91Ó°ÊÓ

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 $R_{\text{s}}$, 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,

${\mathrm{ÒÏ}}_{nu}=\frac{M_{\text{S}}}{V}$

Here, $V$is collapsed volume, $M_{\text{S}}$is mass of the sun.

Rearrange above equation for $V$

$V=\frac{M_{\text{S}}}{{\mathrm{ÒÏ}}_{nw}}$

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

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

Here, $r$is radius of the collapsed sun

Equating equations (1) and (2) and rearrange for $r$

$\begin{array}{l}\frac{4}{3}\mathrm{Ï€}{r}^3=\frac{M_{\text{S}}}{{\mathrm{ÒÏ}}_{mu}}\\ r=\sqrt[3]{\frac{3}{4}\frac{M_{\text{S}}}{\mathrm{Ï€}{\mathrm{ÒÏ}}_{nu}}}\end{array}$

Substitute $1.99×{10}^{30}\text{kg}$for $M_{\text{S}},2.3×{10}^{17}\text{kg}/{\text{m}}^3$for ${\mathrm{ÒÏ}}_{n\text{s}}$.

$\begin{array}{l}r=\sqrt[3]{\frac{3}{4}\frac{\left(1.99×{10}^{30}\text{kg}\right)}{3.14\left(2.3×{10}^{17}\text{kg}/{\text{m}}^3\right)}}\\ =12737\text{m}\left(\frac{1\text{km}}{{10}^3\text{m}}\right)\\ =12.7\text{km}\end{array}$

Therefore, the radius of the collapsed sun is$12.7\text{km}$.

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

${\left(I\mathrm{Ó\neg }\right)}_{\text{after}}={\left(I\mathrm{Ó\neg }\right)}_{\text{before}}$

Here, $I$momenta of inertia, $\mathrm{Ó\neg }$is angular speed

Rearrange above equation ${\mathrm{Ó\neg }}_{\text{after}}$

${\mathrm{Ó\neg }}_{\text{after}}=\frac{I_{\text{thatee}}{\mathrm{Ó\neg }}_{\text{befave}}}{I_{\text{after}}}$

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

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

Now the above equation changes as

$\begin{array}{l}\frac{2\mathrm{Ï€}}{T_{\text{after}}}=\frac{I_{\text{befees}}}{I_{\text{after}}}\frac{2\mathrm{Ï€}}{T_{\text{befioe}}}\\ T_{\text{after}}=\frac{I_{\text{after}}}{I_{\text{befiee}}}{T}_{\text{befine}}\end{array}$

Here, $I_{\text{sefuet}}$is moment of inertia of the sun, $I_{\text{after}}$is momenta of inertia of the neutron star

Substitute $\frac{2}{5}{M}_{\text{s}}{R}_{\text{s}}^2$for $I_{\text{tefine}},\frac{2}{5}{M}_{\text{s}}{r}^2$for $I_{\text{after}}$.

$T_{\text{after}}=\frac{\frac{2}{5}{M}_{\text{s}}{r}^2}{\frac{2}{5}{M}_{\text{S}}{R}_{\text{s}}^2}{T}_{\text{skfiom}}$

$T_{\text{after}}=\frac{r^2}{R_{\text{s}}^2}{T}_{\text{trfies}}$

Here, $T_{\text{befrec}}$is rotational period of the sun

Substitute $12737\text{m}$for$r,6.96×{10}^8\text{m}$for $R_{\text{s}}$and 27 days for $T_{\text{befre}}$in the equation

$\begin{array}{l}{T}_{\text{afer}}=\frac{{(12737\text{m})}^2}{{\left(6.96×{10}^8\text{m}\right)}^2}\left(27\text{days}\right)\left(\frac{86400\text{s}}{1\text{day}}\right)\\ =780×{10}^{−6}\text{s}\left(\frac{1\mathrm{μ}\text{s}}{{10}^{−6}\text{s}}\right)\\ =780\mathrm{μ}\text{s}\end{array}$

Therefore, the rotational period after it collapses is $780\mathrm{μ}\text{s}$.