What is the ratio of radii for a black hole and steel ball in equilibrium?

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SUMMARY

The discussion centers on the equilibrium conditions between a black hole and a steel ball, specifically the ratio of their radii (R for the steel ball and r for the black hole). It concludes that the radius r corresponds to the event horizon of the black hole, while the stability of the steel ball depends on its mechanical properties and the mass of the black hole. The gravity of the combined system remains unchanged from its initial state, determined by the masses of both the black hole and the steel ball.

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JohnBarlow
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My first post here, hope to have more...

I have an idea/question probably brought about by watching too much Discovery Channel and I don't know who to ask, but I found this place so here goes.

Consider a black hole of radius r at the centre of a steel ball of radius R, where R>>r. I am not considering how it got there or whether steel is important - its just the way it is.

Now I am led to believe that the interior of the steel ball would be consumed by the black hole, until one of two things happen; either
a) the steel ball is completely consumed or
b) a state of equilibrium emerges where the interior is hollowed out leaving the black hole at the centre of a steel shell which resists further consumption because of the compressive strength of the ball and the distance of its thick shell from the black hole.

my question is - what is the ratio of R to r for condition b) to arise and what kind of gravity would the steel shell have compared to the earth?
 
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Hi, JohnBarlow,

Welcome to PF!

JohnBarlow said:
my question is - what is the ratio of R to r for condition b) to arise and what kind of gravity would the steel shell have compared to the earth?

The radius r would really be the radius of the event horizon, which I wouldn't call the radius of the black hole. A black hole in equilibrium has all its mass concentrated at a point of zero size.

The condition on R for the ball to remain stable would depend on the mass of the black hole and the mechanical properties of steel. All that can be guaranteed by simple relativistic arguments is that any part of the ball inside r will be consumed.

The gravity of the ball+hole would be the same as it was before you brought them together. It would depend on the mass of the black hole and the mass of the steel ball.
 
bcrowell said:
The radius r would really be the radius of the event horizon, which I wouldn't call the radius of the black hole. A black hole in equilibrium has all its mass concentrated at a point of zero size.

HUH? What does the event horizon have to do with answering this question? If I have it right, this paragraph even contradicts the one directly under it, which IS a correct statement of the result.
 

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