A. [tex]4.64\cdot 10^{11}m[/tex]
The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:
[tex]v=\frac{2\pi r}{T}[/tex]
where we have:
[tex]v=30,000 km/s = 3\cdot 10^7 m/s[/tex] is the orbital speed
r is the orbital radius
[tex]T=27 h \cdot 3600 =97,200 s[/tex] is the orbital period
Solving for r, we find the distance of the clumps of matter from the centre of the black hole:
[tex]r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m[/tex]
B. [tex]6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s[/tex]
The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:
[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex]
where
m is the mass of the clumps of matter
G is the gravitational constant
M is the mass of the black hole
Solving the formula for M, we find the mass of the black hole:
[tex]M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg[/tex]
and considering the value of the solar mass
[tex]M_s = 2\cdot 10^{30}kg[/tex]
the mass of the black hole as a multiple of our sun's mass is
[tex]M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s[/tex]
C. [tex]9.28\cdot 10^9 m[/tex]
The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by
[tex]R=\frac{2MG}{c^2}[/tex]
where M is the mass of the black hole and c is the speed of light.
Substituting numbers into the formula, we find
[tex]R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m[/tex]
