If an object is in orbit around a star, the object has gravitational potential energy that could possibly be extracted. For example, when we perform gravitational slingshots around Jupiter, our spacecraft speeds up and Jupiter drops to a slightly lower orbit -- we convert some of its gravitational potential energy into kinetic energy which we use for our purposes. There is only so much energy that can be extracted before Jupiter would drop so low that it would get too close to the Sun for this to work (e.g. hitting the Sun's surface).

Imagine if the planet were orbiting a black hole instead of a star. In this case the process could be continued for longer. How much energy could be recovered? Can this be expressed as some fixed fraction of the lowered object's rest-mass? Is this the same as what you would get by slowly lowering the object directly into the black hole on an idealised rope connected to a turbine?


lemon gave an excellent explanation of how the gravitational potential energy difference from lowering a mass, $m$, from far away down to the event horizon is equal to half its mass-energy. This would seem to provide an upper bound for a more realistic answer. Here are a couple of additional things I'd like to see taken into account:

(1) The freed energy would get redshifted as it was brought back away from near the black hole. How does this change things?

(2) Below 1.5 times the Schwarzchild radius, there are no circular orbits (circular travel requires outward thrust). This presumably causes a lot of trouble for my method beyond that stage since I had a mass in orbit that I was slingshotting things around.

(3) Below 1.5 times the Schwarzchild radius, ballistic objects cannot escape the black hole unless they are angled away from it (e.g. light starting travelling perpendicular to the centre of the black hole just spirals in). Presumably this causes trouble for my approach since as far as I understand the slingshotting mainly involves ballistic motion roughly perpendicular to the centre of the black hole.

(4) Lower orbits are also faster, so it would seem that some gravitational potential energy goes into speeding up the lowered object, increasing its kinetic energy. This could be a very large amount if it is lowered close to the event horizon, as the orbital speed there is $c$. Thus assuming you can get all the gravitational potential energy out could be a serious over-estimate. Does anyone know how to adjust for that?

This is not a duplicate of How much energy does lowering an object into a black hole generate? as that question is not concerned at all with lowering via orbits, so for instance, my questions (1), (2), (3), and (4) don't apply there. In addition, the brief answer given there appears to be incorrect, as it contradicts the best answer here so far.

  • $\begingroup$ A simple attempt to deal with (2) and (3) is to just stop the process at 1.5 times the Schwarzchild radius. Using the formula lemon gave, this means you would get a third the mass energy out instead of a half. If there is some partial way to keep doing the process below this point, you'd get something in between a third and a half. $\endgroup$
    – Toby Ord
    Commented May 10, 2016 at 13:12
  • $\begingroup$ It certainly sounds from the titles like they are duplicates, but I am concerned mainly with a particular method which that post doesn't cover, and that question was mainly concerned with a speculative interaction with Hawking radiation. In addition, it didn't get a great answer, as its answer appears to contradict lemon's one here (by a factor of 2) and to be incorrect. $\endgroup$
    – Toby Ord
    Commented May 10, 2016 at 18:18
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    $\begingroup$ See also: physics.stackexchange.com/q/175602, physics.stackexchange.com/q/20813, and especially physics.stackexchange.com/q/251820 which this is also largely a duplicate of. $\endgroup$ Commented May 10, 2016 at 18:37
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    $\begingroup$ Hi Toby Ord. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented May 11, 2016 at 8:32

1 Answer 1


Note that this is an incomplete answer.

Imagine an object of mass $m$ at a distance $r$ from the centre of a black hole of mass $M$. The gravitational potential energy is

$$ U(r)=-G\frac{Mm}{r} $$

This has its highest value when $r=\infty$ and its lowest value when $r$ is at the event horizon of the black hole, i.e. the Schwarzschild radius

$$ R = \frac{2MG}{c^2} $$

So the maximum amount of energy extractable is

$$ U(\infty)-U(R) = \frac{mc^2}{2} $$

which is an enormous amount of energy even for a small $m$. Note that this answer is independent of the mass/size of the black hole or even the strength of gravity itself.

If you lowered ~1 car into a black hole and extracted all of the energy then that energy would be enough to fuel the entire planet for a year.

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    $\begingroup$ The sun converts 1.5 million tons of mass-energy per second and the sun is a rather mediocre star. So much for the meaning of the word "enormous". $\endgroup$
    – CuriousOne
    Commented May 10, 2016 at 11:13
  • $\begingroup$ @CuriousOne which is just as well, since the realization that gravitational contraction was insufficient to fuel the luminosity of the Sun was more or less how fusion was discovered. $\endgroup$
    – Kyle Oman
    Commented May 10, 2016 at 11:20
  • $\begingroup$ @KyleOman: The amount of potential energy that a regular star can convert is limited by its large size. The escape velocity of the sun seems to be around 618km/s, which is a far cry from $c$, of course. A much larger star, on the other hand, can become a core collapse supernova, with a much larger conversion efficiency. $\endgroup$
    – CuriousOne
    Commented May 10, 2016 at 11:26
  • $\begingroup$ Which is why an advanced civilization will preferentially build (very small) Dyson spheres around Black Holes and not regular stars $\endgroup$
    – user56903
    Commented May 10, 2016 at 12:14
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    $\begingroup$ I think you meant "it has its lowest value when r=∞" since the term is divided by r. Also, if it has its highest (negative) value at r=R, you need to plug in the extra factor √(1-R/r), otherwise it has its highest value (-∞) at r=0. If you meant that 0>-∞ you are right, but the formulation can be confusing. $\endgroup$
    – Yukterez
    Commented May 10, 2016 at 20:01

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