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Gravity of a black hole (split from How large would a black hole need to be to overcome inflation and pull all matter one day into a big crunch?


Charles 3781

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30 minutes ago, MigL said:

As Joigus says, the gravity produced by a BH is no different than from any other equivalent mass.
If expansion can overcome the gravity of galaxy clusters, it can similarly overcome the gravity of a BH composed of the masses of the equivalent number of stars in that galaxy cluster.

There is no upper limit on BH size.
There is only a limit to how much you can feed them.
Once they 'eat' all close by mass via their accretion disc, they can't overcome farther out stable orbiting material, and stop growing.
Direct collapse, however,without going through star lifetimes, is a totally different mechanism.

I'd like, if I may, to pick up on one issue..  MigL citing a previous post from Joigus,  makes this statement:

"The gravity produced by a BH is no different from any other equivalent mass". 

Now, I wonder whether this takes into account,  the Black Hole's extremely small  physical dimensions. 

 Wouldn't these small dimensions cause a drastic reduction in the BH's ability to exert gravity.  Following the diminishing inverse-square law from its outer surface?

 

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5 minutes ago, Charles 3781 said:

I'd like, if I may, to pick up on one issue..  MigL citing a previous post from Joigus,  makes this statement:

"The gravity produced by a BH is no different from any other equivalent mass". 

Now, I wonder whether this takes into account,  the Black Hole's extremely small  physical dimensions. 

 Wouldn't these small dimensions cause a drastic reduction in the BH's ability to exert gravity.  Following the diminishing inverse-square law from its outer surface?

 

MigL also said "Once they 'eat' all close by mass via their accretion disc, they can't overcome farther out stable orbiting material, and stop growing."

His point being that anything close enough to be readily consumed by the BH has already been consumed. Nothing else is close enough to be impacted by  the extreme gravity found in the near vicinity of a BH.

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2 minutes ago, zapatos said:

MigL also said "Once they 'eat' all close by mass via their accretion disc, they can't overcome farther out stable orbiting material, and stop growing."

His point being that anything close enough to be readily consumed by the BH has already been consumed. Nothing else is close enough to be impacted by  the extreme gravity found in the near vicinity of a BH.

Yeah, but surely "Gravity" doesn't depend on "Mass" but also "Distance" - Inversely Squared.  Therefore if a Black Hole  gets smaller in size as it contracts, its gravitational pull diminishes.

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12 minutes ago, Charles 3781 said:

Yeah, but surely "Gravity" doesn't depend on "Mass" but also "Distance" - Inversely Squared.  Therefore if a Black Hole  gets smaller in size as it contracts, its gravitational pull diminishes.

From any given distance to a black hole the 'gravitational pull' does not change at all as a BH gets smaller. If our sun were to suddenly become a BH there would be no change to its gravitational pull on us.

The gravity gets stronger when you move closer to a center of mass. Imagine the strength of gravity on the surface of the sun. That is the strongest you can feel the sun's gravity.  But if the sun turned into a BH, you could get close to its center of mass since its 'surface' is now closer to its center of mass. THAT is where you would feel a stronger gravitational pull.

Edited by zapatos
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29 minutes ago, zapatos said:

From any given distance to a black hole the 'gravitational pull' does not change at all as a BH gets smaller. If our sun were to suddenly become a BH there would be no change to its gravitational pull on us.

The gravity gets stronger when you move closer to a center of mass. Imagine the strength of gravity on the surface of the sun. That is the strongest you can feel the sun's gravity.  But if the sun turned into a BH, you could get close to its center of mass since its 'surface' is now closer to its center of mass. THAT is where you would feel a stronger gravitational pull.

Thanks zapatos.  After reading your post, I couldn't get it at first.  But now mulling over what you said,  I think I've got it - a Black Hole is so intensely concentrated towards its centre,  that the pull of the  outer layers (if there are any) don't matter.

 

Edited by Charles 3781
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16 hours ago, Charles 3781 said:

Thanks zapatos.  After reading your post, I couldn't get it at first.  But now mulling over what you said,  I think I've got it - a Black Hole is so intensely concentrated towards its centre,  that the pull of the  outer layers (if there are any) don't matter.

 

No, that’s not it.

The pull at a distance r depends only on the mass inside of r (assuming spherical symmetry) If we compare the sun to a black hole if the same mass, as zap suggested, the gravity at the location of e.g. the earth would be the same, because mass and distance are the same 

You would have to be inside the sun to notice a difference, because then some of the mass would be outside of r, and mass outside of that sphere does not contribute to the gravitational pull

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On 10/12/2020 at 1:59 AM, Charles 3781 said:

Following the diminishing inverse-square law from its outer surface?

Just two things to note here:

1. Black holes don’t have a surface, they only have horizons of various kinds
2. The inverse square law doesn’t really apply in the immediate vicinity of a black hole; this is very much a relativistic scenario, so you’d have to use the full machinery of GR to correctly describe it

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On 10/13/2020 at 3:49 PM, Markus Hanke said:

Just two things to note here:

1. Black holes don’t have a surface, they only have horizons of various kinds
2. The inverse square law doesn’t really apply in the immediate vicinity of a black hole; this is very much a relativistic scenario, so you’d have to use the full machinery of GR to correctly describe it

Thanks Markus. 

Do you mean that the "Inverse Square" law is only applicable to mild flat Euclidean space, which is essentially 2-dimensional, as squares are.  Whereas in the immediate vicinity of a Black Hole", the BH's intense gravity pulls space from 2-D  into the full 3 dimensions,  where cubic effects take over in mathematical calculations. 

Is that why the maths of GR are so notoriously complicated, that they are very difficult to solve?

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17 hours ago, Charles 3781 said:

Is that why the maths of GR are so notoriously complicated, that they are very difficult to solve?

Gravity (in the universe we live in) is always a property of 4-dimensional spacetime. If you are far enough away from the black hole, the Newtonian inverse-square law will be a pretty good approximation; but the closer you get, the less accurate it will be, since the geometry of spacetime deviates more and more from being flat.

The maths of GR are more difficult because it is a highly non-linear theory, and you are dealing with a (potentially large) system of coupled partial differential equations; so this is a completely different ballgame, compared to the simple vector calculus of Newtonian gravity.

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