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

Answer 1

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