Chapter 24, Problem 13Q

(a)

To determine

The maximum luminosity generated by the accretion disk of the black hole and compare it with the luminosity of the Milky Way.

Answer to Problem 13Q

The total luminosity of accretion disk of black hole is $45.53\times {10}^{43}\text{erg}/\text{s}\text{}$ and is $4.74$ times more luminous than the Milky Way galaxy.

Explanation of Solution

Given:

The mass of the black hole is, $M=3.7\times {10}^6{\text{M}}_{\odot }$.

The total luminosity of the Milky Way is, $L=2.5\times {10}^{10}{\text{L}}_{\odot }$.

Formula used:

The expression for maximum luminosity is given as,

$L_{\text{Edd}}=3.2\times {10}^4\left(\frac{M}{M_{\text{Sun}}}\right)L_{\text{Sun}}$

Calculation:

The mass of the sun is $M_{\text{Sun}}=2\times {10}^{30}\text{kg}$.

The luminosity of the sun is $L_{\text{Sun}}=3.84\times {10}^{33}\text{}\text{erg}/\text{s}$.

The maximum luminosity is calculated as,

$\begin{array}{c}{L}_{\text{Edd}}=\left(3.2\times {10}^4\right)\left(\frac{M}{M_{\text{Sun}}}\right)L_{\text{Sun}}\\ =\left(3.2\times {10}^4\right)\left(\frac{3.7\times {10}^6{\text{M}}_{\odot }}{2\times {10}^{30}\text{kg}}\right)\left(3.84\times {10}^{33}\text{erg}/\text{s}\right)\\ =\left(3.2\times {10}^4\right)\left(\frac{\left(3.7\times {10}^6\right)\left(2\times {10}^{30}\text{kg}\right)}{\left(2\times {10}^{30}\text{kg}\right)}\right)\left(3.84\times {10}^{33}\text{erg}/\text{s}\right)\\ =45.53\times {10}^{43}\text{erg}/\text{s}\text{}\end{array}$

The comparison between the luminosity of accretion disk of black hole with Milky Way is calculated as,

$\begin{array}{c}\text{Comparison of luminosity}=\left(\frac{L_{\text{Edd}}}{L}\right)\\ =\left(\frac{45.53\times {10}^{43}\text{}\text{erg}/\text{s}}{2.5\times {10}^{10}{L}_{\odot }}\right)\\ =\frac{\left(45.53\times {10}^{43}\text{}\text{erg}/\text{s}\right)}{\left(2.5\times {10}^{10}\right)\left(3.84\times {10}^{33}\text{}\text{erg}/\text{s}\right)}\\ =4.74\end{array}$

Conclusion:

The total luminosity of accretion disk of black hole is $45.53\times {10}^{43}\text{erg}/\text{s}\text{}$ and is $4.74$ times more luminous than the Milky Way galaxy.

(b)

To determine

The change in the sight when the center of Milky Way galaxy becomes an active galactic nucleus with luminosity as calculated above.

Explanation of Solution

Introduction:

AGN (Active Galactic Nucleus) is a compact region located at the center of the galaxy having a high luminosity over a portion of the electromagnetic spectrum. The active galactic nucleus is present in the center of the galaxy which is considered as the central engine of the galaxies.The active galactic nucleus usually feeds on the cosmic material present in the galaxy. The active galactic nucleus shows a high luminosity in the electromagnetic spectrum due to the presence of glowing ionized gas in the center of the nucleus.

The galaxy containing the active galactic nucleus is termed as an active galaxy. Milky Way does not contain an active galactic nucleus in its center. If it contains an active galactic nucleus in the center with the luminosity as calculated above, then the luminosity of the Milky Way would be more, and the black hole in the center of the Milky Way galaxy will start feeding on the cosmic material present in the galaxy. Due to the presence of the active galactic nucleus, the black hole will start emitting cosmic jets from its center.

Conclusion:

Therefore, the existence of an active galactic nucleus in the center of our galaxy will result in the existence of a black hole with extreme inhalation power and high luminosity.