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.
Question:

A star $2.5$ times the mass of the sun and collapsed to a size of $12 \mathrm{~km}$ 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 $=2 \times 10^{30} \mathrm{~kg}$ ).

Solution:

Yes

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_{\mathrm{g}}=\frac{\mathrm{G} M m}{R^{2}}$

Where,

$M=$ Mass of the star $=2.5 \times 2 \times 10^{30}=5 \times 10^{30} \mathrm{~kg}$

$m=$ Mass of the body

$R=$ Radius of the star $=12 \mathrm{~km}=1.2 \times 10^{4} \mathrm{~m}$

$\therefore f_{\mathrm{g}}=\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^{4}\right)^{2}}=2.31 \times 10^{11} \mathrm{~m} \mathrm{~N}$

Centrifuaal force, $f_{c}=m r \omega^{2}$

$\omega=$ Angular speed $=2 \pi v$

$v=$ Angular frequency $=1.2$ rev $s^{-1}$

$f_{\mathrm{c}}=m R(2 \pi v)^{2}$

$f_{\mathrm{c}}=m R(2 \pi v)^{2}$

$=m \times\left(1.2 \times 10^{4}\right) \times 4 \times(3.14)^{2} \times(1.2)^{2}=1.7 \times 10^{5} \mathrm{mN}$

Since $f_{g}>f_{c}$, the body will remain stuck to the surface of the star.

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