What is the gravity at the event horizon of a black hole?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

The answer in terms of escape velocity $v_e=\sqrt{\frac{2GM}{r}}$ is not correct and this is easy to demonstrate:

If we consider launching a projectile from the surface of the Earth, then if its initial velocity is equal to the escape velocity of the surface of the Earth, then it can head off to infinity and even if it has slightly less velocity than the escape velocity of the Earth the projectile can go a long way from Earth before reaching an apogee and falling back.

Particles at the event horizon of a black hole can not leave the event horizon, not even a little bit, unlike the projectiles on Earth and even photons that technically have the required escape velocity (c) that implies they can go all the way to infinity, can not leave the event horizon even by one millimeter.

The proper acceleration experienced by an observer hovering just outside a black hole is given by:

$\frac{d^2 r'}{d\tau ^2} = -\frac{m}{r^2\sqrt{1-2m/r}}$,

Clearly when r = 2m the proper acceleration is infinite. Ref Mathpages.

Inertial observers that are free falling from outside to inside across the event horizon do not measure any proper force at all. However a tall observer falling feet first, will experience a greater acceleration pulling their feet, than on their head and they will stretched out in an process called spaghettification, which is as unpleasant as it sounds.

For a really massive black hole with a really large event horizon radius, this tidal effect is much less noticeable and a free falling observer might not even notice they are in the vicinity of an event horizon by measuring proper force or tidal force.

If you are not familiar with General Relativity, then it worth knowing GR does consider the motion of free falling objects to be due to the "force of gravity" and the motion is explained by objects following a geodesic in curved spacetime.

If we lower a ladder as close as possible to an event horizon with rungs spaced out such that they represent radial distance from the the black hole as measured by the Schwarzschild observer a infinity, then a free falling observer can measure his acceleration using the rungs and his proper time and the acceleration he would measure would be:

$\frac{d^2 r}{dt^2} = -\frac{m}{r^2}$, Ref: Eq (6) Math Pages.

where r is the Schwarzschild radius from the centre of the black hole and m is the mass of the black hole. Note that as he passes the event horizon (r = 2m), he measures his acceleration to be:

$\frac{d^2 r}{dt^2} = -\frac{m}{4m^2} = -\frac{1}{4m}$,

which is finite and is smaller for black holes with greater mass.

In summary, there is no finite minimum force of gravity required to prevent a photon leaving the event horizon. In the limit, the proper force acting on a particle stationary in the gravitational field approaches infinite as the distance from the event horizon approaches zero and this apples to any Schwarzschild black hole of any size.