Chapter 22, Problem 39Q

To determine

(a)

To calculate:

The Schwarzschild radius of a super massive black hole of mass $4.1\times {10}^6{\text{M}}_{\odot }$ at the center of the galaxy in both kilometers and astronomical units.

Answer to Problem 39Q

Radius of a super massive black hole of mass kilometers = $1.2\times {10}^7\text{ km}\text{.}$

Radius of a super massive black hole of mass astronomical units = $8\times {10}^{-2}\text{ au}\text{.}$

Explanation of Solution

Given:

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

Formula used:

The Schwarzschild radius can be calculated using the following:

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

Where, G is the gravitational constant, M is mass, c is the speed of light.

$1\text{ au = 1}\text{.5}\times {\text{10}}^8\text{ km}$

$\text{1 solar mass = 2}\text{.0}\times {10}^{30}\text{ kg}$

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

$c=3.0\times {10}^8\text{ m}\cdot {\text{s}}^{-1}$

Calculation:

Substitute the values:

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

$\begin{array}{l}{r}_s=\frac{2\times 6.7\times {10}^{-11}\times 4.1\times {10}^6\times \text{2}\text{.0}\times {10}^{30}}{{\left(3.0\times {10}^8\right)}^{{}^2}}\\ r_s=\frac{109.88\times {10}^{25}}{9\times {10}^{16}}\\ r_s=12.2\times {10}^9\text{ m}\\ r_s=1.22\times {10}^{10}\text{ m}\\ r_s=1.22\times {10}^7\text{ km}\\ \text{Rounding off to the first decimal place,}\\ r_s=1.2\times {10}^7\text{ km}\end{array}$

$\begin{array}{l}\text{In astronomical units}\\ r_s=\frac{1.2\times {10}^7}{1.5\times {10}^8}\text{ au}\\ r_s=0.8\times {10}^{-1}\text{ au}\\ r_s=8\times {10}^{-2}\text{ au}\end{array}$

Conclusion:

The radius of a supermassive black hole of mass kilometers = $1.2\times {10}^7\text{ km}$

The radius of a supermassive black hole of mass astronomical units = $8\times {10}^{-2}\text{ au}$

To determine

(b)

To calculate:

The angular diameter in arcseconds of the black hole at a distance of $8\text{ kpc}$, the distance from Earth to the galactic center.

Answer to Problem 39Q

Angular diameter in arcseconds of the black hole at a distance of $8\text{ kpc}$ = $2.0\times {10}^{-5}\text{ arcsec}\text{.}$

Explanation of Solution

Given:

Distance from Earth to the galactic center = $8\text{ kpc}\text{.}$

Formula used:

$d=2\times r_s$

$\delta =206265\times \frac{d}{D}$

$1\text{ kpc = 3}\text{.1}\times {\text{10}}^{16}\text{ km}$

Calculation:

$\begin{array}{l}\delta =206265\times \frac{d}{D}\\ \delta =206265\times \frac{2\times 1.2\times {10}^7}{8\times \text{3}\text{.1}\times {\text{10}}^{16}}\\ \delta =206265\times \frac{2.4\times {10}^7}{24.8\times {\text{10}}^{16}}\\ \delta =206265\times 0.097\times {10}^{-9}\\ \delta =20007.705\times {10}^{-9}\\ \delta =2.0007705\times {10}^{-5}\text{ arcsec}\\ \text{Rounding off to the first decimal place,}\\ \delta =2.0\times {10}^{-5}\text{ arcsec}\end{array}$

Conclusion:

Angular diameter in arcseconds of the black hole at a distance of $8\text{ kpc}$ = $2.0\times {10}^{-5}\text{ arcsec}\text{.}$

To determine

(c)

To calculate:

The angular diameter in arcseconds of the black hole when seen from a distance of $45\text{ au}$.

Whether or not, it will be discernible to the naked eye.

Answer to Problem 39Q

The angular diameter of the black hole at a distance of $45\text{ au}$ = $7.4\times {10}^2\text{ arcsec}\text{.}$

Yes, the naked eye can discern the black hole.

Explanation of Solution

Given:

Distance from which the black hole is observed = $45\text{ au}$

The field of view of a naked eye = $60\text{ arcsec}$

Formula used:

$d=2\times r_s$

$\delta =206265\times \frac{d}{D}$

$1\text{ au = 1}\text{.5}\times {\text{10}}^8\text{ km}$

$1\text{ kpc = 3}\text{.1}\times {\text{10}}^{16}\text{ km}$

Calculation:

$\begin{array}{l}\delta =206265\times \frac{d}{D}\\ \delta =206265\times \frac{2\times 1.2\times {10}^7}{45\times \text{1}\text{.5}\times {\text{10}}^8}\\ \delta =206265\times \frac{2.4\times {10}^7}{67.5\times {\text{10}}^8}\\ \delta =206265\times 0.036\times {10}^{-1}\\ \delta =7425.54\times {10}^{-1}\\ \delta =7.42554\times {10}^2\text{ arcsec}\\ \text{Rounding off to the first decimal place,}\\ \delta =7.4\times {10}^2\text{ arcsec}\end{array}$

As $7.4\times {10}^2\text{ arcsec}\gg 60\text{ arcsec,}$ the naked eye can detect the black hole when the person is in Sagittarius A.

Conclusion:

Angular diameter of the black hole at a distance of $45\text{ au}$ = $7.4\times {10}^2\text{ arcsec}$

As $7.4\times {10}^2\text{ arcsec}\gg 60\text{ arcsec}$, the naked eye can detect the black hole.