Fast Stars Near the Galactic Center (APOD 14 Jan 2007)

[Confused]

The APOD description says "over one million times the mass of our Sun is somehow confined to a region less than a fifth of a light-year across". How does the object (black hole or whatever) at the center of our galaxy compare in size to our Sun? I assume that our Sun is much, much smaller than a fifth of a light-year across.

[astro_uk]

Well technically the BH has no size whatsoever, it is a singularity, however people usually define a BH as having the same size as its event horizon (its Schwarzchild radius), the radius at which nothing
not even light can escape.

Rs = 2 x G x m / c^2

So in this case

Rs = 2 x (6.67x10^(-11))x 10^6 x 2x10^30 / (3x10^8 )^2 = 3x10^9m according to Google calculator.

So if I have typed that in correctly its about 5 times bigger than the Sun, seems a bit small, ill check it when I have a calculator I can trust. Though of course we are not comparing apples with apples.

[Andy Wade]

Well technically the BH has no size whatsoever, it is a singularity, however people usually define a BH as having the same size as its event horizon (its Schwarzchild radius), the radius at which nothing
not even light can escape.

Rs = 2 x G x m / c^2

So in this case

Rs = 2 x (6.67x10^(-11))x 10^6 x 2x10^30 / (3x10^8)^2 = 3x10^9m according to Google calculator.

So if I have typed that in correctly its about 5 times bigger than the Sun, seems a bit small, ill check it when I have a calculator I can trust. Though of course we are not comparing apples with apples.

When you say 5 times bigger, do you mean in diameter, or volume, or something else?

Or would it be better described as say a radius from the centre of the sun to say, the orbit of Venus?
I need things explained in very simple terms if possible as, well, I'm very simple...

[astro_uk]

I mean its radius is 5 times larger, the sun has a radius of about 7 x 10^8 m. Though I still have my doubts until I have a look at the numbers with a real calculator.

The gravitational force would still be huge at much greater distances than this number though, still plenty to keep those stars whizzing about.

[cosmo_uk]

I would assume all they mean is that based on the radius of the orbiting stars all they can say is that the mass is within some radius within this upper limit not that the black hole has this radius

[orin stepanek]

http://antwrp.gsfc.nasa.gov/apod/ap070114.html
Now I'M confused. It takes sunlight about 8 min to reach the Earth. now a fifth of a light year would take light a little over two months to traverse. Or is that going back to the size of the event horizon?
Orin

[StACase]

I like the animation, (Who doesn't like things that move) anyway a couple of thoughts:

1. Are those stars following a standard orbital path?

2. Is anyone going to complete the animation via computer model and maybe even do it in 3-D?

[kovil]

The photo is about one light year wide (in the description), = 8-1/2" on my screen, the center star moving at the yellow cross is 3/8" dia. That's 1/22.6667 of the photo -- 1LY =6 trillion miles / 22.6667 = 264.7 billion miles. The photo's motion is over 8 years, the center star moves one diameter in the animation, that = 33 billion miles in one year, which = 1049 miles per second or 0.56% speed of light. Ball park of course, but do the stars really appear as 265 billion miles wide?? that doesn't make sense, but maybe the magnification and time exposure make the suns appear larger than life. Considering our solar system is 6 billion miles in diameter apx, that's 1/1000 th of a light year. (*1LY = 5.88 T miles oops ! This is thumb at arms length anyway) Our solar system is only 2.5% of the suns diameter in the photo, which again doesn't make sense.


Gravitational attraction of 1 million solar masses at 1/10 of a light year.
I'm having trouble finding an equation for this question, Wikipedia is not helping, and gravity is not my strong suit in math.

[Confused]

I like the animation

I am limited by Windows and don't get animation. Of course, it might not be Windows; perhaps all other Windows users get the animation.

[astro_uk]

The size they talk about is, I think, the size that can be inferred by the motion of the stars, this is not necessarily related to any physical scale of the BH, if there were stars that were closer to the BH, they could pin down its size more accurately.

Ball park of course, but do the stars really appear as 265 billion miles wide?? that doesn't make sense, but maybe the magnification

The stars are unresolved, they are not magnified, the size they appear on the image is due to the optics of the system. In reality they should appear as points, but due to the effect of the telescope optics and the Earths atmosphere the stars light gets smeared over several pixels making them seem much larger.

Gravitational attraction of 1 million solar masses at 1/10 of a light year.
I'm having trouble finding an equation for this question, Wikipedia is not helping, and gravity is not my strong suit in math.

I think the equation you want is the standard Newtonian one, it should be good enough in this case.

F = (G x M1 x M2) / R^2

F = (6.67x10^(-11) * (10^6 * 2*10^30) * 1 / (0.1 x 9.4605284 × 10^15)^2

Giving a force of 1.5 x 10^(-4) N per kg at 1/10 lyr.

[BMAONE23]

The Star at the 9:00 position appears to have a black spot in the middle of it and the central core looks like it is slowly spinning. It is most likely bright light burnout but eerily resembles a black hole accretion disk with a centrally spinning event horizon vortex.

[ta152h0]

has that Jan 14, 2007 APOD been impounded ? Can't seem to get it displayed ?

[orin stepanek]

If you log onto APOD and click on to January; Sunday; the 14th box. You should get it. even though the box don't show it. http://antwrp.gsfc.nasa.gov/apod/calendar/ca0701.html
try it!
Orin

[ta152h0]

worked, thank you. now I can practice my math

[iamlucky13]

Well technically the BH has no size whatsoever, it is a singularity, however people usually define a BH as having the same size as its event horizon (its Schwarzchild radius), the radius at which nothing
not even light can escape.

Rs = 2 x G x m / c^2

So in this case

Rs = 2 x (6.67x10^(-11))x 10^6 x 2x10^30 / (3x10^8 )^2 = 3x10^9m according to Google calculator.

So if I have typed that in correctly its about 5 times bigger than the Sun, seems a bit small, ill check it when I have a calculator I can trust. Though of course we are not comparing apples with apples.

That sounds reasonable to me. Keep in mind that if a black hole with the mass of our sun would have a radius of only 3 km (so sayest the great Wikipedia). For a million times the mass you got a million times the radius, and indeed, the radius is linear with respect to the mass. Your math looks solid.

1. Are those stars following a standard orbital path?

They may be in orbit around Sgr A, but they don't have to be. If the stars have enough velocity, they will simply fly by, with their trajectories being bent by the pull of the black hole. They probably are orbiting Sgr A, but you'd probably have to do quite a bit of modeling to be sure.

[makc]

Is anyone going to complete the animation via computer model

It is quite possible, you have set of points and the center, you could feet it to ellipse or parabola, and then expand animation the way you like, but... ain't gonna do it I am quite busy with other thing, you know.

[DavidLeodis]

If you log onto APOD and click on to January; Sunday; the 14th box. You should get it. even though the box don't show it. http://antwrp.gsfc.nasa.gov/apod/calendar/ca0701.html
try it!
Orin

Thanks Orin for providing a link to the APOD of January 14th 2007, as I cannot find that APOD on the APOD archive list that I've seen. Odd that it does not appear, as the APOD for the 15th is in the archive!

[kovil]

"For a million times the mass you got a million times the radius, and indeed, the radius is linear with respect to the mass. "

4/3 pi Radius ^3 = volume of a sphere , yes?

or, the cube root of one million times the radius, you mean?
or 100 times the radius of 3 kilometers = 300 km radius of the sphere of 1 million solar masses, would that equal the event horizon ??

I have some homework to do as the units of G (6.67 x 10 (-11)) aren't familiar.

[Martin]

I could have missed something but I always thought that when a BH's mass is mentioned, that it is just a rough comparison to an observable star's mass. For a BH cannot have "mass" per say.

For example:

If we observe objects that are near enough to an alleged BH we can roughly calculate what the mass of the BH would be, if it were a star, in order to gravitationally interact with those objects from "that distance".

So, ultimately we cannot calculate mass for something that is unobservable. We can only calculate its gravitational effect on surrounding matter and compare it to the mass of what we can observe -like stars.

A crude laymen’s explanation I am sure -but is my understanding flawed?

[iamlucky13]

"For a million times the mass you got a million times the radius, and indeed, the radius is linear with respect to the mass. "

4/3 pi Radius ^3 = volume of a sphere , yes?

or, the cube root of one million times the radius, you mean?
or 100 times the radius of 3 kilometers = 300 km radius of the sphere of 1 million solar masses, would that equal the event horizon ??

I have some homework to do as the units of G (6.67 x 10 (-11)) aren't familiar.

That's for correlating the volume to the radius, yes. If it has a uniform density, the mass also. However, the Schwarzchild radius defines the event horizon, which is dependent on the strength of the gravitational field.

Note that mass of a uniform volume of matter in normal space scales according to the radius-cubed, while gravitational acceleration falls off according to the radius-squared. If you divide, you're left with one factor of the radius. I've never gone through the steps of deriving the equation, but I suspect this fact shows up somewhere in the steps.

The units of the gravitational constant here are m^3 / (kg*s^2) or N*m^2 / kg^2


Martin, I think there's some confusion coming from methods of determining mass. With nothing to see for a black hole, you can't just measure it's radius. There's also no light intensity for you to measure and infer it's mass by the types of stars that radiate that much of various wavelengths. So you are correct that the mass has to be determined by the gravitational effect on nearby objects.

But, the black hole does have mass. Some quantity of matter went into the formation of the black hole (unless Sgr. A isn't really a black hole, but it is generally accepted to be), so at least in this sense it has mass, even if it's collapsed down to a singularity or some form of degenerate matter hidden behind the event horizon.

[FreebirdsWB]

Is this not the projected orbit of the stars?
http://www.mpe.mpg.de/ir/GC/index.php

Also, curious... should we expect to see some (any?) lensing?

[iamlucky13]

Hey, that is a nifty animation you found Freebirds. Thanks.

If an object passes behind the black hole, there can be lensing, but I'm not sure how significant it would be unless it passed exactly behind. Given that we're talking about a volume of space ~10^18 times bigger than the black hole, that's probably very unlikely in the time spans we have the luxury of watching.

[makc]

should we expect to see some (any?) lensing?

we can hardly see anything at all there, yet any objects behind the black hole.

[StACase]

Thanks Freebirds that is exactly what I was thinkig of (-:

[harry]

Hello All

Hello Martin

You said

If we observe objects that are near enough to an alleged BH we can roughly calculate what the mass of the BH would be, if it were a star, in order to gravitationally interact with those objects from "that distance".

So, ultimately we cannot calculate mass for something that is unobservable. We can only calculate its gravitational effect on surrounding matter and compare it to the mass of what we can observe -like stars.

I fully agree with you.