Chapter 21, Problem 60Q

(a)

To determine

The Schwarzschild radius of a black hole having mass equal to that of Earth. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

Answer to Problem 60Q

Solution:

$R_{\text{Sch}}=8.89\times {10}^{-3}\text{ m}$; $\rho =2.03\times {10}^{30}{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

A black hole having mass equal to that of Earth.

Formula used:

The expression for Schwarzschild radius of an object is given as,

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Here, $R_{\text{Sch}}$ is the Schwarzschild radius, $G$ is the universal gravitational constant, $M$ is the mass of the object and $c$ is the speed of light.

The expression for density of an object is given as,

$\rho =\frac{M}{V}$

Here, $\rho$ is the density, $M$ is the mass and $V$ is the volume.

Explanation:

The mass of black hole is equal to that of earth, which is equal to $5.976\times {10}^{24}\text{ kg}$.

Schwarzschild radius of the black hole is calculated as follows:

Recall the expression for Schwarzschild radius of an object.

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Upon substituting $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.976\times {10}^{24}\text{ kg}$ for $M$ and $3\times {10}^8{\text{ m/s}}^2$ for $c$,

$\begin{array}{c}{R}_{\text{Sch}}=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.976\times {10}^{24}\text{ kg}\right)}{{\left(3\times {10}^8{\text{ m/s}}^2\right)}^2}\\ =8.89\times {10}^{-3}\text{ m}\end{array}$

Density of the black hole is calculated by considering it to be a sphere.

Recall the expression for density of an object.

$\rho =\frac{M}{V}$

Upon substituting $\frac{4}{3}\pi R^3$ for $V$ and $5.976\times {10}^{24}\text{ kg}$ for $M$,

$\rho =\frac{\left(5.976\times {10}^{24}\text{ kg}\right)}{\left(\frac{4}{3}\pi R^3\right)}$

Substitute $8.89\times {10}^{-3}\text{ m}$ for $R$,

$\begin{array}{c}\rho =\frac{5.976\times {10}^{24}\text{ kg}}{\frac{4}{3}\pi {\left(8.89\times {10}^{-3}\text{ m}\right)}^3}\\ =2.03\times {10}^{30}{\text{ kg/m}}^3\end{array}$

Conclusion:

Hence, Schwarzschild radius of the black hole is $8.89\times {10}^{-3}\text{ m}$ and its density is $2.03\times {10}^{30}{\text{ kg/m}}^3$.

(b)

To determine

The Schwarzschild radius of a black hole having mass equal to that of Sun. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

Answer to Problem 60Q

Solution:

$R_{\text{Sch}}=2.93\times {10}^3\text{ m}$; $\rho =1.87\times {10}^{19}{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

A black hole having mass equal to that of Sun.

Formula used:

The expression for Schwarzschild radius of an object is given as,

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Here, $R_{\text{Sch}}$ is the Schwarzschild radius, $G$ is the universal gravitational constant, $M$ is the mass of the object and $c$ is the speed of light.

The expression for density of an object is given as,

$\rho =\frac{M}{V}$

Here, $\rho$ is the density, $M$ is the mass and $V$ is the volume.

Explanation:

The mass of black hole is equal to that of Sun, which is equal to $1.989\times {10}^{30}\text{ kg}$.

Schwarzschild radius of the black hole is calculated as follows:

Recall the expression for Schwarzschild radius of an object.

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Upon substituting $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $1.989\times {10}^{30}\text{ kg}$ for $M$ and $3\times {10}^8{\text{ m/s}}^2$ for $c$,

$\begin{array}{c}{R}_{\text{Sch}}=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.989\times {10}^{30}\text{ kg}\right)}{{\left(3\times {10}^8{\text{ m/s}}^2\right)}^2}\\ =2.94\times {10}^3\text{ m}\end{array}$

Density of the black hole is calculated by considering it to be a sphere.

Recall the expression for density of an object.

$\rho =\frac{M}{V}$

Upon substituting $\frac{4}{3}\pi R^3$ for $V$ and $1.989\times {10}^{30}\text{ kg}$ for $M$,

$\rho =\frac{\left(1.989\times {10}^{30}\text{ kg}\right)}{\left(\frac{4}{3}\pi R^3\right)}$

Substitute $2.94\times {10}^3\text{ m}$ for $R$,

$\begin{array}{c}\rho =\frac{1.989\times {10}^{30}\text{ kg}}{\frac{4}{3}\pi {\left(2.94\times {10}^3\text{ m}\right)}^3}\\ =1.87\times {10}^{19}{\text{ kg/m}}^3\end{array}$

Conclusion:

Hence, Schwarzschild radius of the black hole is $2.94\times {10}^3\text{ m}$ and its density is $1.87\times {10}^{19}{\text{ kg/m}}^3$.

(c)

To determine

The Schwarzschild radius of a black hole having mass equal to that of a supermassive black hole in NGC4261. Also calculate the density if matter is uniformly spread throughout the volume of the event horizon.

Answer to Problem 60Q

Solution:

$R_{\text{Sch}}=3.54\times {10}^{12}\text{ m}$; $\rho =12.9{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

A black hole having mass equal to that of a supermassive black hole in NGC4261.

Formula used:

The expression for Schwarzschild radius of an object is given as,

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Here, $R_{\text{Sch}}$ is the Schwarzschild radius, $G$ is the universal gravitational constant, $M$ is the mass of the object and $c$ is the speed of light.

The expression for density of an object is given as,

$\rho =\frac{M}{V}$

Here, $\rho$ is the density, $M$ is the mass and $V$ is the volume.

Explanation:

The mass of black hole is equal to that of a supermassive black hole in NGC4261, which is equal to $2.39\times {10}^{39}\text{ kg}$.

Calculate Schwarzschild radius of a black hole.

Recall the expression of Schwarzschild radius of an object.

$R_{\text{Sch}}=\frac{2GM}{c^2}$

Upon substituting $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $2.39\times {10}^{39}\text{ kg}$ for $M$ and $3\times {10}^8{\text{ m/s}}^2$ for $c$,

$\begin{array}{c}{R}_{\text{Sch}}=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(2.39\times {10}^{39}\text{ kg}\right)}{{\left(3\times {10}^8{\text{ m/s}}^2\right)}^2}\\ =3.54\times {10}^{12}\text{ m}\end{array}$

Density of the black hole is calculated by considering to be a sphere.

Recall the expression for density of an object.

$\rho =\frac{M}{V}$

Upon substituting $\frac{4}{3}\pi R^3$ for $V$ and $5.976\times {10}^{24}\text{ kg}$ for $M$,

$\rho =\frac{\left(2.39\times {10}^{39}\text{ kg}\right)}{\left(\frac{4}{3}\pi R^3\right)}$

Substitute $3.54\times {10}^{12}\text{ m}$ for $R$,

$\begin{array}{c}\rho =\frac{2.39\times {10}^{39}\text{ kg}}{\frac{4}{3}\pi {\left(3.54\times {10}^{12}\text{ m}\right)}^3}\\ =12.9{\text{ kg/m}}^3\end{array}$

Conclusion:

Hence, Schwarzschild radius of the black hole is $3.54\times {10}^{12}\text{ m}$ and its density is $12.9{\text{ kg/m}}^3$.