Chapter 21, Problem 63Q

To determine

To prove:

The density of matter needed for a black hole is inversely proportional to the square of the mass of the black hole.

The mass of a substance having a density equal to water ( $1000\text{ kg}\cdot {\text{m}}^{-3}$ ) needed to make a black hole.

Answer to Problem 63Q

$\text{Density of matter needed to make a black hole }\propto \frac{1}{{\left(\text{mass}\right)}^2}$

Mass of substance with a density of $1000\text{ kg}\cdot {\text{m}}^{-3}$ needed to make a black hole = $3.01\times {10}^{21}\text{ kg}$

Explanation of Solution

Given:

The density of the substance used to make the black hole = $1000\text{ kg}\cdot {\text{m}}^{-3}$.

Formula used:

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

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

Calculation:

$$

$\begin{array}{l}{r}_s=\frac{2GM}{c^{{}^2}}\\ {\left(r_s\right)}^3={\left(\frac{2GM}{c^{{}^2}}\right)}^3\\ \frac{4}{3}\pi \times {\left(r_s\right)}^3=\frac{4}{3}\pi \times {\left(\frac{2GM}{c^{{}^2}}\right)}^3\\ \text{As V}=\frac{4}{3}\pi \times {\left(r_s\right)}^3\\ \text{V}=\frac{4}{3}\pi \times {\left(\frac{2GM}{c^{{}^2}}\right)}^3\\ \text{As }\rho \text{ = }\frac{M}{V}\\ V=\frac{M}{\rho }\\ \therefore \frac{M}{\rho }=\frac{4}{3}\pi \times {\left(\frac{2GM}{c^{{}^2}}\right)}^3\\ \rho =\frac{M}{\frac{4}{3}\pi \times {\left(\frac{2GM}{c^{{}^2}}\right)}^3}\\ \rho =\frac{3\times c^{{}^2}\times M}{4\pi \times {\left(2GM\right)}^3}\\ \rho =\frac{3\times c^{{}^2}\times M}{4\pi \times {(2G)}^3\times M^3}\\ \rho =\frac{3\times c^{{}^2}}{4\pi \times {(2G)}^3\times M^2}\\ \text{Let }k\text{=}\frac{3\times c^{{}^2}}{4\pi \times {(2G)}^3}\\ \therefore \rho =\frac{k}{M^2}\\ \text{As }c,G\text{ and }\pi \text{ are constants, }k\text{ is a constant}\text{.}\\ \therefore \rho \propto \frac{1}{M^2}\end{array}$

When density of the substance used to make the black hole = $1000\text{ kg}\cdot {\text{m}}^{-3}$ ,

$\begin{array}{l}\rho =\frac{3\times c^{{}^2}}{4\pi \times {(2G)}^3\times M^2}\\ 1000=\frac{3\times {\left(3\times {10}^8\right)}^2}{4\pi \times {(2\times 6.67\times {10}^{-11})}^3\times M^2}\\ M^2=\frac{3\times {\left(3\times {10}^8\right)}^2}{4\pi \times {(2\times 6.67\times {10}^{-11})}^3\times 1000}\\ M^2=\frac{3\times 9\times {10}^{16}}{4\pi \times {(13.34\times {10}^{-11})}^3\times 1000}\\ M^2=\frac{27\times {10}^{16}}{4\pi \times 2373.93\times {10}^{-33}\times 1000}\\ M^2=\frac{27\times {10}^{16}}{29831.68\times {10}^{-30}}\\ M^2=9.05\times {10}^{-4}\times {10}^{46}\\ M^2=9.05\times {10}^{42}\\ M=\sqrt{9.05\times {10}^{42}}\text{ kg}\\ M=3.01\times {10}^{21}\text{ kg}\end{array}$

Conclusion:

$\text{Density of matter needed to make a black hole }\propto \frac{1}{{\left(\text{mass}\right)}^2}$

Mass of substance with a density of $1000\text{ kg}\cdot {\text{m}}^{-3}$ needed to make a black hole = $3.01\times {10}^{21}\text{ kg}\text{.}$