0
$\begingroup$

What if the Relativistic Heavy Ion Collider at Brookhaven created an unstable blackhole; could you survive this and how?

$\endgroup$

closed as off-topic by Renan, Alex2006, Ender Look, Frostfyre, sphennings Feb 25 at 20:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – Renan, Alex2006, Ender Look, Frostfyre, sphennings
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Welcome to Worldbuilding. If you have a minute, please take the tour and have a look at the help center. Please observe our [Code of Conduct]. Writing in capital letters is the equivalent of screaming and considered rude. $\endgroup$ – Elmy Feb 25 at 17:38
  • 1
    $\begingroup$ A: I’m not entirely sure that ‘unstable black hole’ is a real concept. B: What mass black hole are we talking here? C: The ‘and how’ part of the question is pretty broad, and very much depends on a host of secondary properties of the black hole, the planet, how the black hole forms and the preparedness of the person doing the escaping. $\endgroup$ – Joe Bloggs Feb 25 at 17:49
  • 1
    $\begingroup$ (1) That's the Relativistic HIC, not "Realistic". (2) What's an "unstable" black hole? (3) Since $m = E / c^2$, one can easily compute the maximum possible mass for whatever kind of hole of whatever color the Brookhaven machine could have produced. Hint: it's not large. RHIC draws about 77 MW; one full day of operation consumes 6.7 TJ, corresponding to a mass of 74 milligrams (0.04 avoirdupois drams in Ancient American). Such a tiny black hole will have a very very very short life... $\endgroup$ – AlexP Feb 25 at 18:01
  • $\begingroup$ If the energy and speed of the collision are high enough (I suspect that the RHIC at BNL would have to be substantially upgraded for this to happen) it is theoretically possible for tiny black holes to be produced, Hower their existence would be fleeting. In the sense that they are fleeting (their Hawking Radiation is sufficient to 'deplete' them before they can gather additional mass) perhaps that could classify them as 'unstable'. $\endgroup$ – Justin Thyme the Second Feb 25 at 18:21
  • $\begingroup$ If you do some googling, you will find some quite well thought out arguments regarding what would happen if a particle accelerator created a black hole. The general consensus is that there is no threat at all, unless some "new" physics is discovered at such high energies, but we can barely even speculate on what would happen if new physics came around beacuse it's...well.. not known. $\endgroup$ – Cort Ammon Feb 26 at 22:42
6
$\begingroup$

The government website says it can't, but we say it can! Unless scientists can somehow bring an incredible amount of mass to earth, the answer is yes, easily. Configuring particles to form an event horizon does not somehow increase their mass or gravity. You could conceivably pack lots of mass close enough to form an event horizon, but it would be incredibly small and incredibly brief. At any appreciable distance, that black hole's gravity would be exactly the same as what those particles had in their previous configuration.

$\endgroup$
0
$\begingroup$

What if the Realistic Heavy Ion Collider at Brookhaven created an unstable black hole; could you survive this and how?

That is a bit strange, not sure what is an unstable black hole, so I will rephrase it to:

What if suddenly a black hole appeared on Earth; could you survive this (gravity)?

Also, I will only take into account gravity and not other effects as time dilatation nor Hawking radiation.

It depends on the mass of the black hole. But for a particle collider, of course.

A black hole doesn't differ in other objects about its gravity. You may need to know about the Surface Gravity. On Earth, that is around 9.8 m/s 2, obviously easy to escape.

Given the following formula:

$$g = \frac{GM}{r^2}$$

Where $M$ is the mass of the black hole in kilograms, $r$ the radius (or distance between you and the body mass) in meters, and $G$ is the gravitational constant, (i.e: $6.67408(31)\times10^{-11}\text{m}^3\text{kg}^{-1}\text{s}^{-2}$) you can calculate the gravity.

With this value, you can calculate the gravitational force of an object (even a black hole) of any given mass to any given distance from it. It is to your choice to determine if you are able to escape or not with your current technology level (remember that if $g >= c$, where $c$ is the speed of light ($299,792,458 \text{ m/s}$) you can never escape, no matter how much you try).

Remember, a black hole of a few kilograms won't produce any relevant gravitation well after a few millimeters from it (instead, you should be careful about its Hawking radiation, not gravity).

A black hole produced in a particle collider would only have the mass of a few atoms, it would be harmless since its Schwarzschild radius (and gravity) will be incredibly small, it will literally die of hunger since particles won't fit on its "mouth" nor on its gravity well!

Another formula that may be useful for you is the Schwarzschild radius mentioned above, also know as gravitational radius, a distance from a black hole where nothing can escape. That is:

$$r_s = \frac{2GM}{c^2}$$

If you want to know the mass of a black hole enough strong to pull anything from a given range use:

$$M = \frac{c^2}{2G}R_s$$

If we resolve it with the radius of Earth which is $6.371\text{ km}$ do:

$$M = \frac{c^2}{2G}6.371.000\text{ m}$$ $$M = 4.28 \times 10^{33} \text{ kg} = 4,289,704,786.36 \text{ Yg}$$

Or $2,157.28 \text{ M}_\odot$, in solar masses.

Other Facts

As said in a comment, I ignored the Hawking radiation. Well, not anymore.

Hawking radiation is a process where the black hole emits energy, thus lowering its mass and shrinking. Due to this radiation, black holes has a limited lifespan and tend to evaporate. However, the evaporation is inversely proportional to the mass of the black hole, so a normal one will die after the thermal death of the universe.

Lifetime

The lifetime of a black hole is calculated given the following formula:

$$t = M^3\frac{5120 \pi G^2}{\hbar c^4}$$

Where $\hbar$ is the reduced Planck constant which is $\hbar = \frac{h}{2\pi}$, where $h$ is the Plack constant equal to $6.62607015 \times 10^{-34} \text{Js}$.

For a black hole of the mass of two hydrogen atoms (since you want to make it in the collider) would be very small. An hydrogen atoms has an atomic weight of 1.008 u. Where $u = 1.66053904020 \times 10^{-27} \text{ kg}$, very small. So its mass will have $3.3476467050432 \times 10^{-27} \text{ kg}$.

$$t = (3.3476467050432 \times 10^{-27})^3\frac{5120 \pi G^2}{\hbar c^4}$$ $$t = 3,155\times 10^{-95} \text{ seconds}$$

As you can see, very little. At that time the black hole will disappear in less time than the photons will take to reach the nearest atom of air in their surroundings.

But, let's do the opposite, we will try with the big black hole from above:

$$t = (4.28 \times 10^{33})^3\frac{5120 \pi G^2}{\hbar c^4}$$ $$t = 6.639 \times 10^{84} \text{ s}$$

Or, $2.105 \times 10^{77} \text{ years}$

Heat

The emission of Hawking radiation produces heat. As everything related to Hawking radiation, it is inversely proportional to the mass of the black hole. The formula is:

$$T = \frac{1}{M}\frac{\hbar c^3}{8 \pi k_b G}$$

Where $T$ is the temperature in degrees kelvin and $k_b$ is the Boltzmann constant equal to $k = 1.3806485279 \times 10^{-23} \text{ Jk}^{-1}$

$$T = \frac{1}{3.3476467050432 \times 10^{-27}}\frac{\hbar c^3}{8 \pi k_b G}$$ $$T = 3.365 \times 10^{49} \text{ k}$$

That is very hot! More that the hottest stars!

And with the big black hole:

$$T = \frac{1}{4.28 \times 10^{33}}\frac{\hbar c^3}{8 \pi k_b G}$$ $$T = 2.86 \text{ k}$$

... very little. Almost nothing. That is why big black holes last longer, they emit very little energy.

Luminosity

The emission of Hawking radiation produces light. As everything related to Hawking radiation, it is inversely proportional to the mass of the black hole. The formula is:

$$L = \frac{1}{M^2}\frac{\hbar c^6}{15360 \pi G^2}$$

Small black hole:

$$L = \frac{1}{(3.3476467050432 \times 10^{-27})^2}\frac{\hbar c^6}{15360 \pi G^2}$$ $$L = 1.063 \times 10^{59} \text{ W}$$

Like you can see, both the temperature and luminosity are incredible big, but since the duration of the black hole is very small, it won't cause any damage. For been exactly, the energy emitted during it lifetime is based on the mass and energy equivalence:

$$E = mc^2$$

So: $$E = 3.3476467050432 \times 10^{-27}c^2$$ $$E = 3.008 \text{ J}$$

Like I said, harmless.

Or $1.063 \times 10^{35} \text{ YW}$

More that the brightest stars or quasars, but just for an instant.

Big black hole: $$L = \frac{1}{(4.28 \times 10^{33})^2}\frac{\hbar c^6}{15360 \pi G^2}$$ $$L = 0.082 \text{ W}$$

P.D: Other facts math take from Hawking Radiation Calculator.

$\endgroup$
  • 1
    $\begingroup$ A reasonable answer, but you're ignoring black hole evaporation, which becomes entirely applicable for holes this small. $\endgroup$ – WhatRoughBeast Feb 26 at 19:10
  • $\begingroup$ @WhatRoughBeast done $\endgroup$ – Ender Look Feb 26 at 22:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.