Chapter 21, Problem 62Q

To determine

The density to which the matter of a dead $8-{\text{M}}_{\odot }$ star must be compressed for the star to disappear inside its event horizon and comparison of this value with the density at the center of a neutron star about $3\times {10}^{18}\text{ kg}\cdot {\text{m}}^{-3}$.

Answer to Problem 62Q

Density to which the matter of a dead $8-{\text{M}}_{\odot }$ star must be compressed = $\text{2}\text{.7}\times {10}^{17}\text{ kg}\cdot {\text{m}}^{-3}$

Density of the compressed star $\ll$ density at the center of the neutron star

Explanation of Solution

Given:

Mass of the dead star in solar mass = $8-{\text{M}}_{\odot }$

Density at the center of a neutron star = $3\times {10}^{18}\text{ kg}\cdot {\text{m}}^{-3}$

Formula Used:

$r_s=\frac{2GM}{c^{{}^2}}$

$\begin{array}{l}\text{Volume of a sphere}=\frac{4}{3}\pi r^3\\ \text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \end{array}$

Calculation:

$\text{8 solar mass = 8}\times \text{2}\times {10}^{30}\text{ kg}$

$G=6.7\times {10}^{-11}{\text{ m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$

$c=3\times {10}^8\text{ m}\cdot {\text{s}}^{-1}$

If the dead star is to be compressed for it to disappear inside the event sphere, it’s radius must be equal to the Schwarzschild radius. Hence, it’s radius can be found as follows.

$\begin{array}{l}{r}_s=\frac{2GM}{c^{{}^2}}\\ r_s=\frac{2\times 6.7\times {10}^{-11}\times 8\times 2\times {10}^{30}}{{(3\times {10}^8)}^2}\\ r_s=23.8\times {10}^3\text{ m}\\ \text{Rounding off to the first decimal place}\\ r_s=2.4\times {10}^4\text{ m}\end{array}$

Assuming when the dead star is a perfect sphere when it is compressed.

Volume of the compressed star

$\begin{array}{c}\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times {(}^{2.4}{\text{ m}}^3\\ =\frac{88}{21}\times 15.625\times {10}^{12}{\text{m}}^3\\ =5.8\times {10}^{13}{\text{m}}^3\end{array}$

$\begin{array}{c}\text{Density =}\frac{\text{Mass}}{\text{Volume}}\\ \text{=}\frac{\text{8}\times \text{2}\times {10}^{30}\text{ kg}}{5.8\times {10}^{13}{\text{m}}^3}\\ \text{=2}\text{.7}\times {10}^{17}\text{ kg}\cdot {\text{m}}^{-3}\end{array}$

Conclusion:

Density to which the matter of a dead $8-{\text{M}}_{\odot }$ star must be compressed = $\text{2}\text{.7}\times {10}^{17}\text{ kg}\cdot {\text{m}}^{-3}.$

Density at the center of a neutron star = $3\times {10}^{18}\text{ kg}\cdot {\text{m}}^{-3}$.

As $\text{2}\text{.7}\times {10}^{17}\text{ kg}\cdot {\text{m}}^{-3}\ll 3\times {10}^{18}\text{ kg}\cdot {\text{m}}^{-3}$

Density of the compressed star $\ll$ density at the center of the neutron star