Question

In 2005 astronomers announced the discovery of large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every$\mathbf{27}\mathbf{hours}$ and moving at$\mathbf{30}\mathbf{,}\mathbf{000}\mathbf{km}\mathbf{/}\mathbf{s}$ . (a) How far these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilogram and as a multiple of sun’s mass. (c) What is the radius of event horizon?

Short answer

(a)The distance of the clumps from the center of the black hole is$4.64\times {10}^{11}\text{m}$.

(b)The mass of the black hole as a multiple of the mass of the sun$\mathrm{M}=3.15\times {10}^6{\mathrm{M}}_{\mathrm{s}}$.

(c) The radius of the event horizon is$9.28\times {10}^9\text{m}$.

Step-by-step solution

Identification of the given data

The given data is listed below as,

• The time period of the galaxy Markarian 766 is $T=\text{27\hspace{0.17em}h}=\text{27\hspace{0.17em}h}\times \left(\frac{\text{3600\hspace{0.17em}s}}{\text{1\hspace{0.17em}h}}\right)=97200\text{\hspace{0.17em}s}$,
• The velocity of the galaxy Markarian 766 is,$\mathrm{v}=30,000\text{\hspace{0.17em}km/s}=30,000\text{\hspace{0.17em}km/s}\times \left(\frac{\text{1000\hspace{0.17em}m}}{\text{1\hspace{0.17em}km}}\right)=3\times {10}^7\text{m/s}$
• Mass of the sun is,${\mathrm{M}}_{\mathrm{s}}=1.99\times {10}^{39}\text{\hspace{0.17em}kg}$.

Significance of the time period

The time period is described as the time taken by a particle to move a complete cycle of a wave for passing a point. The time period is directly proportional to the distance and inversely proportional to the velocity.

(a) Determination of the distance of the clumps from the center of the black hole

The expression for the time period is given as:

$\begin{array}{c}\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}\\ \mathrm{r}=\frac{\mathrm{Tv}}{2\mathrm{\pi }}\end{array}$

Here,$T$ is the time period, $r$is the distance,$v$and is the velocity.

Substitute the values in the above equation.

$\begin{array}{c}r=\frac{(97,200\text{\hspace{0.17em}s}\times 3\times {10}^7\text{\hspace{0.17em}m/s})}{(2\times 3.14)}\\ =\frac{(2.916\times {10}^{12}\text{\hspace{0.17em}m})}{\left(6.28\right)}\\ =4.64\times {10}^{11}\text{\hspace{0.17em}m}\end{array}$

Thus, the distance of the clumps from the center of the black hole is$4.64\times {10}^{11}\text{\hspace{0.17em}m}$.

(b) Determination of the mass of the black hole

The expression for the velocity is expressed as:

Here, $G$is the gravitational constant with value $(6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}})$ and $M$is the mass of the black hole.

$\begin{array}{c}M=\frac{{(3\times {10}^7\text{m}/\text{s})}^2(4.64\times {10}^{11}\text{m})}{(6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}})}\\ =\frac{(9\times {10}^{14}{{\text{m}}^2/\text{s}}^2)(4.64\times {10}^{11}\text{m})}{(6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}})}\\ =\frac{(4.176\times {10}^{26}{\text{\hspace{0.17em}m}}^3/{\text{s}}^{\text{2}})}{(6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}})}\\ =6.26\times {10}^{36}\text{\hspace{0.17em}kg}\\ \end{array}$

The equation of the mass of the black hole as a multiple of the Sun’s mass is expressed as:

$M=a{M}_s$

Here,$M$ is the mass of the black hole,$a$is the multiple of the mass of the sun and$M_s$is the mass of the sun.

Substitute the values in the above equation.

$\begin{array}{c}\frac{M}{M_s}=\frac{6.26\times {10}^{36}\text{\hspace{0.17em}kg}}{1.99\times {10}^{30}\text{\hspace{0.17em}kg}}\\ M=3.15\times {10}^6{M}_s\end{array}$

Thus, the mass of the black hole as a multiple of the mass of the sun$M=3.15\times {10}^6{M}_s$

(c) determination of the radius of event horizon

The expression for the radius of event horizon is expressed as:

${\mathrm{R}}_{\mathrm{s}}=\frac{2\mathrm{GM}}{{\mathrm{c}}^2}$

Here,$M$is the mass of black hole,$G$is the gravitational constant and$c$is the speed of the light.

Substitute the values in the above equation.

$\begin{array}{c}{R}_s=\frac{(2\times 6.67\times {10}^{-11}{\text{\hspace{0.17em}m}}^{\text{3}}\text{/kg}\cdot {\text{s}}^{\text{2}}\times 6.26\times {10}^{36}\text{\hspace{0.17em}kg})}{{(3\times {10}^8\text{\hspace{0.17em}m/s})}^2}\\ =\frac{(2\times 4.17\times {10}^{26}{\text{\hspace{0.17em}m}}^{\text{3}}{\text{/s}}^{\text{2}})}{9\times {10}^{16}{\text{\hspace{0.17em}m}}^2{\text{/s}}^2}\\ =\frac{(8.34\times {10}^{26}{\text{\hspace{0.17em}m}}^{\text{3}}{\text{/s}}^{\text{2}})}{9\times {10}^{16}{\text{\hspace{0.17em}m}}^2{\text{/s}}^2}\\ =9.28\times {10}^9\text{\hspace{0.17em}m}\end{array}$

Thus, the radius of the event horizon is $9.28\times {10}^9\text{m}$.