Question

Stars are powered by nuclear reactions that fuse hydrogen into helium. The fate of many stars, once most of the hydrogen is used up, is to collapse, under gravitational pull, into a neutron star. The force of gravity becomes so large that protons and electrons are fused into neutrons in the reaction p+ + e- S n + n. The entire star is then a tightly packed ball of neutrons with the density of nuclear matter. a. Suppose the sun collapses into a neutron star. What will its radius be? Give your answer in km. b. The sun’s rotation period is now 27 days. What will its rotation period be after it collapses? Rapidly rotating neutron stars emit pulses of radio waves at the rotation frequency and are known as pulsars

Short answer

(a) The radius of the collapsed sun is $12.7km$

(b) The rotational period after it collapses is $780\mu s$

Step-by-step solution

Subpart (a) step 1:

Calculate the radius of collapsed sun by using the relation between the volume, mass and density of the sun.

Let us assume initial radius of the sun is R_s after collapses, the entire star is then tightly packed ball of neutrons with the density of nuclear matter, let the final radius of the collapsed star be r. In the collapses process the sun's mass is unchanged, so, the density of nuclear matter is,

$p_{nu}=\frac{M_s}{v}$

Here, V is collapsed volume , M_s is mass of the sun

Rearrange above equation for V

$V=\frac{M_s}{p_{nu}}.........\left(1\right)$

Since the volume of the collapsed sun is spherical shape, the volume of the sun is

$V=\frac{4}{3}{\mathrm{\pi r}}^3..........\left(2\right)$

Here, r is radius of the collapsed sun

Equating equations (1) and (2) and Rearrange for r

$$\begin{gathered} \frac{4}{3}{\mathrm{\pi r}}^3=\frac{{\mathrm{M}}_{\mathrm{s}}}{{\mathrm{p}}_{\mathrm{nu}}} \\ \mathrm{r}=3\sqrt{\frac{3}{4}}\frac{{\mathrm{M}}_{\mathrm{s}}}{{\mathrm{\pi p}}_{\mathrm{nu}}} \end{gathered}$$

Substitute $1.99\mathrm{x}{10}^{30}\mathrm{kg}$for $M_{\mathrm{s}}2.3\mathrm{x}{10}^{17}\mathrm{kg}/{\mathrm{m}}^3$for $p_{\mathrm{nu}}$

$$\begin{gathered} \mathrm{r}=3\sqrt{\frac{3}{4}\frac{\left(1.99\mathrm{x}{10}^{30}\mathrm{kg}\right)}{3.14\left(2.3\mathrm{x}{10}^{17}\mathrm{kg}/{\mathrm{m}}^3\right)}} \\ =12737\mathrm{m}\left(\frac{1\mathrm{km}}{{10}^3\mathrm{m}}\right) \\ =12.7\mathrm{km} \end{gathered}$$

Subpart (b)  step 2:

Apply the law of conservation of angular momentum to initial radius of the sun to final radius of the collapsed sun as,

${\left(I\omega \right)}_{after}={\left(I\omega \right)}_{before}$

Here, I momenta if inertia, $\omega$is angular speed

Rearrange above equation${\mathrm{\omega }}_{after}$

${\omega }_{after}=\frac{I_{before}{\omega }_{after}}{I_{after}}$

The relation between angular speed and period of time can be expressed as

$\omega =\frac{2\mathrm{\pi }}{T}$

Noe the above equation changes as

$$\begin{gathered} \frac{2\mathrm{\pi }}{{\mathrm{T}}_{\mathrm{after}}}-\frac{{\mathrm{I}}_{\mathrm{before}}}{{\mathrm{I}}_{\mathrm{after}}}\frac{2\mathrm{\pi }}{{\mathrm{T}}_{\mathrm{before}}} \\ {\mathrm{T}}_{\mathrm{after}}=\frac{{\mathrm{I}}_{\mathrm{before}}}{{\mathrm{I}}_{\mathrm{after}}}{\mathrm{T}}_{\mathrm{before}} \end{gathered}$$

Here, $I_{\mathrm{before}}$is moment of inertia of the sun,${\mathrm{I}}_{\mathrm{after}}$is momenta of inertia of the neutron star

Calculation

Substitute$\frac{2}{5}{\mathrm{M}}_{\mathrm{s}}{\mathrm{R}}_{\mathrm{s}}^2\mathrm{for}{\mathrm{I}}_{\mathrm{before}},\frac{2}{5}{\mathrm{M}}_{\mathrm{s}}{\mathrm{r}}^2\mathrm{for}{\mathrm{I}}_{\mathrm{after}}$

$$\begin{gathered} {\mathrm{T}}_{\mathrm{after}}=\frac{{\displaystyle \frac{2}{5}{\mathrm{M}}_{\mathrm{s}}{\mathrm{r}}^2}}{{\displaystyle \frac{2}{5}{\mathrm{M}}_{\mathrm{s}}{\mathrm{R}}_{\mathrm{s}}^2}}{\mathrm{T}}_{\mathrm{before}} \\ {\mathrm{T}}_{\mathrm{after}}=\frac{{\mathrm{r}}^2}{{\mathrm{R}}_{\mathrm{s}}^2}{\mathrm{T}}_{\mathrm{before}} \end{gathered}$$

Here, $T_{\mathrm{before}}$is rotational period of the sun

Substitute $12737\mathrm{m}$for $r6.96\mathrm{x}{10}^8\mathrm{m}$for $R_{\mathrm{s}}$and $27$days for $T_{\mathrm{before}}$in the equation

$$\begin{gathered} {\mathrm{T}}_{\mathrm{after}}=\frac{{\left(12737\mathrm{m}\right)}^2}{{\left(6.96\mathrm{x}{10}^8\mathrm{m}\right)}^2}\left(27\mathrm{days}\right)\left(\frac{86400\mathrm{s}}{1\mathrm{day}}\right) \\ =780\mathrm{x}{10}^{-6}\mathrm{s}\left(\frac{1\mathrm{\mu s}}{{10}^{-6}\mathrm{s}}\right) \\ =780\mathrm{\mu s} \end{gathered}$$