3
$\begingroup$

I've purchased a miniature black hole on ebay. I've heard they make excellent kitchen disposal units. Unfortunately the seller says it is for collection only.

Will I be able to move it?

Anti-gravity mats are the latest thing - I've already got one and tried it out. It's great fun.

Question

Even if I take my anti-gravity mat with me, will I be able to move the mini black hole? I'll have to push it parallel to the Earth's surface and I'm concerned I'll never get it moving because of inertia.

Are there any other snags I should be aware of?

Note: Anti-gravity mats work rather like a magic carpet but not quite. If you place a heavy weight on top of them, they behave like an immensely strong table. The advantage is that they can be moved sideways with little or no friction.

Correction It seems that I was wrong. Anti-gravity mats are impervious to gravity passing through them. Apparently this is similar to how a Faraday cage works.

Improve this question
$\endgroup$
11
  • 2
    $\begingroup$ That's a worryingly powerful anti-gravity mat if it can levitate a stable black hole. I believe the low-end estimate for the amount of mass is something like 1.5x the mass of Sol. Unless you're referring to a micro black hole, but those only last for a very brief amount of time. $\endgroup$
    – Halfthawed
    Jul 22, 2020 at 20:16
  • 1
    $\begingroup$ @Halfthawed what if I feed it well? $\endgroup$
    – John Dvorak
    Jul 22, 2020 at 20:16
  • 1
    $\begingroup$ What's the mass, approximately, of the black hole? That's the key parameter. $\endgroup$
    – HDE 226868 ♦
    Jul 22, 2020 at 20:26
  • 1
    $\begingroup$ @HDE - It has to be big enough to last for at least a year in my kitchen. I don't know how to calculate this. $\endgroup$
    – chasly - supports Monica
    Jul 22, 2020 at 20:28
  • 4
    $\begingroup$ @JohnDvorak there are problems with feeding small black holes, as paradoxical as that sounds. Black holes evaporate, the smaller they are the faster they evaporate. In the last few seconds of a black hole's lifespan that "evaporation" equals nuclear blast levels of energy, and you have to fire matter through that blast to feed it. There's also the problem that these black hole event horizons would be smaller than the atoms you are firing at it, so you are limited in how much matter goes in per second, the energy required to feed it and the big boom blowing your face off $\endgroup$
    – Demigan
    Jul 22, 2020 at 20:41

2 Answers 2

Reset to default
5
$\begingroup$

In the comments, you say:

It has to be big enough to last for at least a year in my kitchen. I don't know how to calculate this.

I don't know either because I'm the wrong kinda nerd for this, but the right kind of nerds have created online calculators for us. I used this one to estimate a mass of 7.2 $\times$ 107 kilograms. That's seventy thousand metric tons, or 2.5 times the weight of Lady Liberty.

The problem is not the mass but the density. Still according to the calculator, the black hole would have a radius of 10-19 meters. I know no material that could withstand that much weight on such a small point. If you magic mat can withstand it, though, a few locomotives working together or a really big cargo ship could maybe be able to move it.

Also, its surface gravity would be almost half of Earth's. You've got quite a beast there. But the kicker is that it would would be 1.7 quadrillion Kelvins hot, so even with your magic mat the air around it will accrete into a disc and emit enough x-ray radiation to cause a planetary catastrophe. You might consider keeping your black hole away from the solar system instead, for safety.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ Still using the same calculator: if you want a room temperature black hole, you don't need to go too much heavier, actually. Just 3.3 orders of magnitude, really. No, wait, scratch that. 13.3 orders of magnitude. Never mind. The good news is though, that "4e20 kg" is quite easy to remember. And it's still smaller than Vesta, if only by 33%. $\endgroup$
    – John Dvorak
    Jul 22, 2020 at 21:10
  • $\begingroup$ How would it have such high gravity? The statue of liberty 2 times in weight doesn't generate a gravity of half the earth? I understand that because everything is closer you can get closer to each atom, so it'll be a bit more powerful in that regard, but half the earth? Does it really grow in power that quickly near the surface? $\endgroup$
    – Trioxidane
    Jul 22, 2020 at 21:13
  • 3
    $\begingroup$ @Trioxidane the "surface" of a black hole is really tiny. Oh yeah, that the black hole "surface" is actually an all-devouring region of spacetime that projects no actual force (gravity doesn't count) is also an issue when you're trying to keep it in place, but let's assume the magic carpet has that little detail figured out. $\endgroup$
    – John Dvorak
    Jul 22, 2020 at 21:17
  • 1
    $\begingroup$ @trioxidane yes. But if you stand 1km away from it you get the same pull as you get from Lady Liberty while standing 1km from her. $\endgroup$
    – The Square-Cube Law
    Jul 22, 2020 at 21:19
  • $\begingroup$ @John Dvorak thanks. I know it's tiny and can do weird things after the schwarzschild collaps thing, but it still boggles the mind how much of a difference... $\endgroup$
    – Trioxidane
    Jul 22, 2020 at 21:26
0
$\begingroup$

If you precharge your black hole you may be able to hold it in a penning trap. Black holes conserve charge, so if you feed in a bunch of electrons then you can end up with a negatively charged black hole. A big issue is that it gets harder and harder to push new electrons in the more negatively charged it gets, and how do you hold your black hole at the endpoint of a huge particle accelerator before it is charged. I imagine you would do it in space with many accelerators shooting at it from all sides and active feedback to modulate the strength of the accelerators to keep the black hole centered.

Of course then you have the issue of getting the black hole to earth and moving a penning trap the size of a city block.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .