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Question

# Suppose that the Sun runs out of nuclear fuel and suddenly collapses to form a so-called white dwarf star, with a diameter equal to that of the Earth. Assuming no mass loss, what would then be the new rotation period of the Sun, which cur- rently is about $25$ days? Assume that the Sun and the white dwarf are uniform spheres.

Solution

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In this problem we'll consider Sun the solid sphere which rotational inertia is going to be:

$I_1=\frac{2}{5}MR_s^2$

$M$ is mass of sun, so $R_s$ is radius of the sun. Also the rotational inertia after Sun has shrinked to size of earth is going to be:

$I_2=\frac{2}{5}MR_e^2$

Here the point is that angular momentum is going to reamin unchanged, so we can write:

$L_1=L_2$

$\omega_1I_1=\omega_2I_2$

$\frac{2\pi}{T_1}I_1=\frac{2\pi}{T_2}I_2$

$\frac{I_1}{T_1}=\frac{I_2}{T_2}$

$T_2=\frac{I_2 T_1}{I_1}=\frac{\displaystyle\frac{2}{5}MR_e^2 T_1}{\displaystyle\frac{2}{5}MR_s^2}$

$T_2=\frac{R_e^2 T_1}{R_s^2}$

We know that one full rotation of Sun takes 25 days so the peirod is $T_1=600\ \text{h}$. Radius of Earth is 6370 km an radius of Sun is 696.340 km.

$T_2=\frac{{(6370 km)}^2 \cdot 600\ \text{h}}{{(696.340 km)}^2}$

$T_2=0.05 \ \text{h}$

$\boxed{T_2=3\ \text{min}}$

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