Chapter 21, Problem 60Q

To determine

(a)

The Schwarzschild radius and the density of a black hole which is having a mass equal to the planet Earth.

Answer to Problem 60Q

The Schwarzschild radius of the black hole which is having the mass of the planet Earth is $8.84\times {10}^{-3}\text{m}\text{.}$

The density of the black hole is $5,496.07\text{kg}{\text{m}}^{\text{-3}}\text{.}$

Explanation of Solution

Given:

Mass of the Earth is $5.97\times {10}^{\text{24}}\text{kg}\text{.}$

Radius of the Earth is $6,378\text{km}$

Formula Used:

Schwarzschild radius of an object can be found by using the formula

$R_s=\frac{2GM}{C^2}$

Where,

$\begin{array}{l}{R}_s=\text{Schwarzschild radius of the object}\\ G=\text{gravitational constant}\\ M=\text{mass of the object}\\ C=\text{speed of the light}\end{array}$

Density of an object can be found by using the formula

$Density=\frac{Mass}{Volume}$

Calculation:

$\begin{array}{c}{R}_s=\frac{\text{2×6}{\text{.67×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{×5}{\text{.97×10}}^{\text{24}}\text{kg}}{{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}\\ =8.84\times {10}^{-3}\text{m}\end{array}$

Therefore, the radius of the black hole to the event horizon

$6,378\times {10}^3\text{m}+8.84\times {10}^{-3}\text{m}\cong 6,378\text{km}$

Density of the black hole,

$\begin{array}{c}Density=\frac{\text{5}{\text{.97×10}}^{\text{24}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{π}{\left({\text{6,378×10}}^{\text{3}}\right)}^{\text{3}}{\text{m}}^{\text{3}}}\\ =\frac{\text{5}{\text{.97×10}}^{\text{24}}\text{kg}}{\frac{4}{3}\times 3.14\times {\left(6,378\times {10}^3\right)}^{\text{3}}{\text{m}}^3}\\ \text{=5,496}\text{.07}\text{kg}{\text{m}}^{\text{-3}}\end{array}$

Conclusion:

Therefore, the Schwarzschild radius of the black hole which is having the mass of the planet Earth is $8.84\times {10}^{-3}\text{m}$ and the density of the black hole is $5,496.0\text{7}\text{kg}{\text{m}}^{\text{-3}}\text{.}$

To determine

(b)

The Schwarzschild radius and the density of a black hole which is having a mass equal to the Sun.

Answer to Problem 60Q

The Schwarzschild radius of the black hole which is having the mass equal to the Sun is $2,964.44\text{m}.$

The density of the black hole is $1,418.\text{71}\text{kg}{\text{m}}^{\text{-3}}\text{.}$

Explanation of Solution

Given data:

Mass of the Sun is $2\times {10}^{30}\text{kg}\text{.}$

Radius of the Sun is $6.957\times {10}^8\text{km}\text{.}$

Formula used:

Schwarzschild radius of an object can be found by using the formula

$R_s=\frac{2GM}{C^2}$.

Where,

$\begin{array}{c}{R}_s=\text{Schwarzschild radius of the object}\\ G=\text{gravitational constant}\\ M=\text{mass of the object}\\ C=\text{speed of the light}\end{array}$

Density of an object can be found by using the formula

$Density=\frac{Mass}{Volume}$

Calculation:

$\begin{array}{c}{R}_s=\frac{\text{2×6}{\text{.67×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}{\text{×2×10}}^{\text{30}}\text{kg}}{{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}\\ =2,964.44\text{m}\end{array}$

$$

Therefore, the radius of the black hole to the event horizon

$6.{\text{957×10}}^{\text{8}}\text{m+2,964}\text{.44}\text{m}\cong \text{6}{\text{.957×10}}^{\text{8}}\text{m}$

Density of the black hole

$\begin{array}{c}Density\text{=}\frac{{\text{2×10}}^{\text{30}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{π}{\left(\text{6}{\text{.957×10}}^{\text{8}}\right)}^{\text{3}}{\text{m}}^{\text{3}}}\\ \text{=}\frac{{\text{2×10}}^{\text{30}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{×3}\text{.14×}{\left(\text{6}{\text{.957×10}}^{\text{8}}\right)}^{\text{3}}{\text{m}}^{\text{3}}}\\ =1,418.7\text{1}\text{kg}{\text{m}}^{\text{-3}}\end{array}$

Conclusion:

Therefore, the Schwarzschild radius of the black hole which is having the mass equal to the Sun is $2,964.44\text{m}$ and the density of the black hole is $1,418.71\text{kg}{\text{m}}^{\text{-3}}\text{.}$

To determine

(c)

The Schwarzschild radius and the density of a black hole which is having a mass equal to the supermassive black hole in NGC 4261.

Answer to Problem 60Q

The Schwarzschild radius of the black hole which is having the mass of the supermassive black hole in NGC 4261 is $3.55\times {10}^{12}\text{m}\text{.}$

The density of the black hole is $6.77\times {10}^{-10}\text{kg}{\text{m}}^{\text{-3}}\text{.}$

Explanation of Solution

Given:

Mass of the supermassive black hole in NGC 4261 is $1.2\times {10}^9{M}_{\odot }.$

Radius of the supermassive black hole is $400\text{light years}\text{.}$

$1\text{light year}=9.{\text{46×10}}^{\text{15}}\text{m}\text{.}$

$1M_{\odot }=2\times {10}^{\text{30}}\text{kg}$

Formula used:

Schwarzschild radius of an object can be found by using the formula,

$R_s=\frac{2GM}{C^2}$

Where

$\begin{array}{c}{R}_s=\text{Schwarzschild radius of the object}\\ G=\text{gravitational constant}\\ M=\text{mass of the object}\\ C=\text{speed of the light}\end{array}$

Density of an object can be found by using the formula,

$Density=\frac{Mass}{Volume}$

Calculation:

$\begin{array}{c}{R}_s=\frac{\text{2×6}{\text{.67×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{×1}{\text{.2×10}}^{\text{9}}{\text{×2×10}}^{\text{30}}\text{kg}}{{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}\\ \text{=3}{\text{.55×10}}^{\text{12}}\text{m}\end{array}$

$$

Therefore, the radius of the black hole to the event horizon

$9.46\times {10}^{15}\text{m}+3.55\times {10}^{12}\text{m}\cong 9.46\times {10}^{15}\text{m}$

Density of the black hole,

$\begin{array}{c}Density\text{=}\frac{\text{1}{\text{.2×10}}^{\text{9}}{\text{×2×10}}^{\text{30}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{π}{\left(\text{9}{\text{.46×10}}^{\text{15}}\right)}^{\text{3}}{\text{m}}^{\text{3}}}\\ =\frac{\text{1}{\text{.2×10}}^{\text{9}}{\text{×2×10}}^{\text{30}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{×3}\text{.14×}{\left(\text{9}{\text{.46×10}}^{\text{15}}\right)}^{\text{3}}{\text{m}}^{\text{3}}}\\ =6.77\times {10}^{\text{-10}}\text{kg}{\text{m}}^{\text{-3}}\end{array}$

Conclusion:

Therefore, the Schwarzschild radius of the black hole which is having the mass of the supermassive black hole in NGC 4261 is $3.55\times {10}^{12}\text{m}$ and the density of the black hole is $6.77\times {10}^{-10}\text{kg}{\text{m}}^{\text{-3}}\text{.}$