91Ó°ÊÓ

The given data is listed below as,

• The time period of the galaxy Markarian 766 is $T=\text{27 h}=\text{27 h}×\left(\frac{\text{3600 s}}{\text{1 h}}\right)=97200\text{ s}$,
• The velocity of the galaxy Markarian 766 is,$\mathrm{v}=30,000\text{ k³¾/²õ}=30,000\text{ k³¾/²õ}×\left(\frac{\text{1000″¾}}{\text{​1 k³¾}}\right)=3×{10}^7\text{m/s}$
• Mass of the sun is,${\mathrm{M}}_{\mathrm{s}}=1.99×{10}^{39}\text{ k²µ}$.

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.

The expression for the time period is given as:

$\begin{array}{c}\mathrm{T}=\frac{2\mathrm{Ï€°ù}}{\mathrm{v}}\\ \mathrm{r}=\frac{\mathrm{Tv}}{2\mathrm{Ï€}}\end{array}$

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

Substitute the values in the above equation.

$\begin{array}{c}r=\frac{(97,200\text{ s}×3×{10}^7\text{″¾/²õ})}{(2×3.14)}\\ =\frac{(2.916×{10}^{12}\text{″¾})}{\left(6.28\right)}\\ =4.64×{10}^{11}\text{″¾}\end{array}$

Thus, the distance of the clumps from the center of the black hole is$4.64×{10}^{11}\text{″¾}$.

The expression for the velocity is expressed as:

Here, $G$is the gravitational constant with value $(6.67×{10}^{−11}{\text{″¾}}^{\text{3}}\text{/kg}â‹\dots {\text{s}}^{\text{2}})$ and $M$is the mass of the black hole.

$\begin{array}{c}M=\frac{{(3×{10}^7\text{m}/\text{s})}^2(4.64×{10}^{11}\text{m})}{(6.67×{10}^{−11}{\text{″¾}}^{\text{3}}\text{/kg}â‹\dots {\text{s}}^{\text{2}})}\\ =\frac{(9×{10}^{14}{{\text{m}}^2/\text{s}}^2)(4.64×{10}^{11}\text{m})}{(6.67×{10}^{−11}{\text{″¾}}^{\text{3}}\text{/kg}â‹\dots {\text{s}}^{\text{2}})}\\ =\frac{(4.176×{10}^{26}{\text{″¾}}^3/{\text{s}}^{\text{2}})}{(6.67×{10}^{−11}{\text{″¾}}^{\text{3}}\text{/kg}â‹\dots {\text{s}}^{\text{2}})}\\ =6.26×{10}^{36}\text{ k²µ}\\ \end{array}$

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.

$\begin{array}{c}\frac{M}{M_s}=\frac{6.26×{10}^{36}\text{ k²µ}}{1.99×{10}^{30\text{ }}\text{ k²µ}}\\ M=3.15×{10}^6{M}_s\end{array}$

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

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.

$\begin{array}{c}{R}_s=\frac{(2×6.67×{10}^{−11}{\text{″¾}}^{\text{3}}\text{/kg}â‹\dots {\text{s}}^{\text{2}}×6.26×{10}^{36}\text{ k²µ})}{{(3×{10}^8\text{″¾/²õ})}^2}\\ =\frac{(2×4.17×{10}^{26}{\text{″¾}}^{\text{3}}{\text{/s}}^{\text{2}})}{9×{10}^{16}{\text{″¾}}^2{\text{/s}}^2}\\ =\frac{(8.34×{10}^{26}{\text{″¾}}^{\text{3}}{\text{/s}}^{\text{2}})}{9×{10}^{16}{\text{″¾}}^2{\text{/s}}^2}\\ =9.28×{10}^9\text{″¾}\end{array}$

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