When a star like our sun runs out of nuclear fuel, it collapses to a white dwarf star the size of the Earth, but with the same mass the star

Question

When a star like our sun runs out of nuclear fuel, it collapses to a white dwarf star the size of the Earth, but with the same mass the star had before the collapse. If a star initially rotates once every 25 days, like our sun, what will its rotation rate be after it becomes a dwarf

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Neala 2 months 2021-07-29T22:31:34+00:00 1 Answers 0 views 0

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    2021-07-29T22:33:00+00:00

    Answer:

    T = 1.8 10² s

    Explanation:

    This is an exercise of conservation of angular momentum,

    Initial instant. Before collapsing

             L₀ = I₀ w₀

    Final moment. After the collapse

             L_f = I w

    as the system is isolated, the moment is conserved

             L₀ = L_f

             I₀ w₀ = I w

             w = \frac{I_o}{I}  \ w_o

    indicates that the rotation period is 25 day, let’s reduce to the SI system

            T = 25 day (24 h / 1 day) (3600 s / 1 h) = 2.16 10⁶ s

         Angular velocity and period are related

           w = 2π/ T

           

    The moment of inertia of a sphere is

            I = 2/5 M R ²

    in this case the moment before and after the collapse is

           I₀ = 2/5 M R₀²

           I = 2.5 M R²

    the radius of a star like the sun is

          R₀ = 6.96 10⁸ m

    the radius of the earth

          R = 6.371 10⁶ m

    we substitute

         w = \frac{I_o}{I}  \ w_o  

         

         \frac{2\pi }{T}  =( \frac{R_o}{R} )^2 \ \frac{2\pi }{T_o}

          T = (\frac{R}{R_o} )² T₀

    let’s calculate

        T = (6.371 10⁶ / 6.96 10⁸ )² 2.16 10⁶

        T = 1.8 10² s

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