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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 2525 days? Assume that the Sun and the white dwarf are uniform spheres.

Solution

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Answered 2 years ago
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In this problem we'll consider Sun the solid sphere which rotational inertia is going to be:

I1=25MRs2I_1=\frac{2}{5}MR_s^2

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

I2=25MRe2I_2=\frac{2}{5}MR_e^2

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

L1=L2L_1=L_2

ω1I1=ω2I2\omega_1I_1=\omega_2I_2

2πT1I1=2πT2I2\frac{2\pi}{T_1}I_1=\frac{2\pi}{T_2}I_2

I1T1=I2T2\frac{I_1}{T_1}=\frac{I_2}{T_2}

T2=I2T1I1=25MRe2T125MRs2T_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}

T2=Re2T1Rs2T_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 T1=600 hT_1=600\ \text{h}. Radius of Earth is 6370 km an radius of Sun is 696.340 km.

T2=(6370km)2600 h(696.340km)2T_2=\frac{{(6370 km)}^2 \cdot 600\ \text{h}}{{(696.340 km)}^2}

T2=0.05 hT_2=0.05 \ \text{h}

T2=3 min\boxed{T_2=3\ \text{min}}

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