Question

th core of a star is the size of our sun with mass 2 times as great as the sun and is rotating at a frequency of 1 rev evru 100 days. If it were to undergo gravitational collapse to a neutron star of radius 10 kn what would its rtation frequency be

Answer 1

Answer:

3522.7 rad/s

Explanation:

From the law of conservation of angular momentum,

I₁ω₁ = I₂ω₂

I₁, ω₁ and I₂, ω₂ are the rotational inertia and angular speed of the sun and neutron star respectively.

ω₂ = I₁ω₁/I₂     I₁ = ²/₅M₁R₁² and I₂ = ²/₅M₂R₂² where M₁,M₂ and R₁,R₂ are the masses and radii of star and neutron star respectively.

ω₂ = I₁ω₁/I₂ = ²/₅M₁R₁²ω₁÷²/₅M₂R₂² = (M₁/M₂)(R₁/R₂)²ω₁

M₁ = M₂, R₁ = radius of sun = 6.96 10⁵ km and R₂ = 10 km, ω₁ = 1 rev/100days = 2π/(100 × 24 × 60 × 60 s) = 7.272 × 10⁻⁷ rad/s

ω₂ = (M₁/M₂)(R₁/R₂)²ω₁ = 1 × ( 6.96 10⁵ km/10 km)² × 7.272 × 10⁻⁷ rad/s

     = 3522.7 rad/s