Question

Most stars maintain an equilibrium size by balancing two forces - an inward gravitational force and an outward force resulting from the star's nuclear reactions. When the star's fuel is spent, there is no counterbalance to the gravitational force. Whatever material is remaining collapses in on itself. Imagine a star of mass 3.25⋅10³⁰ kg and radius 8.67⋅10⁸m that rotates once in 29.0 days. Suppose this star undergoes gravitational collapse to form a neutron star of radius 11.3 km. Determine its rotation period.

Answer 1

To calculate the neutron star's rotation period after stellar collapse, we use the conservation of angular momentum. The initial angular momentum of the star with known mass and original rotation period is equated with the final angular momentum post-collapse to determine the new rotation period of the much smaller neutron star.

Step-by-step explanation:

To determine the rotation period of a neutron star formed from a stellar collapse, we use the principle of conservation of angular momentum. When a star of mass 3.25⋅10³⁰ kg with an original radius of 8.67⋅10¸ m rotates once every 29.0 days and then collapses into a neutron star with a radius of 11.3 km (11.3×10³ m), its rotation speed increases significantly due to the principle that states 'if an object gets smaller, it can spin more rapidly'.

The angular momentum of the original star is given by L = Iω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for a sphere is I = (2/5)MR², and ω can be obtained from the rotation period. Since angular momentum is conserved, L_initial = L_final, and we can find the final angular velocity (ω_final) and thus the rotation period of the neutron star.