In my world, people have powers based on the four fundamental forces.

One of my characters can generate spontaneous gravity to crush his opponents. He's also resistant to his own power (so I ignore his own mass).

I want to create a twist where he breaks his limit and generates a black hole. How many "g" should I put?

  • The size of the black hole may varies (according to the drama I need to create). Let's take the two following scenarios A- Ten meters diameter (melee combat) B- Two hundreds and fifty meters diameters (let's rip battleships combat)
  • The gravitational field he generates his constant (eg. 10 g everywhere). The character's mass doesn't count. The gravity don't crush people on him, but on the ground. It's more like an added g to the environment.
  • 4
    $\begingroup$ How much natural gravity is required to make a black hole? I bet Google knows!! $\endgroup$
    – RonJohn
    Apr 28, 2018 at 16:36
  • $\begingroup$ With relativistic effects, things get really complicated really quickly. I suspect there is no value for "G" that is set in stone, but rather a lambda equation based on the amount of mass and the volume it is being compessed into. The good news is that a black hole that small is likely to dissipate quickly due to Hawking radiation. The bad news is that the math is beyond me, which is why this is a comment not an answer. $\endgroup$
    – pojo-guy
    Apr 28, 2018 at 16:40
  • 1
    $\begingroup$ G is a gravitational constant. You’re thinking of g (acceleration measured in multiples of 9.8m/s/s, not to be confused with g for grams) or possibly f_g which is force due to gravity. $\endgroup$
    – Joe Bloggs
    Apr 28, 2018 at 16:41
  • $\begingroup$ I use G (earth = 1 G) so readers can follow easily the power curve. I've search on Google and didn't found anything except for mass and singularity so it's hard for me to evaluate. If I ignore mass and singularity (since I consider the power optimize it) I need a G value. $\endgroup$ Apr 28, 2018 at 16:45
  • 1
    $\begingroup$ @Demigan The G in that equation is not Earth's gravity acceleration. It's the gravitation constant (6.67408 × 10-11 m3 kg-1 s-2). F in that equation is equal to mg, where m is the mass that is not the Earth's mass. So $g_E = \frac{G*m_E}{r_E^2} $ $\endgroup$
    – Brythan
    Apr 28, 2018 at 17:33

3 Answers 3


The formula for escape velocity is

$$v_e = \sqrt{\frac{2Gm}{r}}$$

$v_e$ is the escape velocity.

$G$ is the gravitational constant, approximately equal to $6.67×10^{−11} m^3\cdot {kg}^{−1}\cdot s^{−2}$.

$m$ is the mass of the object from which we are trying to escape (the black hole).

$r$ is the distance between that object and the object that is trying to escape.

The formula for the magnitude of gravitational acceleration is

$$g = \frac{Gm}{r^2}$$

$g$ is the acceleration due to gravity.

Now, we know $G$ (a constant), $v_e$ (greater than the speed of light), and $r$ (either 10 or 250 meters: given in your question). We do not know $g$, although we know that it is at most fifteen times the Earth's acceleration due to gravity at sea level. We do not know $m$.

So let's rewrite the first equation in terms of $m$.

$$m = \frac{{v_e}^2r}{2G}$$

Now we substitute that into the second equation.

$$g = \frac{{v_e}^2}{2r}$$

Now, substitute in the speed of light and the distance. We'll use 10 meters as the distance.

$$g = 4.5\cdot 10^{15} \frac{m}{s^2}$$

That's roughly $4.5\cdot 10^{14}g_E$, where $g_E$ is the acceleration due to gravity at sea level on the Earth. That's much higher than fifteen.

We can calculate things the other way. A 15g acceleration at the event horizon of a black hole implies a black hole that is less than $3\cdot 10^{-13}$ meters in size. This is tiny. For comparison, a proton is about $10^{-15}$ meters. So this is bigger than a nucleus but a lot smaller than an atom.

I have ignored relativistic effects, but I don't think they would matter much. We start in the same inertial frame. There might be a black hole, but it would be smaller than an atom.

You could improve on this by making the initial acceleration dependent on distance as well. For example, if he could make an acceleration equal to fifteen g at a distance of ten meters from the center of attraction, then it would be a lot larger at one meter and larger still at one millimeter.

Part of the confusion may be that a black hole still has gravitational attraction outside its event horizon. The event horizon is the place where the black hole won't release light. But you'd have to moving close to the speed of light to escape if you came close to the event horizon.

Your fifteen g of acceleration is confusing in that it doesn't mention the distance to the center of attraction. The presumption is that it is being created at the accelerated object. But gravity doesn't work that way. It accelerates towards a center of mass somewhere.

It might make more sense to think of him as always creating a black hole. The stronger he gets, the farther away that he can pull an object. The black hole might get visible at sufficient strength.

For example, consider if he can create a fifteen g acceleration at 250 meters from the point of attraction.

Let 250 meters be $d$ and solve the second equation for $m$.

$$m = \frac{gd^2}{G}$$

Now solve the first equation for $r$.

$$r = \frac{2Gm}{v_e^2}$$


$$r = \frac{2gd^2}{c^2}$$

$c$ is the velocity of light.

This gives us about a $10^-10$ meters black hole. That's similar to the size of an atom, so still far below visibility.

TL;DR: fifteen g is not enough to make a visible black hole, even operating at a distance of 250 meters.

  • 1
    $\begingroup$ @Eligauthier: For the record this is what a better answer looks like! $\endgroup$
    – Joe Bloggs
    Apr 28, 2018 at 19:00
  • $\begingroup$ I intend to put more than fifteen g in my next novel. That's why I asked to know the "breaking point". The power is to "stack" gravity (he augment gravity toward a surface) From what I understand, I should justify it by saying he uses an already attracting center to multiply the it's influence (eg. a planet or centrifugal habitat), becoming a catalyst of that center. Since he would use that gravity from a distance and increase it, it would be impossible to generate fusion or a black hole (because he distorts gravity but doesn't incarnate the mass himself). $\endgroup$ Apr 28, 2018 at 19:23
  • 1
    $\begingroup$ I don't actually know enough of the math to write an answer I'd be confident in, but I would still caution that it's probably going to give some fairly inaccurate results to manipulate classical equations (specifically those for g) in order to try to reach a conclusion about a inherently relativistic phenomenon. For an accurate answer, you'd probably have to consider the metric tensor, and the math would get nasty pretty quickly. $\endgroup$ Apr 29, 2018 at 2:22
  • $\begingroup$ Although I'm not sure offhand, I suspect it's not reasonable to say that relativistic effects don't matter much. A black hole is exactly the type of situation where relativistic effects completely overwhelm their non-relativistic counterparts. $\endgroup$
    – David Z
    Apr 29, 2018 at 5:20

There is no real way to define such a value.

Black holes are so called because at any radius below the Event Horizon the escape velocity (the speed you need to go to not fall back in) is equal to the speed of light.

The problem is that the formula for escape velocity includes M - a mass. To be a black hole you must be massive (as in have some mass).

Your induced ‘black hole’ doesn’t have mass: it’s just a force being created by the guy.

So: what force could be considered to be ‘black holey’ enough?

The first option is choosing a force great enough to overcome neutron degeneracy pressure. This pressure is what stops neutron stars from collapsing to form black holes, so if your hero creates a force great enough that any mass he catches will be pulled in with more force than the forces keeping its neutrons apart then it will very quickly make a teeny tiny black hole. The exact value of this force depends on the properties (temperature, spin, charge, what have you) of the mass being dragged in. The lowest bound mass required to beat neutron degeneracy pressure (given that real physics doesn’t have induced gravitational fields) is the wonderfully named Tolman-Oppenheimer-Volkoff limit (2.17 solar masses), but it’s safest to say your man will have to be throwing down a point of gravity equal to that of 2-3 solar masses. The exact ‘g-force’ felt will vary depending on distance from that point, but it’s pretty intense.

For example: a kilometre away you’d feel ~13000000000000 g...

The second (worse) option comes from the equivalence of mass and energy. Theoretically ramming enough energy in one spot is equivalent to having a mass in that same place (hooray for string theory). Your ‘hero’ could create an energetic enough event that the energy involved just spawns a black hole, but it would require more energy than I can even ballpark and would probably lead to the term ‘naked singularity’ being used more than once.

Really I think you don't really want black holes unless you also want to be tearing up the planet you're standing on: It would be better to work out what kind of force you want to exert over what distance and work backwards from there. Even a gravitational point with mass equal to that of the Earth would crush you to indefinable paste if it were right next to you instead of 6300 km under your feet.

TLDR: don’t define it in terms of g forces, define it in terms of what mass of object your guy is simulating the gravity of. Then it’s about 2-3 solar masses.


Black holes don't have a size. They are singularities: Thoroughly unpleasant mathematical constructs where all mass is concentrated into a single dimensionless point. The event horizon (where you can't see past) can have a radius.

If we use Event Horizon as the measure of size then since this gravity-bender is making gravitational fields without any mass he will be making a sphere in open space that light cannot escape. Everything he does is creating ludicrously small black holes.

Making a black hole big enough for the event horizon to be visible requires a large enough 'mass' that it will destroy absolutely everything nearby, leaving nobody around to see it.

Needless to say this is Not A Good Idea.

EDIT FOR 'added g':

If all your hero is doing is increasing gravity locally then he is doing one of three things:

1: Telekinetically pushing people towards the ground (not really gravity based)

2: 'Multiplying' gravity worldwide

3: 'Multiplying' gravity locally

I'll ignore 1. It's boring.

2 would have consequences for everyone on the globe all the time, and I'm going to assume your hero isn't going to apply bone crushing force to innocents, so let's discount that.

3 has some very interesting consequences before you even get to black hole status (pushing the Earth out of orbit, for one), but lets ignore those and try focus on the question:

The question doesn't make sense any more. Black holes are points. If this power is a gravity multiplier (which is functionally the same as adding gravity) then the only 'point' a black hole could start to form is the centre of the Earth. Essentially the ground would have to collapse in a column centred on wherever this man was exercising his power, with more and more earth falling into the hole as it progressed. For a black hole to happen our 'Hero' would have to crush the entire mass of the Earth into a ball no larger than it's Schwarzschild Radius, which stands at a stupendously tiny 1.4 x 10-27m, in the process destroying everything he ever loved.

Oddly though, the moon, satellites and folks in the ISS would carry on orbiting as normal, so that's something.

  • $\begingroup$ The naked singularity would be a great explanation for aesthetically reasons (the battle scenes could be seen). But generating 2-3 solar mass... I can't think how I'm supposed to justify that. (The hero can generates about 10-15 g at the end of the novel, which is terribly strong) $\endgroup$ Apr 28, 2018 at 17:31
  • $\begingroup$ @EliGauthier: 10-15g where, though? 10-15 g is just another way of saying '98-147N of force', or 'reasonably telekinetic'. If he's making gravitational fields then closer to the source of the gravitational field that force will be stronger, further away it will be weaker and everything in-between will feel the force too. $\endgroup$
    – Joe Bloggs
    Apr 28, 2018 at 17:39
  • 1
    $\begingroup$ @EliGauthier And don't forget that gravity acts in all directions and doesn't stop for anything. If he's exerting 10g of force on someone 20m away (trying to pull them towards him) he's also exerting 10g of force on the person standing right next to him (pulling them away from him). Gravity is a fickle mistress, unless you just treat it like telekinesis and lose all the intricacy. $\endgroup$
    – Joe Bloggs
    Apr 28, 2018 at 17:42
  • $\begingroup$ Ok, I forgot to explain something. The gravitational field he generates his constant (eg. 10 g everywhere). The mass of the character doesn't count. The gravity don't crush people on him, but on the ground. It's more like an added g to the environment. $\endgroup$ Apr 28, 2018 at 17:44
  • $\begingroup$ @EliGauthier: So always pointed towards the centre of the Earth? If that's the case then the concept of black holes just doesn't make sense... $\endgroup$
    – Joe Bloggs
    Apr 28, 2018 at 17:53

A micro black whole is theoretically possible it would look like a distortion of light and the center would be the tip of a pin. It would do mass devastation if present for a fraction of a millisecond leaving a perfect circle in the Earth. Like a inward sucking bomb. The sudden appearance and disappearance would splatter anything organic near it. Then there is the electromagnetic field with a black hole.


  • $\begingroup$ I like the image :) But would it also need a 2-3 solar mass? (I've reed your link and it's seems so). In that case it's difficult to find how the hero could do such a feat. $\endgroup$ Apr 28, 2018 at 17:49
  • 1
    $\begingroup$ The Planck mass is the smallest possible unit of mass for a real object in our universe. A 10 meter radius black hole would be disruptive to the solar system, and Earth would have a really Bad Day as it shattered and became an accretion disk. Oh, and a lot of people would be very upset. $\endgroup$
    – pojo-guy
    Apr 28, 2018 at 18:24
  • 1
    $\begingroup$ @pojo-guy The Planck Mass is /not/ the smallest possible mass for a real object in our universe. As you mention in your previous comment, it's about 22 micrograms. There are objects visible to the naked eye (such as a human egg cell, ~4 micrograms) with less mass than that. Planck Units are about derivation, and there's usually no boundary on the universe set by their value. $\endgroup$
    – notovny
    Apr 29, 2018 at 3:01
  • 1
    $\begingroup$ @Muze Theoretically? Yeah, much smaller. An event horizon 10 meters in radius is about three Jupiters, and you /really/ don't want something like that near the Earth, as Pojo-Guy mentions. One the mass of the Earth gets an event horizon of about 0.8 cm radius. Black holes with human-scale masses have other issues, though; they wind up rapidly evaporating from Hawking Radiation, and turn their mass into energy in very short and unpleasant disorder. $\endgroup$
    – notovny
    Apr 29, 2018 at 3:38
  • 1
    $\begingroup$ Hey Muze, just wanted to let you know that we have a Sandbox on Meta where you can post question drafts first before posting on the main site. Some regulars try to help people there who sometimes struggle with their posts on main or who simply feel that their draft still needs some work. Just to let you know that this exists and that it might help you. $\endgroup$
    – Secespitus
    May 1, 2018 at 17:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .