How long until a small black hole makes the sun fail?
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?
1 answer
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.
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