Question

A star 2.5 times the mass of the sun and collapsed to a size of radius 12 km rotates with a speed of 1.2 revolution 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? ($\text{Mass of the sun}$ = $2 \times {10^{30}}kg$).

Answer 1

Hint: If the gravitational force of the star exceeds than the centrifugal force then the object placed on its equator remains stuck to its surface otherwise if it becomes greater, the object will not remain on the surface.

Step by step solution:
Given: Mass of star, $M = 2.5 \times 2 \times {10^{30}}kg = 5 \times {10^{30}}kg$
Radius of star, $r = 12000m$
Angular velocity, $\omega = 2\pi \times 1.2 = 7.54 rev/sec$

We know that for a body to remain stuck on the surface of the star the gravitational force should be greater than its centrifugal force that is opposite to the gravitational force in direction.

Gravitational force of a body at the surface is given by the formula,
${F_g} = \dfrac{{GMm}}{{{r^2}}}$
where $m$= mass of the object

On putting the values we get
${F_g} = \dfrac{{6.67 \times {{10}^{ - 11}} \times 5 \times {{10}^{30}} \times m}}{{{{\left( {12000} \right)}^2}}}$$ = 2.31 \times {10^{12}}mN$

Centrifugal force of a rotating body is given by,
${F_c} = m{\omega ^2}r$
${F_c} = m \times {\left( {7.54} \right)^2} \times 12000 = 6.82 \times {10^5}mN$

Here ${F_g} \gg {F_c}$ therefore the object will remain stuck to the surface of the star.
Hence yes, an object placed on its equator will remain stuck to its surface due to gravity.

Note: We can also find the answer by comparing the only values of g on star and ${\omega ^2}r$. After finding it we will get the value of g on the star will be more than ${\omega ^2}$ therefore the body will not remain stuck to the surface of the star.