I know that a typical stellar black hole would spaghettify someone who crosses its event horizon.

Is this also true for a hypothetical tiny black hole with a small mass (the mass of an apple)? Would someone touching such a black hole spiral into it and get dead?


5 Answers 5


This is just a quick calculation that shows what would happen to a black hole with a mass equal to the mass of an apple: It is shown that not only the builder of this black hole but also the whole city around that person will probably be destroyed.

Let first compute the initial temperature of a hypothetical black hole with a mass equal to an apple, e.g., $M=0.2$ $\rm{kg}$. According to the Hawking radiation formula, for the temperature of a static (Schwarzschild) black hole, we have

$${T_{BH}} \approx 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{\rm{K} },$$

where $M_\odot \approx 2×10^{30} \rm{kg}$ is the solar mass (the mass of sun). For your black hole with the mass of an apple, one obtains

$${T_{BH}} \approx 6 \times 10^{23} \, {\rm{K} }.$$

This is extremely huge. Now, let's compute the lifetime of this black hole. This can be calculated by the use of the Stefan-Boltzmann law and assuming that the black hole initial mass ($M$) will eventually evaporate to nothing. For static black holes, this results in

$${\tau _{{\rm{life}}}} \approx (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3}.$$

Therefore, for your hypothetical black hole, the lifetime is

$${\tau _{{\rm{life}}}} \approx 10^{-19}-10^{-18} \, \rm{second},$$

which is extremely small.

Now, let's evaluate the relativistic energy of this hypothetical black hole by use of the Einstein's mass-energy equivalence, $E=Mc^2$, where $c$ is the speed of light. This amount of energy will be released during evaporation. It is straightforwardly obtained as

$$E \approx 1.8 \times 10^{16} \rm{J}.$$

It is huge. Comparing this amount with a nuclear bomb is interesting: A nuclear bomb typically has an explosion of about $10-10000$ kilotons of TNT and one kiloton of TNT is equivalent to one trillion Joules ($10^{12}$ $\rm{J}$) of energy (see WikiPedia). So, our hypothetical black hole released an amount of energy as much as a nuclear weapon!

These calculations show that if you could create a black hole with a mass equal to an apple, it will evaporate suddenly (${\tau _{{\rm{life}}}} \approx 10^{-19}-10^{-18} \, \rm{second}$) and release a big explosion equivalent to a nuclear bomb. So not only the builder of this black hole but also the whole city around that person will probably be destroyed.

And, finally, I think the problem of gravity and the event horizon radius of this hypothetical black hole is not important here. The radius of the event horizon, in this case, is extremely small (about $10^{-28}$ $\rm{m}$), much much smaller than the radius of the proton which is $10^{-15}$ $\rm{m}$! In addition, the gravitational field outside the black hole's event horizon is still the same as the gravity of an apple (spaghettification due to the tidal forces needs a very strong non-homogeneous gravitational field). On the other hand, the tidal forces act at the level of elementary particles of your body and there is no chance for any spaghettification in comparison with the black hole's lifetime. So, it's not important here.

Explosion of an apple-mass black hole

Here, a precise comparison with a nuclear bomb is discussed.

$1$ kilotons of TNT are exactly equivalent to $4.184 \times 10^{12}$ $\rm{J}$ (joules) of energy (See TNT equivalent convention). So, an apple-mass black hole release $4300$ kilotons of energy (See the computation in this link, using Wolfram alpha).

The first nuclear weapon which used in warfare exploded with an energy of approximately $15$ kilotons of energy (see WikiPedia). This amount of energy is equivalent to $698$ milligrams relativistic mass $m$ from Einstein’s formula $E=mc^2$ (See this computation in this link using Wolfram alpha). Comparing this ($15$ kilotons of energy = $698$ milligrams relativistic mass) with an apple-mass black hole ($4300$ kilotons of energy = $200$ grams) shows that the evaporation of apple-mass black hole in the laboratory would actually be pretty catastrophic, as it would release about as much energy as a nuclear bomb.

Furthermore, in comparison with a nuclear explosion, the blast radius (and the other relevant data) of an apple-mass black hole explosion with 4300 kilotons of energy has been presented in this link, again, by use of Wolfram alpha.

You may be interested in comparing the data of apple-mass black hole explosion with Tsar Bomba data as the most powerful nuclear weapon ever created and tested on our planet which yielded around $50$ megatons of TNT ($2.092 \times 10^{17}$ joules). See this link that I prepared using Wolfram alpha.

I have checked this calculation and improved them using Wolfram Alpha to be more accurate (thanks to everybody for your comments). The Hawking temperature and the lifetime formulas (including all the physical constants) can be found in Ref. [1]. This calculation is just an estimate and the order of magnitude is important here ... and I should emphasize that there may be other details that could probably change the final result and we have not considered them ... I try to discuss further another aspect in the next section.

Further technical information

You can skip this part. In this section, the relationships used in this answer together with three discussions (Hawking radiation contents, massive particle emission, and the graybody factor) are presented.

The Hawking (radiation) temperature formula [1]: $${T_{BH}} = \frac{{\hbar {c^3}}}{{8\pi G{M_ \odot }{k_B}}}\left( {\frac{{{M_ \odot }}}{M}} \right) = 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{{\rm{K}}} \tag{1}.$$

The black hole lifetime formula [1]:

$${\tau _{{\rm{life}}}} = \frac{{256{\pi ^3}k_B^4}}{{3G\sigma {\hbar ^4}}}{(GM)^3} = (2.095 \times {10^{67}}\,{\rm{year}})\,{\left( {\frac{M}{{{M_ \odot }}}} \right)^3} \tag{2}.$$

The rate at which the static black hole radiates energy [1]:

$$\,\frac{{dE}}{{dt}} = - \frac{{dM{c^2}}}{{dt}} = 4\pi r_s^2\sigma {\left( {\frac{{\hbar {c^3}}}{{8\pi GM{k_B}}}} \right)^4} \tag{3}.$$

$G=$ gravitational constant, $\hbar=$ Planck constant, $c=$ speed of light, $\sigma=$ Stefan–Boltzmann constant, $k_B=$ Boltzmann constant and ${r_s} = \frac{{2GM}}{{{c^2}}}=$ Schwarzschild radius.

On the contents of Hawking radiation:

The contents of Hawking radiation are fermions and bosons (both particles and anti-particles). By implementing the quantum field theory in the black hole background (i.e., by quantizing fermionic/bosonic fields in the curved black hole background), one finds the same Hawking temperature, Eq. $(1)$, for each fermionic/bosonic quantum field. However, for bosonic fields, one always obtains the Bose-Einstein statics while the Fermi-Dirac statics is obtained for fermionic fields. So, the Hawking temperature is the same for every type of particle (e.g., according to the surface gravity definition, ${T_{BH}} = \frac{1 }{{2\pi }}\sqrt { - \frac{1}{2}({\nabla _\mu }{\xi _\nu })({\nabla ^\mu }{\xi ^\nu })}$), and the number of species only adds a numerical factor to the Stefan-Boltzmann law (and consequently to the black hole's lifetime) but, still, the order of magnitude of the final answer is approximately valid.

On the emission of massive particles:

In addition, at the final stages of black hole evaporation, the production of pair massive particles will become relevant. In fact, there is a threshold $E > mc^2 $ for the production of massive quanta and the requirements needed for this to happen are simply provided at the final stages of black hole evaporation. And, perhaps more interestingly, it's been shown that massive particles tunnel across the horizon at the same rate as massless particles of the same energy $E$ even with $E < mc^2$, meaning that in such cases massive particles (with $E < m$) do not reach infinity, but they could be detectable at any finite distance (see Ref. [4]). So, in our example, we expect to see the emission of massive particles as well.

On the graybody factor ($\Gamma (\Omega)$):

In a more realistic situation, only a fraction of the emitted radiation reaches the asymptotic observer located at infinity. This means the particle (or the associated wave) scaping from the black hole needs to pass through the black hole potential and this change the resulting spectrum by a greybody factor ($Γ(Ω)<1$). Here, we assumed that $Γ(Ω)=1$, which is reasonable since the lifetime of a black hole is too small so particles can scape the black hole's potential.


[1] T.A. Moore, A General Relativity Workbook, University Science Books (2012)

[2] S.M. Carroll, Spacetime and geometry: an introduction to general relativity, Addison-Wesley, San Francisco, USA (2004)

[3] Wolfram Alpha

[4] G. Jannes, Hawking radiation of $E< m$ massive particles in the tunnelling formalism, JETP letters 94 (2011) 18-21.

[5] V. Mukhanov and S. Winitzki, Introduction to quantum effects in gravity, Cambridge University Press (2007)

[6] Thanks to @Andrew for his friendly suggestion about the radius of a proton.

[7] A suggestion: This handy calculator (for computing the Hawking radiation of static black holes) is really interesting and it is easy to work with (Thank you, @PM2Ring).

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    May 6, 2021 at 20:29
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    $\begingroup$ note that complete evaporation of a black hole assumes that the back-reaction of the hawking radiation with the semiclassical GR background continues to make sense as the size of the black hole gets arbitrarily small. This is likely to be a bad assumption. I think it is fair to say "this is a high temperature object likely to emit a lot of radiation", but saying what evolution the object will take is highly speculative absent a theory of quantum gravity. $\endgroup$ May 6, 2021 at 20:29
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    $\begingroup$ Simply put, the BH in question is an atomic bomb with a yield of 4.3 megatons TNT equivalent. Such a weapon is enough to turn a large city into a smoking crater. Doesn't matter if part of the radiation is particle radiation, it's equal parts of matter and antimatter anyways. As such, the antimatter particles will annihilate with matter around the BH, turning their mass into photons, with the result that virtually the entire weight of the BH is turned into the heat of a growing plasma ball, and the 4.3 megaton yield figure is precise. $\endgroup$ May 7, 2021 at 9:32
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    $\begingroup$ Tiny nitpick: the gravitational field at least ~5 cm outside the black hole's event horizon is still the same as the gravity of an apple. Significantly closer than that it's not, because the gravity field of an extended body (such as an apple) deviates from the inverse square law inside the body (and, if the body isn't perfectly spherically symmetric, also near but outside it). Thus your apple-mass black hole will in fact produce much greater tidal forces at microscopic distances (for the faction of an attosecond until it evaporates) than an apple would. $\endgroup$ May 7, 2021 at 9:55

No. A black hole created from something like an apple will still have the gravitational pull of an apple, so it's not going to suck you into it. If you swiped it it would pass through your hand, probably without doing any harm (maybe tidal forces might cause a little damage).

The bigger problem is that a small black hole like that will evaporate fairly quickly, producing a huge burst of Hawking radiation. When that happens it would probably kill you (and people for quite a radius around you).

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    May 6, 2021 at 20:29
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    $\begingroup$ I love the simplicity of "A black hole created from something like an apple will still have the gravitational pull of an apple". Apparently this question is the good old children's riddle 'what is heavier: a kilogram of lead or a kilogram of feathers?' all over again. I have known this riddle for almost 40 years so I can hardly believe that I fell for it, but somehow the allure of blackholes make me go all 'hmmm interesting question'. $\endgroup$
    – Vincent
    May 7, 2021 at 13:17
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    $\begingroup$ @Vincent The lead $\endgroup$ May 7, 2021 at 15:42

Black holes aren't actually any sort of super-strong gravitational field. They have basically the same gravitational field as a non-black-hole object of the same mass.

A stellar black hole is no more gravitationally dangerous than a star. What is different about a black hole is that it is tiny compared to a star. We often simplify gravitational calculations by treating planets and stars as point-masses; when using that approximation a black hole's gravity behaves identically to that of an object of the same mass. The special thing about black holes isn't that they have stronger gravity, it's that they are so tiny that you can get close enough to them that their ordinary-strength gravitational field can result in arbitrarily high attraction. IF the Sun were a point-mass, it would have an event horizon too... it's just that the radius where that "event horizon" would be is deep inside the surface of the Sun, so other things would have killed us on a dive into the Sun long before we would worry about spaghettification from the Sun's gravitational field.

So that suggests that an apple-mass black hole should be no more gravitationally dangerous than an apple.

It is true, you can get your finger closer to it than you could to an apple, although the event horizon would be subatomic1. But there is simply no way to get you compressed into a small enough space that the gravitational strength of an apple could force you to "spiral into it". Perhaps the tidal forces could rip some electrons or atoms out of you, maybe that would be enough to cause some detectable local damage to your body where it passed through; I honestly have no idea. But there would be zero effect on the rest of your body.

As a quick calculation, even assuming we could get a 100kg point-mass human at 1mm distance from a 200g point-mass apple, the gravitational attraction is only just over 1 millinewton of force. That is the weight of a few grains of rice (in Earth's gravity), and it was assuming that a human fit in a small enough space that its entire mass was pretty close to that 1mm radius away from the black hole. Regardless of what your actual mass is, given that the overwhelming majority of your body would be more than 1mm away from the apple-mass black hole2, I think it is safe to say that muscle power would be enough to overcome the force of gravity3 and avoid being dragged towards it.

1 So the notion of "touching it" doesn't really make sense, any more than you can "touch" individual electrons. But lets assume by "touch" you mean something like: it is located somewhere within one of the atoms making up the outer edge of your skin.

2 I am making assumptions about your body shape here, but in this particular instance I feel comfortable doing that.

3 Of course, if we consider other processes than just gravitational attraction, an apple-mass black hole is not at all a safe object to be around, and not stable enough for you to try any of this anyway, as other answers have explained.

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    $\begingroup$ Is it odd that I want to touch it now? $\endgroup$ May 7, 2021 at 16:21

I know that the answers here are complete but I’d like to explain differently about tidal forces in a very tiny black hole when we touch it.

In this problem, tidal forces have no chance to cause any damage. The black hole with the apple mass could only interact through gravitational force with the elementary particles of your body’s atoms. The interaction is well explained by Newton’s gravitational law as $F=Gm_1m_2/r^2$. For neutron ($m_n \sim 10^{-27}$ kg), proton ($m_p \sim 10^{-27}$ kg) and electron ($m_e \sim 10^{-30}$ kg), we have

$$F_{electron} \sim 10^{-19} N, \qquad (1)$$ $$F_{proton \, or \, neutron} \sim 10^{-16} N, \qquad (2)$$ where $G \sim 6 \times 10^{-11} $ $m^3kg^{-1}s^{-2}$ and $r=10^{-10}$ $m$ (atom size). Thus, for this case, it is about $10^{ - 19}$ N $\le F \le $ $10^{ - 16} $ N. The tidal force formula depends on the body's size and its distance from the another body. Therefore, we conclude that you cannot get into the black hole and before that it will explode.

Hope this helps.

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    $\begingroup$ What value(s) of $r$ are you using to compute the force? I guess $m_1$ is the mass of the black hole; does $m_2$ take a range of different values for some choice of elementary particles or atoms, and which values? Also technically the $1/r^2$ force is just an attractive force, the leading order part of the tidal force (the difference in the $1/r^2$ force over the length of the body) scales as $1/r^3$. Thanks! $\endgroup$
    – Andrew
    May 6, 2021 at 21:25
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    $\begingroup$ For $r$, I used the atom size and for the elementary particle masses I considered proton, neutron and electron. Thanks. $\endgroup$
    – mathLover
    May 6, 2021 at 22:06
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    $\begingroup$ I can't get your math to work out: even if I plug in 0.2 kg (a very large apple) for $m_1$, 1 dalton (~mass of a proton or a neutron) for $m_2$ and 1 ångström (~diameter of a hydrogen atom) for $r$, I get a force of only about $2 \times 10^{-18}$ newtons. Your general conclusion still seems correct, though. $\endgroup$ May 7, 2021 at 19:07
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    $\begingroup$ Anyway, for cases like this where $m_1 \gg m_2$, it's often more useful to consider the acceleration $a = F/m_2 = Gm_1/r^2$ experienced by the smaller body, since then the dependence on its mass cancels out. For a particle 1 Å away from a 0.2 kg black hole, I get $a ≈ 1\ {\rm Gm/s^2}$, so there would certainly seem to be some tidal disruption even before the black hole evaporates and blows up. But it would be very localized — the distance at which the black hole's gravity exceeds $9.81\ {\rm m/s^2}$ is only about one micrometer. $\endgroup$ May 7, 2021 at 19:14
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    $\begingroup$ @IlmariKaronen. Sorry for my mistake (corrected now). Thank you. $\endgroup$
    – mathLover
    May 8, 2021 at 7:46

The answer depends on its angular momentum and (thus) whether the singularity has an event horizon or not. Naked singularities don't evaporate.

The cosmic censorship conjecture makes the claim that they're all clothed, but it is nothing but a conjecture, and it's an area of active research to determine whether a clothed singularity can be stripped; e.g. https://link.springer.com/article/10.1140/epjc/s10052-019-6851-z

Loop quantum gravity, for instance, permits the formation of naked singularities. Actual observations haven't found any (yet), but there some close calls - like GRS 1915+105 (https://en.wikipedia.org/wiki/GRS_1915%2B105).

More importantly is what already holds true at the level of fundamental particles. A Kerr-Newman solution with the same mass, gauge charge and angular momentum as any fundamental particle is a naked Kerr-Newman singularity - possibly except the right neutrino and left anti-neutrino, if they exist, unless they have a gauge charge with respect to a "5th force" (in which case, them too).

Although it is tempting to just put one's head in the sand and issue the standard "classical physics doesn't apply at that scale" reply (which begs the question of the existence of a quantized theory of gravity), the fact remains that we're talking about scales in the 10^{-18} to 10^{-15} meter range. So, if these are not Kerr-Newman solutions, and if whatever they are cannot be already accounted for by classical General Relativity (GR), then that ipso facto means that there is a breakdown of GR at this scale, rather than the Planck scale, which prevents GR from being applicable, and prevents us from drawing the conclusion that they are, indeed, Kerr-Newman singularities (and hence: naked singularities) or something similar that is already accounted for by GR.


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