Question

A neutron star is a stellar object whose density is about that of nuclear matter,$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ . Suppose that the Sun were to collapse and become such a star without losing any of its present mass. What would be its radius?

Short answer

The value of the radius of the star is 13 km.

Step-by-step solution

Given data

The density of the nuclear matter,$\mathrm{p}=2\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3$

Mean density of Sun according to Appendix C,${\mathrm{p}}_{\mathrm{s}}=1410\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3$

Mean radius of Sun according to Appendix C,${\mathrm{R}}_{\mathrm{s}}=6.96\times {10}^8\mathrm{m}$

Understanding the concept of density  

As per the given concept, the Sun is to collapse to form into a star whose density is given. Thus, keeping the mass constant and relating this to the density, volume, and mass relation, we can get that the radius' cube value is inversely proportional to the density. Thus, using this proportionality relation, we can get the required value.

Formula:

The density of a spherical object, $p=\frac{m}{{\displaystyle \frac{4}{3}}\pi R^3}.......\left(1\right)$

Calculation of the radius of the star

From equation (1), we get that

$\mathrm{p}\mathrm{\alpha }\frac{1}{{\mathrm{R}}^3}$with all other terms as costant.

$\mathrm{R}\mathrm{\alpha }\frac{1}{{\mathrm{p}}^{1/3}}$

Thus, using this above relation and the given data, we can get the radius of the star as follows:

$$\begin{gathered} \mathrm{R}={\mathrm{R}}_{\mathrm{s}}{\left[\frac{{\mathrm{p}}_{\mathrm{s}}}{\mathrm{p}}\right]}^{1/3} \\ =\left(6.96\times {10}^8\mathrm{m}\right){\left[\frac{\left(1410\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3\right)}{\left(2\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3\right)}\right]}^{1/3} \\ =1.3\times {10}^4\mathrm{m} \\ =13\mathrm{km} \end{gathered}$$

Hence, the value of the radius is 13 km.