Chapter 24, Problem 13Q

(a)

To determine

The maximum luminosity that could be produced by accretion onto a black hole of $3.7\times {10}^6{M}_{\Theta }$. Also, compare the luminosity with the luminosity of the Milky Way, which is about $2.5\times {10}^{10}{L}_{\Theta }$.

Answer to Problem 13Q

Solution:

The maximum luminosity that can be produced due to accretion onto a black hole is $111,000\times {10}^6{L}_{\Theta }$.

As compared to the luminosity of the Milky Way, the luminosity is $4.44$ times more.

Explanation of Solution

Given data:

The size of the black hole found at the center of the Milky Way is $3.7\times {10}^6{M}_{\Theta }$.

The luminosity of the Milky Way is $2.5\times {10}^{10}{L}_{\Theta }$.

Formula used:

The maximum luminosity that can be radiated by accretion around a black hole is:

$L_{Edd}=30,000\left(\frac{M}{M_{\Theta }}\right)L_{\Theta }$

Here, $L_{Edd}$ is the maximum luminosity that can be radiated by accretion around a black hole, $M$ is the mass of the black hole, $M_{\Theta }$ is the mass of the sun, and $L_{\Theta }$ is the luminosity of the sun.

Explanation:

Recall the expression for the maximum luminosity that can be radiated by accretion around a black hole.

$L_{Edd}=30,000\left(\frac{M}{M_{\Theta }}\right)L_{\Theta }$

Substitute $3.7\times {10}^6{M}_{\Theta }$ for $M$.

$\begin{array}{l}{L}_{Edd}=30,000\left(\frac{3.7\times {10}^6{M}_{\Theta }}{M_{\Theta }}\right)L_{\Theta }\\ L_{Edd}=30,000\times \left(3.7\times {10}^6\right)L_{\Theta }\\ L_{Edd}=111,000\times {10}^6{L}_{\Theta }\end{array}$

The ratio of the luminosity of the black hole to the luminosity of the Milky Way will be:

$\frac{L_{Edd}}{L_{Milkyway}}$

Substitute $111,000\times {10}^6{L}_{\Theta }$ for $L_{Edd}$ and $2.5\times {10}^{10}{L}_{\Theta }$ for $L_{Milkyway}$.

$\begin{array}{c}\frac{L_{Edd}}{L_{Milkyway}}=\frac{111,000\times {10}^6{L}_{\Theta }}{2.5\times {10}^{10}{L}_{\Theta }}\\ \frac{L_{Edd}}{L_{Milkyway}}=4.44\\ L_{Edd}=4.44L_{Milkyway}\end{array}$

Conclusion:

Therefore, the maximum luminosity that can be radiated by accretion around a black hole is $111,000\times {10}^6{L}_{\Theta }$ and it is $4.44$ times the luminosity of the Milky Way galaxy.

(b)

To determine

The observation if the center of our galaxy becomes an active galactic nucleus with the luminosity observed in sub-part (a).

Answer to Problem 13Q

Solution:

As the radiation pressure would increase, it would cause the intensity to increase from the direction of the Milky Way’s center.

Explanation of Solution

Introduction:

If the luminosity increases, it would create a lot of radiation pressure. Due to this, the surrounding gas will be pushed outward, rather than falling inward and it would blow away the accreting gas.

Explanation:

Refer the value of luminosity from sub-part (a), that is, 4.44 times the luminosity of the Milky Way galaxy. When the Milky Way galaxy became a galactic active nucleus with this luminosity, the balance between the radiation pushing gas fuel outward and gravity pushing the gas inward would become imbalanced.

This would result in an increase of radiation pressure, due to which the surrounding gas is pushed outward rather than falling inward and it would blow away the accreting gas. So, the mass of the center must be quite large to keep pulling in gases.

Conclusion:

Therefore, the increase in luminosity would cause the intensity to increase from the direction of the Milky Way’s center at all wavelengths.