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DerKrebs [107]
3 years ago
13

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

round 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?
Physics
1 answer:
IRISSAK [1]3 years ago
8 0

A. 4.64\cdot 10^{11}m

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:

v=\frac{2\pi r}{T}

where we have:

v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

r is the orbital radius

T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

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:

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

and considering the value of the solar mass

M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s

C. 9.28\cdot 10^9 m

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

R=\frac{2MG}{c^2}

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

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

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<em />

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v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

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Learn more here: brainly.com/question/22406248

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