Black hole the size of an apple

I was intrigued by this question: Would touching a black hole of a small mass (the mass of an apple) cause you to spiral in and get dead? until several answerers pointed out that a black hole with the mass of an apple would just have the gravitational pull of an apple and so would not be very interesting from a gravitation perspective. (Although it could be interesting in other ways, notably Hawking radiation.)

I felt a bit stupid not thinking of that myself but also realized that subconsciously I (and perhaps the original poster of the original apple question?) was picturing a black hole with the size rather than mass of a black hole. So now I'm stuck with the followup question:

What would it be like to have an apple sized black hole somewhere here on earth, and what happens when you get too close to it?

Using the formula's in the accepted answer to the other question (and some fruit statistics I found online) I figured that the apple size black hole would have a mass of roughly 4.5 earth masses, live extremely long (ca 3.5 times the age of the universe) and be extremely cold (close to absolute zero) so that the horror scenario of that answer would not occur. But does that mean that everything round the apple black hole is nice and cozy, or am I being too optimistic?

• Having something more massive than the Earth on the Earth is not good for us. Jul 1, 2021 at 14:31

Very roughly, the tidal acceleration would be of order $$a_{\rm tidal} \sim 2 \left(\frac{GM_{\rm BH}}{r^3}\right)l\ ,$$ where $$l$$ is the length of your arm and $$r$$ is how far away your elbow is from the black hole.

So say you got within touching distance, with $$l \sim 0.5$$ m and $$r \sim 0.3$$ m, and with $$M_{\rm BH} \sim 2\times 10^{25}$$ kg, then $$a_{\rm tidal} \sim 5 \times 10^{16}$$ m s$$^{-2}$$.

In other words, you would be shredded (spaghettified) by tidal forces before you got anywhere near the black hole.

Furthermore, even if you allow $$r= 6400$$ km and $$l \sim 13800$$ km, one gets a tidal acceleration of 140 m s$$^{-2}$$ (i.e. "14 g"), so the Earth would be tidally disrupted too.

• Wait, how is the tidal acceleration different from the ordinary acceleration caused by the gravitational force? I mean I can see that at my elbow it is different by a factor $2l/r$ but where does this come from? Jul 1, 2021 at 20:28
• Because the tidal acceleration is caused by the gravity gradient - the difference in gravity across an object. That's how things get spaghettified @Vincent I e. $(dg/dr)l$ Jul 1, 2021 at 20:36
• right, so it is the derivative of the accelaration by distance to the black hole? No wait, then it wouldn't have the dimension of acceleration. So what is it then, the difference in acceleration between the two ends of my arm? Jul 1, 2021 at 20:40
• @Vincent it's the difference in gravity across an object $(dg/dr)l$. The bit that's closer to the BH is pulled much harder than the bit that's further away. The same as how the Moon causes tides. Jul 1, 2021 at 20:43
• Thanks! So the extra factor $l$ gets the units back to normal, but apart from that, what is the interpretation of $l$? One thing I do understand now much better thanks to this formula is why the tidal effect of a black hole is much stronger than that for a planet of the same mass because there you couldn't get $r$ so small, so that is progress... Jul 1, 2021 at 20:51

In additon to the tidal disruption described by ProfRob, there would be accretion. Basically, most of Earth will be trying to get into an apple-sized volume. An Earth mass falling under 14g over several thousand km acquires tremendous amount of kinetic energy (hundreds of megajoule per kg or more). That flow and compression will heat things up. A lot.

The result is that the super-heated plasma radiates more and more until it reaches a brightness so the radiation pressure equals the gravity, at which point the mass flow chokes. The flow rate at this point produces roughly 10% of solar luminosity. Given the high temperature the output will be in the hard gamma ray range, converting remaining as yet unaccreted matter into plasma.

The timescale for this is basically the free-fall timescale for an Earth mass, about half an hour. After an hour Earth would mostly be a rotating cloud/disk of iron plasma, irradiated by the core accretion disk.