

g > r \omega^{2}

, therefore the body will remain stuck with the surface of star.

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