A star with the same mass and diaeter as the sun rotates about a central axis with a period of 25 days. Suppose that this star runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. What would be the new rotation period of the white dwarf

182.5 s

Explanation:

From the law of conservation of angular momentum,

I₁ω₁ = I₂ω₂

where I₁,ω₁ are the rotational inertia and angular speed of the star and I₂,ω₂ are the rotational inertia and angular speed of the white dwarf star

I₁ = 2/5MR₁² where M = mass of star and R₁ = radius of star = radius of sun = 696340 km

I₂ = 2/5MR₂² where M = mass of white dwarf star = mass of star and R₂ = radius of white dwarf star = radius of earth = 6400 km

ω₁ = 2π/T₁ where T₁ = period of star = 25 days = 25 × 24 × 60 × 60 s = 2.16 × 10⁶ s

ω₂ = 2π/T₂ where T₂ = period of white dwarf star.

So, I₁ω₁ = I₂ω₂

2/5MR₁² × 2π/T₁ = 2/5MR₂² × 2π/T₂

R₁²/T₁ = R₂²/T₂

T₂ = T₁R₂²/R₁² = 2.16 × 10⁶ s × (6400 km/696340 km)² = 182.5 s

52 / 4 = 13 m/s

Explanation: