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  • 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. Will an object placed on its equator remain stuck to its surface due to gravity ?

Q: 8.23 A star \small 2.5 times the mass of the sun and collapsed to a size of  \small 12 \hspace{1mm}km rotates with a speed of \small 1.2 \hspace{1mm} 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 \small =2 \times 10^3^0 \hspace{1mm}kg).

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A body will get stuck at the star's surface if the centrifugal force of star is less than the gravitational force.

The gravitational force is given by :

                                                        F_g\ =\ \frac{GMm}{r^2}

or                                                             =\ \frac{6.67\times 10^{-11}\times 5\times 10^{30}\times m}{(1.2\times 10^4)^2}\ =\ 2.31\times 10^{12}\ m\ N

The centrifugal force is given by :

                                                          F_c\ =\ mr\omega ^2

or                                                               =\ mr(2\pi v) ^2=\ m\times 1.2\times 10^4\times (2\pi \times 1.2) ^2

or                                                               =\ 6.8\times 10^5\ m\ N

As we can see that the gravitational force is greater than the centrifugal force thus the body will remain at the star.

Posted by

Devendra Khairwa

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