Imagine aliens dropped a small black hole (say, 1% of the moon's mass, so you wouldn't notice the difference in gravitation) into the sun (from the far side of the sun, so nobody on earth can see it). That small black hole would then, of course, start eating the sun, so that it eventually fails.

Now my question is: How long would it take until someone on earth would notice that there's something wrong with the sun? And how much time would they then have to react before the earth would become uninhabitable?

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    $\begingroup$ 'bout eight minutes, well at least ;) $\endgroup$
    – Ghanima
    Commented Dec 12, 2014 at 19:54
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    $\begingroup$ I think that I've heard that the earth would be the size of a golf ball if turned into a black hole, so I think moon would be more like a marble. $\endgroup$
    – bowlturner
    Commented Dec 12, 2014 at 19:57
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    $\begingroup$ You also have to figure out how they move something with the mass of the moon too! $\endgroup$
    – bowlturner
    Commented Dec 12, 2014 at 19:59
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    $\begingroup$ Small black holes evaporate, explosively. I'm not sure what the evaporation rate of a lunar sized black hole is. $\endgroup$
    – Oldcat
    Commented Dec 12, 2014 at 20:00
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    $\begingroup$ Remember the smaller the black hole the faster it decays. So you want it 'created' and shoved into the sun to grow before it detonates $\endgroup$
    – bowlturner
    Commented Dec 12, 2014 at 20:30

3 Answers 3


Good analysis by the others, but I want to add in some math here, because I'm really that nerdy.

We can model the growth of a black hole by the matter it accretes. Normally, a black hole accretes matter via a (surprise, surprise) accretion disk. Analysis of this type of object is nice because it's two-dimensional, for most practical purposes. Here, though, the accretion is decidedly three-dimensional. To analyze this, we have to model a phenomenon known as Bondi accretion.

The accretion rate onto a spherical body of mass $M$ in a medium of density $\rho$, the rate of accretion is $$\frac{dM}{dt}=\frac{4 \pi \rho G^2M^2}{c_s^3}$$ $G$ is the familiar universal gravitational constant, while $c_s$ is the speed of sound in the medium, a quantity that is actually pretty ubiquitous in studying astrophysical mediums.

Anyway, we can then write $$\int_{.01 \times M_{\text{Moon}}}^{M_{\odot}} \frac{1}{M^2} dM=\int \frac{4 \pi \rho G^2}{c_s^3} dt$$ $$\frac{1}{M_{\text{Moon}}}-\frac{1}{.01 M_{\odot}}=\frac{4 \pi \rho G^2}{c_s^3}t$$ and then, solving for $t$, we find $$t=\frac{(.01M_{\odot}-M_{\text{Moon}})(c_s^3)}{M_{\odot} \times .01M_{\text{Moon}} \times 4 \pi \rho G^2}$$ Of course, $.01 \times M_{\text{Moon}}\ll{}M_{\odot}$, but that's okay here.

Now, we know that $M_{\odot}=1.98855±0.00025×10^{30} \text{ kg}$, $V_{\odot}=\frac{4}{3} \pi r_{\odot}^3=1.41 \times10^{18} \text{ km}^3$, and $\rho=0.1403 \text{ kg/m}^3$, and that $M_{\text{Moon}}=7.3477×10^{22} \text{ kg}$. I haven't been able to find any figures for $c_s$, but we can still simplify the above equation to $$t=\frac{(1.98855±0.00025×10^{30}-7.3477×10^{18})(c_s^3)}{1.98855±0.00025×10^{30} \times .01 \times 7.3477×10^{22} \times 4 \pi \times 0.1403 \times 4.4528929 \times 10^{-21}}$$ As per ckersch's link, $c_s \approx 2,500,000 \text{ m/s}$. This means that $$t=\frac{(1.98855±0.00025×10^{30}-7.3477×10^{18})((2500000)^3)}{1.98855±0.00025×10^{30} \times .01 \times 7.3477×10^{22} \times 4 \pi \times 0.1403 \times 4.4528929 \times 10^{-21}}$$ $$=2.709 \times 10^{15} \text{ seconds}$$ $$=85.89 \text{ million years}$$

There are some things that were neglected here. For example, the black hole will lose some mass due to Hawking radiation, and the Sun can fail even if it doesn't lose all (of even the majority of) its mass. Still, though, this analysis should show you that we've got not a lot to worry about if a Moon-sized black hole decides to take a jaunt through the Sun.

Note: There may be an error here somewhere along the line (which I can't find just yet), but it appears to be around where I started plugging stuff in. At any rate, until I'm able to fix this, know that you can use Bondi accretion to figure out how long the Sun has to live.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – user
    Commented Dec 15, 2014 at 13:52

At that size of black hole (1% the size of the moon), the presence of the black hole in the sun wouldn't significantly change the life span of the sun. Using this calculator, we can see that the radius of such a black hole would be 10 micrometers. Not a whole lot of matter would be falling into a black hole that size compared to the rate at which the sun sheds matter, which is on the order of 1.5 million tons per second.

The micro black hole would form a small accretion sphere in the middle of the sun, but wouldn't significantly affect most of the sun, since the forces from pressure and the constant fusion reactions are far greater than the forces generated by the black hole.

  • $\begingroup$ That was my gut feeling - BH is way too small because of required density. +1 $\endgroup$ Commented Dec 12, 2014 at 21:37
  • $\begingroup$ Tiny nitpick: It would form an accretion sphere, not a disk. But your underlying point gets across nonetheless. $\endgroup$
    – HDE 226868
    Commented Dec 12, 2014 at 22:25
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    $\begingroup$ Didn't know accretion spheres were a thing. The more you know... :) $\endgroup$
    – ckersch
    Commented Dec 12, 2014 at 22:45
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    $\begingroup$ @Kevin Not until it eats up the Sun, because no significant amount of mass is added to the system. One lunar mass near the center of mass of the solar system should not affect anything. $\endgroup$
    – HDE 226868
    Commented Dec 13, 2014 at 2:28
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    $\begingroup$ @HDE226868 even if a black hole would eat up the whole sun it wouldn't "throw off the Earth's orbit" - the mass and orbits would stay unchanged, and any significant problems for humanity would be caused by the lack of emmited sunlight. The Earth would be just fine, though; it'd just be a snowball. $\endgroup$
    – Peteris
    Commented Dec 13, 2014 at 6:35

The smallest black hole found so far has 3.8 times the mass of the Sun.

Micro black holes are possible in theory but it is not obvious how they could be created or handled. And they are expected to evaporate (lose mass) via Hawking radiation. My quick skimming of that article showed that evaporation of a micro-BH is faster than matter intake (because of required density, it would be extremely small, atom-sized).

This article says black hole are capable of consuming nearby stars pretty quickly, in less than a million years - but the black hole discussed is pretty big to start with, not as small as OP's - with a mass of less than 15 Suns (which counts as small in BH-land).

The effects of a small Black hole on Earth were debated here

Yes, this danger is yet another reason to invest more into space-faring. Let's go places!

Of course there is no danger from such small black hole - but don't tell the Congress! Tell them is IS dangerous and we need space travel to deal with such eventuality!

  • $\begingroup$ A black hole of 1% moon mass would emit a small fraction of a watt as Hawking radiation, so it would be essentially stable (the formulas are at Wikipedia. I calculated it recently because I had the idea that a small black hole might be a solution for this question about a geocentric system and a black hole of that mass would have had about solar spectrum — but absolutely negligible radiation power). $\endgroup$
    – celtschk
    Commented Dec 12, 2014 at 20:14
  • $\begingroup$ How much will evaporate and how much sun plasma it can consume is deep physics question, I admit I let my my physics skills to atrophy. Maybe you should ask at physics exchange to get scientific answer, not just guts feeling? Do we have real physicists around here? $\endgroup$ Commented Dec 12, 2014 at 20:18
  • $\begingroup$ The negligible evaporation is not a gut feeling, that's a calculation using the formulas in the linked Wikipedia page. $\endgroup$
    – celtschk
    Commented Dec 12, 2014 at 20:22
  • $\begingroup$ Gut feeling is my answer. I have no doubts about your skills. That's why I suggested to ask precise question about rate of evaporation vs rate of mass consumption at physics exchange, because from the questions/answers I got impression that even if few people on this subforum are capable of hard physics, interests of majority is far from it. $\endgroup$ Commented Dec 12, 2014 at 20:28
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    $\begingroup$ Tell Congress and they'll declare war on it! $\endgroup$
    – HDE 226868
    Commented Dec 13, 2014 at 0:34

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