Context: I'm trying to design a massive colony ship (as in, at least 3 billion tonnes unfueled), built by a hyper-advanced K2 civilization, that will be accelerated via laser pushers and will use a black hole drive to decelerate. Since black holes have a similar matter-to-energy rate as antimatter, I figured that I could figure out how much speed a ship of a given size could eventually burn off by treating the black hole like an equal mass of antimatter. (I used the relativistic antimatter rocket equation I found on wikipedia for this). However, the quirk of black hole evaporation means that simply scaling up a ship with the same mass-to-black-hole ratio will make it decelerate much slower.

I came up with a workaround to this problem, which is to make a bigger ship use many small black holes rather than a big one. I won't get into the weeds of how the ship would harness multiple black holes at once (maybe in another question) but basically, if I'm not wrong, I could make a huge ship exert as much or as little thrust as I want by changing the number of black holes as long as they add up to the required mass.

So, the full question. Let's assume this black hole drive has an Isp of, say, 0.69c (which is what the wikipedia article suggests as an example value - I don't know how accurate it is). Let's also assume the drive itself is made of handwavium and will convert the black hole's energy to thrust at the same efficiency no matter how much power the black hole is putting out, and will never melt or be destroyed. Given this, what is the smallest mass of ship for which the g-forces of converting the black hole's final explosion into thrust would not kill a human crew?

Anything that can dampen g-force, such as pusher plate, can be used as well. So I suppose it would be useful to imagine the ship as a huge Orion drive, big enough to use a black hole bomb.

I need to know this because the "usable black holes per kilogram of ship" effectively puts a limit on how quickly any ship could decelerate. Now, obviously a way to deal with this would be to jettison black holes before they explode, but that would complicate the math and I'm not sure if it would actually give more performance for ships that are big enough to withstand the explosion force.


1 Answer 1


That one line of

converting the black hole's final explosion into thrust is very problematic.

The energy output of the last fractional moment before the black hole's final evaporation is the VERY closest approximation of true infinity the physical universe will ever present you.

Example: a quick list of the mass, and matching energy output of smallish black holes. The pattern should be obvious.

100000 tonnes__3.56194E16
10 tonnes______3.56194E24 (100 000 000 times greater)
10 kilograms___3.56194E30
1 gram__________3.56194E38
1 microgram____3.56194E50

You see the trend.. at almost-zero mass, the black hole's evaporation rate is almost-infinity.
That last figure is higher wattage than the effect of a million quasars ramming into each other. Sure it does not last very long at all, but for that one infinitestimal fraction of second, the power output is a very good approximation of infinity.

There is just no ways you will ever build a mechanism that usefully utilizes the last few seconds of a dying black hole's evaporation. The steady-state energy release for masses of as small as 10 000 tonnes, yes, although even that is very scary. But the final gasp just blows off the scales and tries to invent its own branch of mathematics just to be measured.

Don't let your back hole(s) die. Keep them fed, and they can be good energetic pets.

  • $\begingroup$ The thing is, while the power output approaches infinity, the time interval also approaches infinitely small. A black hole of 10 tonnes only has a lifetime of 0.0000841072 seconds; a black hole of a gram would only have a lifetime of 8.41072E-26 seconds. So I figured it's sort of a zeno's paradox situation and the extreme brevity of the output would mean that it actually wouldn't be so much in the end. $\endgroup$
    – kt-void
    May 28, 2021 at 18:13
  • 1
    $\begingroup$ Near-zero mass isn't the problem. The power output of a 1ug BH might be 3e50W, but the total energy contained in said black hole is only about 90MJ - similar to the chemical energy in 2 cubic meters of natural gas. That's not much, and could easily be absorbed no matter how quickly it's dumped. Harder to say that about the 10T black hole - it would be releasing more energy than the entire global nuclear arsenal in well under a second. $\endgroup$
    – Gene
    May 28, 2021 at 21:05
  • $\begingroup$ Yeah, and in the last 1 second it releases 2e+22 Joule of energy. Good luck containing the energy of 100 000 tsar bombas, released in one second, in a volume of space significantly smaller than one proton. $\endgroup$
    – PcMan
    May 28, 2021 at 21:38
  • $\begingroup$ I suppose in that light, the original question can be rephrased as "How big does an orion drive have to be to use a 4780 gigatonne nuke" $\endgroup$
    – kt-void
    May 28, 2021 at 21:48
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    $\begingroup$ @kt-void the other problem is that the energy is released in relatively few photons of truly staggering energy levels. Gamma of 10^22 electronvolt and even higher. So energetic that any matter they hit gets shattered. And those pieced shred others in secondary impacts. And tertiary, and.... what is the word for 216-generational radioactive cascading? Anyways, energy levels of many quadrillions of times stronger than the strongest normal cosmic ray. $\endgroup$
    – PcMan
    May 28, 2021 at 21:58

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