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During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen 鈥渇uel鈥� is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse
until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These 鈥減ulsing
stars鈥� were discovered in the 1960s and are called pulsars.
a. A star with the mass M = 2.0 x 10³⁰ kg and size R =7.0 x 10⁸ m of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.
b. What is the speed of a point on the equator of the neutron star? Your answers will be somewhat too large because a star cannot be accurately modeled as a solid sphere. Even so, you will be able to show that a star, whose mass is 10⁶larger than the earth鈥檚, can be compressed by gravitational forces to a size smaller than a typical state in the United States!

Step by step solution

01

Part(a) Step1: Given information

M = 2 x 10³⁰ kg

R =7.0 x 10⁸ m

One rotation is in 30 days

02

Part(a) Step 2: Explanation

First find the moment of inertia. Consider this sphere so

$I=\frac{2M{r}^2}{5}$

Substitute the values we get

$$\begin{gathered} I=\frac{2\left(2.0脳{10}^{鈭�}30\mathrm{kg}\right){\left(7.0脳{10}^{鈭�}8\mathrm{m}\right)}^2}{5} \\ =39.2脳{10}^{46}\mathrm{kg}{\text{m}}^2 \end{gathered}$$

Find angular velocity

$\left({蝇}_1\right)=\frac{2蟺rad}{30\text{days}}=\frac{2蟺}{(30脳24脳3600)}rad/s$

Angular momentum is constant form the momentum conservation law
We can get final angular velocity

$\left({蝇}_2\right)=\frac{2蟺}{0.10\mathrm{s}}=20蟺\mathrm{rad}/\mathrm{s}$

Now find momentum

$$\begin{gathered} L_1=I_1{蝇}_1{L}_1 \\ =\left(39.2脳{10}^{46}\mathrm{kg}路{\mathrm{m}}^2\right)\left(2.42脳\frac{{10}^{-6}\mathrm{rad}}{\mathrm{s}}\right) \end{gathered}$$

From law of momentum conservation

$$\begin{gathered} L_1=L_2 \\ {I}_1{蝇}_1=I_2{蝇}_2 \end{gathered}$$

Substitute values

$$\begin{gathered} \left(39.2脳{10}^{46}\mathrm{kg}路{\mathrm{m}}^2\right)\left(2.42脳\frac{{10}^{-6}\mathrm{rad}}{\mathrm{s}}\right) \\ =\left(I_2\right)20蟺\mathrm{rad}/\mathrm{s} \end{gathered}$$

Moment of inertia is given by $\frac{2M{r}^2}{5}$
Substitute the value and solve for r

$\left(39.2脳{10}^{46}\mathrm{kg}路{\mathrm{m}}^2\right)\left(2.42脳\frac{{10}^{-6}\mathrm{rad}}{\mathrm{s}}\right)=\frac{20蟺\mathrm{rad}}{\mathrm{s}\left(\frac{2M{r}^2}{5}\right)r^2}$

$$\begin{gathered} =\frac{5\left(39.2脳{10}^{46}\mathrm{kg}路{\mathrm{m}}^2\right)\left(2.42脳\frac{{10}^{-6}\mathrm{rad}}{\mathrm{s}}\right)}{2\left(\frac{20蟺\mathrm{rad}}{\mathrm{s}}\right)M}{r}^2 \\ =\frac{5\left(\left(39.2脳{10}^{46}\mathrm{kg}路{\mathrm{m}}^2\right)\right)\left(2.42脳\frac{{10}^{-6}\mathrm{rad}}{\mathrm{s}}\right)}{2\left(\frac{20蟺\mathrm{rad}}{\mathrm{s}}\right)\left(2.0脳{10}^{30\mathrm{kg}}\right)}r \\ =\sqrt{1.88521脳{10}^{10}}r=137.30脳{10}^3\mathrm{m} \\ 鈮�137\mathrm{km} \end{gathered}$$

03

Part(b) Step 1 : Given

M = 2 x 10³⁰ kg

R =7.0 x 10⁸ m

One rotation is in 30 days

04

Part(b) Step2: Explanation

linear velocity of a point on the equator of the neutron star can be calculated as

$$\begin{gathered} v=蝇r \\ v=(20蟺\mathrm{rad}/\mathrm{s})\left(137脳{10}^3\mathrm{m}\right) \\ v=8603600\mathrm{m}/\mathrm{s} \\ v=8.6脳{10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

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