Question

The core of a star collapses during a supernova, forming a neutron star. Angular momentum of the core is conserved, and so the neutron star spins rapidly. If the initial core radius is 5.0×10⁵ km and it collapses to 10.0 km, find the neutron star’s angular velocity in revolutions per second, given the core’s angular velocity was originally 1 revolution per 30.0 days.

a)2×10⁻³rev/s
b)5×10⁻⁵rev/s
c)1×10⁻⁴rev/s
d)3×10⁻⁶rev/s

Answer 1

Angular momentum conservation is applied to a collapsing stellar core, using the formula L = I×ω and taking into account the mass cancellation, to find the neutron star's angular velocity as approximately 2×10^-3 revolutions per second. The correct option is a).

Step-by-step explanation:

The question deals with the conservation of angular momentum in astrophysics, specifically regarding a neutron star formed from the remnants of a collapsed stellar core. The core originally had a radius of 5.0×105 km and an angular velocity of 1 revolution per 30 days, which needs to be calculated for a much smaller radius of 10 km, to find the new angular velocity of the neutron star. To find the angular velocity after the collapse, we apply the principle that angular momentum (L = I×ω, where I is moment of inertia and ω is angular velocity) is conserved.

Given that the moment of inertia for a sphere is I = (2/5)mr2, the conservation of angular momentum can be written as (Iinitial×ωinitial) = (Ifinal×ωfinal). By canceling the mass (m) and solving for ωfinal, we see that it is inversely proportional to the square of the radius, and therefore: ωfinal = ωinitial×(rinitial/rfinal)2. Converting the initial rotation period from days to seconds and plugging in the given values, we find that the angular velocity of the neutron star in revolutions per second is approximately 2×10-3 rev/s, making option a) the correct answer.