Chapter 22, Problem 39Q

(a)

To determine

The Schwarzschild radius for a black hole, which has a mass of $4.1\times {10}^6{\text{ M}}_{\odot }$, in km and astronomical units.

Answer to Problem 39Q

Solution:

The radius is $1.2\times {10}^7\text{ km}$ or $0.08\text{ au}$.

Explanation of Solution

Given data:

Mass of the black hole is $4.1\times {10}^6{\text{ M}}_{\odot }$

Formula used:

The expression for the Schwarzschild radius is:

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

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

The conversion formula for km into au is:

$1.5\times {10}^8\text{ km}=1\text{ au}$

The conversion formula from m to km is:

$\text{1 m}={10}^{-3}\text{ km}$

Explanation:

Consider the value of $G$ as $6.67\times {10}^{-11}\text{ N}{\text{.m}}^2{\text{/kg}}^2$, $c$ as $3\times {10}^8\text{ m/s}$ and ${\text{M}}_{\odot }$ as,

$1.989\times {10}^{30}\text{ kg}$.

Recall the expression for Schwarzschild radius.

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

Substitute $6.67\times {10}^{-11}\text{ N}{\text{.m}}^2{\text{/kg}}^2$ for $G$, $4.1\times {10}^6{\text{M}}_{\odot }$ for $M$, $3\times {10}^8\text{ m/s}$ for $c$ and $1.989\times {10}^{30}\text{ kg}$ for ${\text{M}}_{\odot }$. Also, use the conversion formula to convert m to km.

$\begin{array}{c}{R}_{\text{Sch}}=\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2{\text{/kg}}^2\right)\left(4.1\times {10}^6\right)\left(1.989\times {10}^{30}\text{ kg}\right)}{{\left(3\times {10}^8\text{ m/s}\right)}^2}\\ =12.08\times {10}^9\text{ m}\times \frac{{10}^{-3}\text{ km}}{1\text{ m}}\\ \simeq \text{1}\text{.2}\times {10}^7\text{ km}\end{array}$

To convert the value of radius from km to au, use the conversion formula.

$\begin{array}{c}1.2\times {10}^7\text{ km}=1.2\times {10}^7\text{ km}\times \left(\frac{1\text{ au}}{1.5\times {10}^8\text{ km}}\right)\\ =0.08\text{ au}\end{array}$

Conclusion:

Therefore, the Schwarzschild radius of the black hole is $\text{1}\text{.2}\times {10}^7\text{ km}$ or $0.08\text{ au}$.

(b)

To determine

The angular diameter of a black hole, which is at a distance of $8\text{ kpc}$ between Earth and the center of the galaxy. If the radius of the black hole is the radius calculated in sub-part (a).

Answer to Problem 39Q

Solution:

$2\times {10}^{-5}\text{ arcsec}$.

Explanation of Solution

Given data:

The distance between Earth and the center of the galaxy is $8\text{ kpc}$.

Formula used:

The angular diameter can be calculated by the following expression:

$\alpha =\frac{206265D}{d}$

Here, $D$ is the width or linear size of the object, $d$ is the distance to the object and $\alpha$ is the angular size of the object.

The conversion formula from parsec to au is:

$1\text{ pc}=206265\text{ au}$

Explanation:

Refer the sub-part (a) for the value of radius of black hole that is $0.8\text{ au}$.

The linear size, $D$ is twice the radius. So, the value of $D$ is:

$\begin{array}{c}D=2\times 0.8\text{ au}\\ \text{=1}\text{.6 au}\end{array}$

Recall the expression for angular diameter.

$\alpha =\frac{206265D}{d}$

Substitute $8\text{ kpc}$ for $d$, $0.16\text{ au}$ for $D$. Also, use the conversion formula.

$\begin{array}{c}\alpha =\frac{206265\times \left(0.16\text{ au}\right)}{8000\text{ pc}\times \left(\frac{206265\text{ au}}{1\text{ pc}}\right)}\\ =2\times {10}^{-5}\text{ arcsec}\end{array}$

Conclusion:

The angular diameter is $2\times {10}^{-5}\text{ arcsec}$.

(c)

To determine

The angular diameter of a black hole which is at a distance of $45\text{ au}$ between Earth and the center of the galaxy. If the radius of the black hole is the radius calculated in sub-part (a).

Answer to Problem 39Q

Solution:

$730\text{ au}$.

Explanation of Solution

Given data:

The distance between Earth and the center of the galaxy is $45\text{ au}$.

Formula used:

The angular diameter can be calculated as,

$\alpha =\frac{206265D}{d}$

Here, $D$ is the width or linear size of the object, $d$ is the distance to the object and $\alpha$ is the angular size of the object.

Explanation:

Refer the sub-part (a) for the value of radius of black hole that is $0.8\text{ au}$.

The linear size, $D$ is twice the radius. So, the value of $D$ is:

$\begin{array}{c}D=2\times 0.8\text{ au}\\ \text{=1}\text{.6 au}\end{array}$

Recalling the expression for angular diameter as,

$\alpha =\frac{206265D}{d}$

Substitute $45\text{ au}$ for $d$ and $0.16\text{ au}$ for $D$.

$\begin{array}{c}\alpha =\frac{206265\times \left(0.16\text{au}\right)}{45\text{au}}\\ =730\text{arcsec}\end{array}$

Conclusion:

The angular diameter is $730\text{ arcsec}$, and it would be visible to the naked eyes if it was not a black hole.