# Black hole stardrives/artificial gravity [closed]

I'm a sci-fi world builder, and I was recently reading Charles Stross's sci-fi/space opera novel Singularity Sky for inspiration, and I noticed some interesting but scientifically dubious worldbuilding.

In the novel, ships are built around a electromagnetically charged black hole. These black holes provide power (at least 20 gigawatts) and propulsion, when fed a steady stream of particles, an artificial gravitational field that can be manipulated, and (through technobabble) faster-than-light travel. According to the book, the "kernels" are approximately 8 billion tons and electron sized. My questions are:

Would a black hole of this size and mass be able to provide that much power, and if so, what possible method exists to create one?

This is a bit of an open ended question, so I'm looking for the best answer.

Secondary question: how far away would you need to be from the black hole to experience one g? Would it be possible to construct a ship around the black hole at that distance?

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

## closed as off-topic by Morris The Cat, CaM, Ash, Frostfyre, NosajimikiJul 9 at 15:33

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• This may be useful: youtu.be/ulCdoCfw-bY – Renan Jul 3 at 1:02
• You're asking too many questions. The obvious answer to your first question is: yes, yes of course it does! This is what you get when you rely on a scifi novel for actual science. Also, why did you choose the hard-science tag? This is so far beyond our understanding of science that no one can be reasonably expected to answer. – elemtilas Jul 3 at 5:53
• I don’t think you’d be able to use the gravity of a black hole built into your ship to accelerate relative to external objects- it’d be like trying to lift yourself up by your shoelaces. – nick012000 Jul 3 at 8:24
• Please, we want one question per post, no open ended question and a clear metric to evaluate the answer. – L.Dutch Jul 9 at 10:01

All black holes are theoretically predicted to emit Hawking radiation due to quantum effects, and the black hole's luminosity (the total power of the radiation it's emitting) depends only on its mass, the smaller the mass the greater the luminosity. There is a calculator on this page which allows you to enter in one parameter for a black hole like mass, and see the result in terms of other parameters like the black hole's radius and the luminosity of its Hawking radiation. If you select "metric tons" for the mass units on the menu and "MW" for luminosity, then it seems the numbers you quoted are not quite right. 20 gigawatts would be 20000 MW, and if you type that for luminosity and hit enter, you find that the black hole should have a mass of about 1.33E8 = 133 million tons, whereas 8 billion tons would mean a luminosity of only 5.57 MW. The 133 million ton black hole would have a radius of about $$2 * 10^{-16}$$ meters (about 14 times smaller than the classical electron radius of $$2.8 * 10^{-15}$$ meters), while the 8 billion ton black hole would have a radius of about $$1.19 * 10^{-14}$$ (about 4.25 times larger than the classical electron radius).

I just had a look at Singularity Sky and it didn't actually say the 8 billion ton black hole was the sole source of energy--p. 76 suggests its main use is to generate a "jump field" for FTL travel as well as some idea it could be used to generate ordinary momentum via "complex tunneling interactions" (both ideas assume fictional future physics, presumably). Then the author adds "The kernel had a few other uses: it was cheap source of electricity and radioisotopes". So it could be the ship has other power sources. Also, since power is energy per unit time, even if the black hole was the ultimate power source it could be that it's sometimes used to build up energy in some kind of battery or capacitor, which can then release that energy at a faster rate (and thus greater power) than the black hole, for a short time. The scene mentioning a figure of "twenty gigawatts" was on p. 249 when the ship released an intense laser burst, so there's no suggestion the ship generates this much power routinely.

As for constructing it, a kugelblitz is a black hole created by radiation (as from a gamma laser), not a separate phenomenon. By the no-hair theorem in classical general relativity, the only traces of the matter/energy that formed a black hole are mass, charge, and angular momentum, beyond that there should be no traces of what formed it (in quantum gravity there might be subtle information about what formed it that could be found by detailed measurement of all the particles emitted as Hawking radiation, but this wouldn't matter in terms of broad variables like gravity or luminosity).

As for the question about gravity, I've worked it out below, but you can skip to the bolded sentences if you just want the final results. If you are "hovering" at a constant distance r from a black hole of mass m (in terms of Schwarzschild coordinates, whose physical meaning I discussed here), rather than orbiting or falling or otherwise moving in Schwarzschild coordinates, then the proper acceleration you experience, which corresponds to the gravitational force you would measure in your local region using a scale or accelerometer, would be given by the formula on the bottom of this page:

$$a = \frac{m}{r^2}(1 - 2m/r)^{1/2}$$

$$1/r^2= r^{-2}$$ is equivalent to $$r^{-3/2} r^{-1/2}$$, so we can rewrite this as:

$$a = (m r^{-3/2}) (r (1 - 2m/r))^{-1/2} = (m / r^{3/2}) (r - 2m)^{-1/2}$$

As is common in general relativity textbooks, this is expressed in "geometrized units" (see p. 4 here) where the gravitational constant G and the speed of light c have been defined to equal 1 so are not included, but someone calculates here that the above expression in geometrized units is equivalent to the following non-geometrized expression:

$$a = (G m r^{-3/2}) (r - r_s)^{-1/2}$$

Where $$r_s$$ is the Schwarzschild radius for the black hole, given by $$r_s = \frac{2 G m}{c^2}$$

The above equation can be rearranged as:

$$\frac{a}{Gm} = (r^3 (r - r_s))^{-1/2}$$

and if you put both sides to the power of -2 you get:

$$(\frac{G m}{a})^2 = r^3 (r - r_s) = r^4 - r_s r^3$$

Now if you wish, you can set the desired proper acceleration a on the left side to equal one g, or 9.8 m/second^2, and then with the gravitational constant as 6.674 * 10^-11 meters^3 / (kg * second^2), if you have the black hole have a mass of 133 million metric tons = 133 billion kg (what was needed for a luminosity of 20 gigawatts), the left side of this equation will be equal to 0.82 meters^4. And I already found using the calculator that the 133 million ton black hole would have a Schwarzschild radius of $$2 * 10^{-16}$$ meters, so then to find the radius $$r$$ where we have one g acceleration we just solve for r in this equation:

$$0.82 = r^4 - (2 * 10^{-16}) r^3$$

Using the quartic equation solver here and throwing out the negative solution, this gives an answer of $$r = 0.95$$ meters for the radius at which you'll feel 1 g acceleration from the 133 million metric ton black hole. Any closer than that and the proper acceleration will be greater than 1 g, any further and it'll be less.

On the other hand, if you pick a black hole with a mass of 8 billion metric tons = 8 trillion kg, the left side of the equation will be equal to 2968 meters^4, and the Schwarzschild radius in this case was found earlier to be $$1.19 * 10^{-14}$$ meters giving the equation

$$2968 = r^4 - (1.19 * 10^{-14}) r^3$$

And in this case, this gives an answer of $$r = 7.38$$ meters for the radius at which you'll feel a 1 g acceleration from the 8 billion metric ton black hole.

I checked these numbers against what would be predicted by the Newtonian formula where acceleration a at radius r from a point mass m is given by $$a = G m / r^2$$, and ended up getting the same answers for both black holes, given the number of significant digits I used (if I calculated the answer in both general relativity and Newtonian gravity out to more significant digits there'd of course be some differences eventually). And this makes sense since the general relativity formula $$a = (G m r^{-3/2}) (r - r_s)^{-1/2}$$ would reduce to $$a = (G m r^{-3/2}) r^{-1/2} = G m / r^2$$ in the limit as $$r \gg r_s$$. So, when considering radii much larger than the Schwarzschild radius, the Newtonian formula for gravitational acceleration will work find as a close approximation.

This also means that at distances much larger than the radius of the black hole, you can assume the gravitational pull drops off according to an inverse-square law, i.e. if you know the pull at a given distance than at twice the distance the pull will be 1/4 of that, at three times the distance the pull will be 1/9 of that, etc.