What is the mass of the black hole orbiting the Milky Way?

[aeromat]

Homework Statement

Orbital motions are routinely used by
astronomers to calculate masses. A ring of
high-velocity gas, orbiting at approximately 3.4 × 104 m/s at a distance of 25 light-years
from the centre of the Milky Way, is considered
to be evidence for a black hole at the centre.
Calculate the mass of this putative black hole.
How many times greater than the Sun’s mass
is it?

Homework Equations

Relationship with Kepler's Law and Law of Universal Gravitation equation
Centripetal force
Law of Universal Gravitation

The Attempt at a Solution

I know 25Light years = 2.365 x 10^17m. This is the radius from the milky way to this gas. However, I am not sure how I am going to get the mass because when you make Fc = Fg, you are cancelling out the object that is orbitting and are left with the object being orbitted.

Example: the Earth orbiting around the sun
centripetal force equation:

mE[v^2]
--------
r

law of universal grav. equation:

G*(mE)(mS)
-------------
r^2
in this case when you re-arrange, the mE [mass of the earth; orbitting object] is canceled out.

How do we proceed with this question?

EDIT: What I am trying to do is solve for the milky way's mass, then sub it back into regular law of universal gravitation equation to solve for the black hole mass. Would this work?

 

[gneill]

The ring of gas is in a circular orbit around some mass, M. You have the radius and the velocity. Can you write two expressions for the centripetal acceleration that a particle in the ring would experience?

 

[aeromat]

All I can think of is Fg = Fc

 

[gneill]

You must remember something about circular motion? What's the formula for centripetal acceleration given the velocity and radius?

 

[aeromat]

centripetal acceleration = (v)^2/r

 

[gneill]

centripetal acceleration = (v)^2/r

Yes. Excellent. Now, gravity supplies the force that determines that centripetal acceleration. Do you know a formula for the acceleration due to gravity given M and r?

 

[aeromat]

But how can we use mass when there are two masses; one for the milky way, and the other for the black hole. I don't understand what to do at this point.

 

[gneill]

The mass of the milky way as a whole doesn't enter into the picture. You're only concerned with the mass of the black hole at the center and the particles in orbit in the ring of gas. The individual particles have negligible mass. There is a formula that specifies the acceleration due to gravity given a central mass M, and a distance r.

 

[aeromat]

I found this relationship between centripetal acceleration and the gravitational constant

M*ac = [v^2/r] = [GmM/r^2]
M*ac = [GmM/r^2]
Divide out the big M
ac = [G*m/r^2]

Now I am using this:

v^2/r = G*m/r^2 <--- to solve for the mass of the black hole. Is this correct?

 

[D H]

Correct. You know v, r, and G, so now solve for m.

 

[aeromat]

I did the following:
$$\frac{(3.4*10^4)(4.46*10^15)}{(6.67*10^-11}=M$$

It is actually 4.46*10^15, I don't know why it is going back to regular numbers

I received the answer: 1.64*10^35 kg
However, the back of the book says: 4.1*10^36 kg.
How the heck did they get that?

 

[D H]

Look at your units. Do they match? Don't just use numbers. Always carry the units.

Look at your equation. Did you use it right?

 

[aeromat]

$$\frac{(m^2/s^2)(m)}{(kg)(m/s^2)*(m^2/kg^2)}$$

Simplified and it comes out to be:
$$\frac{kg}{1}$$

So, I know it is not a problem with units. I converted light years to metres to get the radius, and the speed was given, and the gravitational constant is always given. Is this still incorrect?

 

[gneill]

I did the following:
$$\frac{(3.4*10^4)(4.46*10^15)}{(6.67*10^-11}=M$$

It is actually 4.46*10^15, I don't know why it is going back to regular numbers

I received the answer: 1.64*10^35 kg
However, the back of the book says: 4.1*10^36 kg.
How the heck did they get that?

$$\frac{(m^2/s^2)(m)}{(kg)(m/s^2)*(m^2/kg^2)}$$

Simplified and it comes out to be:
$$\frac{kg}{1}$$

So, I know it is not a problem with units. I converted light years to metres to get the radius, and the speed was given, and the gravitational constant is always given. Is this still incorrect?

Your velocity is not squared. Where did the number 4.46*10¹⁵ come from? 25 ly is more like 2.37 x 10¹⁷m. (Always carry units with your calculations).

 

[aeromat]

It is squared
$$(m^2/s^2$$ <-- see numerator portion.

Oops, I think I got the light year to metre conversion. Actually, it is my fault.

 

[aeromat]

Still getting the wrong answer; I get 1.21*10^32 kg taking account the correct lightyear-metre conversion.

 

[gneill]

Still getting the wrong answer; I get 1.21*10^32 kg taking account the correct lightyear-metre conversion.

So you say! Show your numbers if you want someone to check them.