Black Hole Gravity: Mass vs Distance

[scilover89]

Are mass and distance the only factor to affect gravity?
When star become a black hole, the mass remain the same, but the gravity of the black hole is certainly stronger than star. Why?

 

[somy]

Well,
In the case of a black hole you can not talk about the simple Newton's law! In fact I'm not fammiliar with a black hole modeling! but I know in that case, the structure and the interaction of the particles is different. And I think you and me(!) have to know lots of deep concepts inorder to percept the black hole!

 

[scilover89]

Well,
In the case of a black hole you can not talk about the simple Newton's law! In fact I'm not fammiliar with a black hole modeling! but I know in that case, the structure and the interaction of the particles is different. And I think you and me(!) have to know lots of deep concepts inorder to percept the black hole!

How is gravity affected by the interaction of the particles?

 

[Chronos]

A black hole behaves just like any other massive body until you get very close to it. It is simply a gravitating structure that is very small. Once you get extremely close to it you gain a whole new perspective on the power of gravity.

 

[somy]

I'm really unfammiliar with the black hole concept. And I think it needs the knoledge of quantum and general relativity. You can visit these:
http://nbsp.sonoma.edu/resources/teachers_materials/hd/black_holes_activity.pdf
http://www.sciencedirect.com/science?_ob=MImg&_imagekey=B6TVP-4F05RCP-1-2C2&_cdi=5540&_user=1400317&_orig=search&_coverDate=01%2F01%2F2005&_sk=995939997&view=c&wchp=dGLbVzb-zSkWb&md5=2f990ea11f78e76268731657ed0f9532&ie=/sdarticle.pdf
http://www.sciencedirect.com/science?_ob=MImg&_imagekey=B6TJC-4D99BMG-1-45&_cdi=5307&_user=1400317&_orig=search&_coverDate=03%2F01%2F2005&_sk=999459998&view=c&wchp=dGLbVzb-zSkWb&md5=c15a1a98a0ad66b2c699eb065a0e6966&ie=/sdarticle.pdf

 

[kirovman]

For example a mass that of the earth.
If you go inside the Earth's surface, you can only count the gravitational affect from all that mass below you, in a Gaussian sphere(I think). Only the mass within the Gaussian sphere, defined by the radius between you and the centre of mass, only this counts towards the gravitational attraction between you and the earth.

ie When you get to the centre of the earth, r=0, there is no mass in this surface and you are weightless. ie g=0

A black hole is different. Imagine the black hole with the same mass as the Earth. The mass is ALL at the centre of mass, and so the mass within the gaussian surface does not decrease, and the gravitational attraction will build up, and when you get to the centre, g=k/r balloons to infinity. Far away from the black hole, gravity will be the same as the earth, but when you start comparing to the gaussian surface and mass contained within, ie r<R (where R is radius of earth, r is your radius), you will get differences, due to the phenomena described above.

Sorry if I haven't explained that very well. I'm hoping there's no errors, I learned something about that 2 or 3 years ago.

EDIT: I wish I could draw some diagrams...that would be simpler to explain.

 

[scilover89]

Thank you kirovman, but...g=k/r, what does k and r represent? (Sorry, I am just a secondary school student)

 

[kirovman]

Thank you kirovman, but...g=k/r, what does k and r represent? (Sorry, I am just a secondary school student)

Oh sorry, k is just a constant of proportionality (probably equal to G I think)

actually I think I did that formula wrong... it should be g = k/(r)^2 I think.

It's such a long time ago. it's either r or r^2, but anyway the principle is the same. As you get to r=0, g ---> infinity for a black hole, using classical laws.

r is the radius from the centre of mass (of the earth)

 

[Chronos]

A black hole obeys the inverse square law just like a normal object. If one partner in a binary star system were to suddenly collapse forming a black hole, it's companion would continue to orbit it as if nothing had happened. The total attractive force depends solely on mass and distance, as kirovman noted. The field is simply more intense near a black hole due to its infinitesimal volume.

 

[DB]

Are mass and distance the only factor to affect gravity?
When star become a black hole, the mass remain the same, but the gravity of the black hole is certainly stronger than star. Why?

Check this out: http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag2.jpg

As Chronos and kirovman has said, a black hole will have the same force of gravity as a massive star from a safe distance. Too add a little, this safe distance is considered away from the event horizon. Beyond the event horizon there is now return. The reason a black hole has such a strong gravitational force is because it has the same mass as a massive star (the one it used to be) but now this mass has been cramed into like Chronos said infinitesimal volume. So the density of a black hole's singularity (it's centre) is near infinite and today unmeasureable. In space, if an obect keeps it's mass but has higher densities, it curves spacetime more (like in the link), therefore leading to a stronger gravitational force.

 

[awhalen]

Are mass and distance the only factor to affect gravity?
When star become a black hole, the mass remain the same, but the gravity of the black hole is certainly stronger than star. Why?

When a super massive star with a mass of ten or more suns collapses into a black hole gravity is powerful enough to collapse the atom structure its self. With the same amount of gravitational attraction as before but less volume gravity becomes greater in a volume to gravity ratio and grows to infinity within milliseconds after the super nova. So in a sense gravity and mass stay constant the volume to gravity ratio changes so the gravitational ATTRACTION is the only thing that changes

 

[Chronos]

The gravitational force of a black hole is no different than that of an uncondensed matter object - until you get very close.

 

[awhalen]

The gravitational force of a black hole is no different than that of an uncondensed matter object - until you get very close.

the gravity of a black is infinite at the event horizon and very strong for light years out. a black hole is a never ending hole in space and will gravitationally attract matter near the hole.

 

[yenchin]

the gravity of a black is infinite at the event horizon and very strong for light years out. a black hole is a never ending hole in space and will gravitationally attract matter near the hole.

This is not correct. The surface gravity at the event horizon is finite [See e.g. http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole]

 

[awhalen]

This is not correct. The surface gravity at the event horizon is finite [See e.g. http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole]

wikipedia is not a valid source and can not be used as a reference

 

[russ_watters]

wikipedia is not a valid source and can not be used as a reference

It is better than no source, which is what you used...

 

[twofish-quant]

Are mass and distance the only factor to affect gravity?

Yes.

When star become a black hole, the mass remain the same, but the gravity of the black hole is certainly stronger than star. Why?

It's not. If the sun suddenly turned into a black hole, the gravitiation force is the same. The difference is that with a black hole, you get 3km to the center, whereas with the sun, you can't get that close without going into the sun.

 

[awhalen]

It is better than no source, which is what you used...

Try http://www.astronomytoday.com/cosmology/blackholes.html or
http://www.astrophysicsspectator.com/topics/generalrelativity/BlackHole.html

 

[yenchin]

wikipedia is not a valid source and can not be used as a reference

Fine. In that case you should refer to *any* valid *textbook* on general relativity, see e.g. "A Relativist's Toolkit: The Mathematics of Black-Hole Mechanics" or "Spacetime and Geometry: An Introduction to General Relativity". Or Carroll's online lecture notes [http://arxiv.org/PS_cache/gr-qc/pdf/9712/9712019v1.pdf] equation 7.155. Surface gravity is perfectly well-defined, *not* infinite.

 

[russ_watters]

Try http://www.astronomytoday.com/cosmology/blackholes.html or
http://www.astrophysicsspectator.com/topics/generalrelativity/BlackHole.html

1. If you're going to complain that wikipedia isn't a valid source, then you shouldn't be providing peoples' personal websites as your sources.
2. While I don't think they are all that clear on the issue, neither of them agree with your claim.
3. I'll put a finer point on it: You are not correct in your assertions and any reliable source that deals with the issues you brought up directly will confirm that. And on this point, the wiki is accurate.

The only reliable source I have handy (I'm an engineer, not a physicist) is "A Brief History of Time", which says "The star [after collapsing into a black hole] would, however, continue to exert the same gravitational force on the spaceship, which would continue to orbit the black hole."

http://books.google.com/books?id=4Y... gravitational force on the spaceship&f=false

 

[FtlIsAwesome]

the gravity of a black is infinite at the event horizon

No, the event horizon where escape velocity reaches lightspeed. Gravitational strength is not infinite at the event horizon.

and very strong for light years out.

Depends on what you mean by "very strong". A suspected supermassive black hole at the center of the Milky Way causes several stars to orbit it; these stars have orbital distances of around lighthours. I'm not sure of any stars that directly orbit the black hole and have orbital distances of lightyears. (All the stars of the Milky Way orbit the center, but it is more complicated than that because of the gravity of other stars)

But if you think that if the Sun turned into a black hole it would suck up Alpha Centauri, no.

a black hole is a never ending hole in space

In sense, but if you crossed the event horizon you would hit the singularity after a while.

and will gravitationally attract matter near the hole.

Yes, black holes do exert gravitational influence on other objects. But this only gets extreme at close distances; objects far away (such as Earth orbiting the Sun) would not have their orbits affected.

 

[awhalen]

A black hole behaves just like any other massive body until you get very close to it. It is simply a gravitating structure that is very small. Once you get extremely close to it you gain a whole new perspective on the power of gravity.

That "new perspective" on the power of gravity would be no different than the gravity we know now. It is only when we are watching somthing fall in do we notice a big change. i.e. "spagittifaction". If we fall into a black hole your perspctive does not change.

 

[Jack21222]

As you're being torn apart by tidal forces, your perspective doesn't change? Please stop.

Go take a class or read a textbook before continuing to post, awhalen.

 

[George Jones]

I think folks are using a number of different meanings for "gravitational force", including:

1. The gravitational influence of a star on an external object, like an orbiting planet, remains unchanged as the star collapses to form a black hole.

2. The force required to keep an object "stationary" goes to infinity at the event horizon of a Schwarzschild black hole.

3. The tidal force on a spatially extended object causes spaghettification, which happens outside the event horizon of a "small" black hole and inside the event horizon of a "large" black hole.

4.The surface gravity, which is finite, is the force required at infinity to keep a unit-mass object hovering at the event horizon.

 

[K^2]

This is not correct. The surface gravity at the event horizon is finite [See e.g. http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole]

Incorrect. Take the norm of the acceleration vector using metric at the horizon, and you'll see that it is infinite in magnitude, despite all its components being finite.

Gravity at the event horizon is, indeed, infinite. See any GR book for derivation.

 

[George Jones]

Incorrect. Take the norm of the acceleration vector using metric at the horizon, and you'll see that it is infinite in magnitude, despite all its components being finite.

But this not the definition of surface gravity. yenchin is correct; see my previous post.

 

[K^2]

If the tension in the hypothetical cable suspending an object at event horizon is infinite, why would it be finite on the other end of the cable? I'm not sure how measuring force at infinity makes any difference.

Edit: Besides, surface gravity is defined as acceleration experienced by object, which, again, is infinite.

 

[George Jones]

If the tension in the hypothetical cable suspending an object at event horizon is infinite, why would it be finite on the other end of the cable? I'm not sure how measuring force at infinity makes any difference.

Edit: Besides, surface gravity is defined as acceleration experienced by object, which, again, is infinite.

Again, this is not the usual definition of surface gravity for black holes. Look in any standard book. For a Schwarzschild black hole, surface gravity is 1/(4M). For example, see section 5.2.4 on page 141-143 (pdf page 157-159) of Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of black hole Mechanics.

 

[K^2]

That derivation seems screwed up. It basically tells you that the amount of force at two ends of the rope is different, which means that conservation of total momentum is violated. (F=dp/dt regardless of metric.) I'm pretty sure that the problem is with assuming that ds at infinity results in ds at the event horizon. The displacement at the event horizon will be different due to the metric. Take that into account and you should get infinite force required to support the object from infinity as well. (Edit: I do need to think about this a bit more. I might be wrong, but it really seems off to me for reasons stated above.)

If you wish to define surface gravity differently, fine. But that has nothing to do with how the discussion started. Acceleration of an object at rest at the event horizon is infinite. Force required to support it is infinite.