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 xx 10^(30) kg).

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

1.	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 xx 10^(30) kg).
Answer» Solution :`implies` Mass of SUN `= 2 xx 10^(30) kg` Mass of neutroní stars `M = 2.5 xx 2 xx 10^(30) kg` ` = 5 xx10^(30) kg` Radius of neutron stars `R = 12 km = 1.2 xx 10^(4 )m` Frequency revolution `F = 1.5 Hz ` Let the acceleration DUE to gravity on the surface of neutron stars is g, `g=(GM)/(R^2)` `= (6.67xx10^(-11) xx 5 xx 10^(30))/((1.2xx10^(4))^2)` `:. g = (GM)/(R^2) = 2.3 xx10^(12) ms^(-2)` Centrifugalacceleration `a_c` gets in body at equator `:. a_c = v^2/R` `:. a_c = Romega^2""[ :. v = 2Romega]` ` = R xx (2pif)^2` `= R pi^2Rf^2` `= 4xx(3.14)^2 xx1.2 xx10^(14) xx(1.5)^2` `= 1046.48 xx10^8` `:. a_c = omega^(2) R = 1.1 xx 10^(6) ms^(-2)` `g gt gt a_c` inward force at equator is more than centripetal force. So, the body will remain struck to the surface of the star due to gravitational force of star,