Question

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s.

A. How far are 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 kilograms and as a multiple of our sun's mass.
C. What is the radius of its event horizon?

Answer 1

A)
`4.6\cdot 10^(11) m`

The period of the orbit of the clumps around the black hole is

`T=27 h \cdot (3600 s/h)=97,200 s`

While their orbital speed is

`v=30,000 km/s=3.0\cdot 10^7 m/s`

And the orbital speed is equal to the ratio between the circumference of the orbit and the orbital period:

`v=(2\pi r)/(T)`

So re-arranging the equation, we find the radius of the orbit of the clumps:

`r=(vT)/(2\pi)=((3.0\cdot 10^7 m/s)(97,200 s))/(2\pi)=4.6\cdot 10^(11) m`

B)
`6.2\cdot 10^(36)kg, 3.1\cdot 10^6 M_s`

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:

`G(Mm)/(r^2) = m(v^2)/(r)`

where G is the gravitational constant, M the mass of the black hole, m the mass of the clumps.

Solving for M,

`M=(v^2r)/(G)=((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`

And since 1 solar mass is

`M_s = 2.0\cdot 10^(30) kg`

the mass of the black hole in multuple of solar masses is

`M=(6.2\cdot 10^(36)kg)/(2.0\cdot 10^(30) kg)=3.1\cdot 10^6 M_s`

C)
`9.2\cdot 10^9 m`

The radius of the event horizon of a black hole is given by

`R=(2GM)/(c^2)`

where

G is the gravitational constant

M is the mass of the black hole

c is the speed of light

Substituting, we find

`R=(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`