interesting question.
but it seems to me that 10^12 kg is quite heavy, the ping pong ball would collapse even though it is outside of the event horizon. then the water would come in and further feed the black hole.
i think even the bucket that it was in would be crushed. but that might be all. the black hole would sit at the bottom of the bucket slowly sucking away the air and growing until it can crush the bucket then if there is anything else near it they could be sucked in, and the air would continually be sucked in. eventually such a black hole could destroy earth.
umm, as for the buoyancy, the ball would sink, 10^12 kg in the size of a ping pong ball would certainly be more than the density of water, so the ball would sink, assuming nothing collapses.
2007-05-29 08:03:32
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answer #1
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answered by Tim C 5
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The force water exerts is exactly equal to the weight of the water being displaced by the object. So it doesn't matter whether the ping pong ball is empty or weighs 10^12 kg. It displaces 4/3pi(r^3) of water, and water has a density of 1g/cm^3. So it will displace 4/3*pi*8 cm^3. This should also be its mass in grams.
To get its weight in newtons (and therefore find the force of the buoyancy), you need to multiply this by 1000 and the water's acceleration due to gravity (f=ma). This should be G*mass of black hole/(.01 m)^2.(g=Gm/r^2)
So I have no idea what these numbers turn out to be, but that's the method I came up with. I'm kind of assuming that density does NOT change with increased gravity and that it's more of a introductory physics problem. Hope it works out.
Edit: Nevermind, it's 0. The force it's exerting is equal in all directions.
2007-06-05 09:14:07
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answer #2
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answered by razorj06 2
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An interesting thought problem. Given that there is no force acting to suspend the mass of the black hole in the center of the ping pong ball.
One thing you have to do is to remove the influence of other gravitational fields. Otherwise the mass would quickly fall to the bottom of the pail and consume the ball, water and bucket.
In any system the downward pressure of the mass would far exceed the negative effect of buoyancy. You could calculate the displaced mass of water dictated by the volume of the ball. It is insignificant when compared to the mass of the black hole.
2007-06-04 04:21:23
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answer #3
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answered by morgan j 4
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I assume that the water is incompressible and that the gravitational field of the blackhole has no effect on the water. If this assumption is correct, the buoyancy will be exactly the same as the buoyancy exerted on a normal ping-pong ball of the same diameter.
If the density of water has gone up because of the gravitational pull of the mini-blackhole, the buoyanct also goes up. By how much, I have no idea.
2007-05-29 06:20:55
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answer #4
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answered by Swamy 7
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In fact, there would be no buoyancy. Even at that infintesimal amount of mass, the black hole actually consumes the ball and the event horizon grows as the mass expands to encompass the surrounding water.
2007-05-29 06:19:11
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answer #5
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answered by scoffron 2
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confident monkey interior the midsection with a compelled rite to introduce the nut sucking monkey vs. guy mayhem. Or a minimum of that's how i could react to the situation. at the start, wreck the ping pong balls. i'm sorry, yet they must circulate. Secondly, your buddy's automobile is on hearth. positioned that sh*t out with a paddle and a few pumpkin juice. finally, i'm sorry, yet your paraplegic chum, Paula, exceeded directly to the finished beyond mutually as pooping on a piano in the past Pope Peter's ***** social gathering.
2016-11-23 14:43:39
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answer #6
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answered by ? 4
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This is a nonsense question not worth any time spent on it.
2007-06-04 03:56:13
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answer #7
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answered by johnandeileen2000 7
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for me it's impossible (i didn't read the paragraph)
ps. i teach at caltech
2007-05-29 06:18:05
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answer #8
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answered by mhd1995 1
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