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21 April, 22:50

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

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  1. 21 April, 23:20
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    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
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