The molecular clouds circling the active nucleus of the galaxy M106 orbit at a speed of about 1000 km/s, with an orbital radius of 0.49 light-year = 4.8×1015 meters. Part A Use the orbital velocity law Mr=r×v2G to calculate the mass of the central black hole. Give your answer in kilograms.

Respuesta :

Answer:

The mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]

Explanation:

The Universal law of gravitation shows the interaction of gravity between two bodies:

[tex]F = G\frac{Mm}{r^{2}}[/tex]  (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

For this particular case M is the mass of the central black hole and m is the mass of the molecular cloud. Since it is a circular motion the centripetal acceleration will be:

[tex]a = \frac{v^{2}}{r}[/tex]  (2)

Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):

[tex]ma = G\frac{Mm}{r^{2}}[/tex]

By replacing (2) in equation (1) it is gotten:

[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex] (3)

Therefore, the mass of the central black hole can be determined if M is isolated from equation (3):

[tex]M = \frac{rv^{2}}{G}[/tex] (4)

Equation 4 it is known as the orbital velocity law.

Where M is the mass of the central black hole, r is the orbital radius, v is the orbital speed and G is the gravitational constant.

Before replacing the orbital speed in equation 4 it is necessary to make the conversion from kilometers to meters:

[tex]1000 \frac{km}{s} x \frac{1000 m}{1 km}[/tex] ⇒ [tex]1000000 m/s[/tex]

Then, equation 4 can be finally used:

[tex]M = \frac{(4.8x10^{15} m)(1000000 m/s)^{2}}{(6.67x10^{-11} N.m^{2}/Kg^{2})}[/tex]

[tex]M = 7.19x10^{37} Kg[/tex]

Hence, the mass of the central black hole is [tex]7.19x10^{37} Kg[/tex]