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I've been reading and couldn't find any answer to this question. What is the minimum required gravity to prevent light from escaping?

Qmechanic
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Praearcturus
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  • For black holes, it is better to think of light not being able to escape due to the curvature of spacetime rather than some force. – jng224 May 16 '21 at 19:40

3 Answers3

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I’ll answer using Newtonian gravity, since you may not have studied General Relativity.

Nothing can escape when the escape velocity

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

exceeds the speed of light. In terms of the gravitational potential

$$\varphi=-\frac{GM}{r},$$

the condition $v_e=c$ for the event horizon becomes

$$\varphi_\text{horizon}=-\frac12c^2$$

as the condition on the potential.

There is no condition on the gravitational field/force/acceleration because these depend on a different combination of $M$ and $r$, namely $GM/r^2$. For a supermassive black hole the field can be arbitrarily weak at the horizon!

To see this, notice that $r\propto M$ at the horizon. This means that the field is proportional to $1/M$ at the horizon, and this becomes arbitrarily small as the mass becomes arbitrarily large.

To get an actual example number for the gravity at the event horizon, we need to get more specific than just looking at proportionality. The horizon is at

$$r_\text{horizon}=\frac{2GM}{c^2}$$

and the Newtonian gravitational acceleration there is

$$g_\text{horizon}=\frac{GM}{r_\text{horizon}^2}=\frac{c^4}{4GM}.$$

For a one-solar-mass black hole, this is about 1.6 trillion g’s (i.e. 1.6 trillion times the gravitational acceleration we experience at the surface of the Earth). This is of course huge, but the acceleration at the horizon of a black hole still goes to zero as $M$ goes to infinity.

G. Smith
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  • What is the minimum gravitational force for this to occur? – Praearcturus May 16 '21 at 19:46
  • @G. Smith, Hah! you beat me to the punch by approximately 3 milliseconds. -NN – niels nielsen May 16 '21 at 19:49
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    And of course a monster SMBH of around 1.6 trillion solar masses has EH gravity of 1 g. Using https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator the mass is ~1.55 trillion $M_\odot$, and the Schwarzschild radius is ~0.484 lightyears. But I guess such a monster would also have a lot of spin, so it's not well-modeled by Schwarzschild calculations. – PM 2Ring May 16 '21 at 21:04
  • You didn't specify the mass of the black hole in question - but now you have the equations in an answer. The limits would be approaching infinity for the smallest black-hole (Plank-length in diameter) and approaching zero for the biggest (the total mass of the universe.) You decide. @Praearcturus – Jiminy Cricket. May 17 '21 at 02:06
  • According to this, the EH gravity of the M87 black hole (2.4 billion M) would be around 666g, right? – Snowcrash Dec 20 '21 at 17:10
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    Not sure I like this idea of weak field near horizon, since the rope or other structure supporting an observer just above the horizon has a huge tension and will indeed break as the observer approaches the horizon thus supported. Or have I misremembered? – Andrew Steane Feb 01 '23 at 09:13
  • This answer is wrong and misleading. Using Newtonian gravitational acceleration in this case is incorrect. The correct answer is here, https://physics.stackexchange.com/a/647494/155073 . The acceleration is infinite at all event horizons. – zylstra Oct 07 '23 at 06:38
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I like G Smith's answer. Anyone upvoting mine should upvote theirs. This is a less mathy answer for anyone whose brain shuts down at the sight of an equation.

The short answer is there cannot be a minimum required gravity at the surface of a black hole. The reason light doesn't escape is not because the gravity is too strong at the surface, it's because it would take too much energy to escape from the sphere of influence of the black hole.

Consider a small black hole. Let's say one only slightly larger than Sol (that's the name of the Sun, for the uninitiated). From far away, it has around the same gravitational influence as Sol. With it's sphere of influence extending only slightly farther, it would need to have a massive amount of gravity at its surface in order to make the energy you need to escape prohibitively high.

Now consider a super-massive black hole. Something with billions of solar masses. It's sphere of influence is going to be enormous. It can even be galactic in scale. Having to travel so much farther while being decelerated by its gravity all the while, you can imagine that it doesn't need nearly as much gravity at the surface to make the energy required to escape prohibitively high.

This is the general point G Smith makes. Very massive black holes have tiny amounts of gravity at their surface. For a super-massive black hole, you can have the acceleration due to gravity at the event horizon be the same as that on Earth's surface. But it's influence is so much greater than Earth's that it would take an infinite amount of energy to escape. Light sees that and goes "How much energy!? Nuts to that! I'm turning around and taking a nap". As the mass of the black hole increases, the gravity at the surface shrinks even more. The reverse is also true. Low-mass black holes have insane gravity at the event horizon. Its these ones that are so difficult to approach because the gravity can crush you or rip you apart. Super-massive black holes are easy to get close to. Like a walk in the park. A walk where eventually a monster jumps out and eats you. Kind of like Ski-Free (90s video game reference).

Jim
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  • By "surface" you mean event horizon? – zylstra Oct 06 '23 at 21:29
  • @zylstra Correct. The surface of the event horizon. Not a true physical surface, but mathematically a surface – Jim Oct 10 '23 at 12:24
  • In that case, you should not like G. Smith's answer, since it's incorrect. ; ) I have placed a link to a correct answer in my comment to his answer. The correct answer is that acceleration at all event horizons is infinite. – zylstra Oct 11 '23 at 20:44
  • @zylstra That's fair, but I doubt you'll find someone who literally would claim you could use a rocket to just hover a given distance. If you could, then a little more power from the rocket would have you leave the event horizon. We mean some arbitrary surface close to but not exactly at the event horizon. I now see the purpose of your earlier question and that I misunderstood. The intention is to get across that spaghettification isn't strong in supermassive black holes. Acceleration goes to infinity though because in addition to peculiar acceleration, space itself is accelerating inwards. – Jim Oct 16 '23 at 18:04
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Suppose you are an observer at a fixed position near the horizon, above it. Such an observer is sometimes called a "shell observer". You will feel a certain gravity.

The gravity will increase as you try to stay closer to the horizon -- while still staying at a fixed position. It will become infinite if you try to stay at the horizon.

But of course you can fall through the horizon, in which case you will feel no gravity, just like someone in free fall on Earth or in any field locally flat.

Andy
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