A) [tex]4.6\cdot 10^{11} m[/tex]
The period of the orbit of the clumps around the black hole is
[tex]T=27 h \cdot (3600 s/h)=97,200 s[/tex]
While their orbital speed is
[tex]v=30,000 km/s=3.0\cdot 10^7 m/s[/tex]
And the orbital speed is equal to the ratio between the circumference of the orbit and the orbital period:
[tex]v=\frac{2\pi r}{T}[/tex]
So re-arranging the equation, we find the radius of the orbit of the clumps:
[tex]r=\frac{vT}{2\pi}=\frac{(3.0\cdot 10^7 m/s)(97,200 s)}{2\pi}=4.6\cdot 10^{11} m[/tex]
B) [tex]6.2\cdot 10^{36}kg, 3.1\cdot 10^6 M_s[/tex]
The mass of the black hole can be found by equalizing the gravitational attraction between the black hole and the clumps to the centripetal force:
[tex]G\frac{Mm}{r^2} = m\frac{v^2}{r}[/tex]
where G is the gravitational constant, M the mass of the black hole, m the mass of the clumps.
Solving for M,
[tex]M=\frac{v^2r}{G}=\frac{(3.0\cdot 10^7 m/s)^2(4.6\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.2\cdot 10^{36}kg[/tex]
And since 1 solar mass is
[tex]M_s = 2.0\cdot 10^{30} kg[/tex]
the mass of the black hole in multuple of solar masses is
[tex]M=\frac{6.2\cdot 10^{36}kg}{2.0\cdot 10^{30} kg}=3.1\cdot 10^6 M_s[/tex]
C) [tex]9.2\cdot 10^9 m[/tex]
The radius of the event horizon of a black hole is given by
[tex]R=\frac{2GM}{c^2}[/tex]
where
G is the gravitational constant
M is the mass of the black hole
c is the speed of light
Substituting, we find
[tex]R=\frac{2(6.67\cdot 10^{-11})(6.2\cdot 10^{36}kg)}{(3.0\cdot 10^8 m/s)^2}=9.2\cdot 10^9 m[/tex]
