# The bigger the mass of a B.H. the smaller the acceleration at the E.H. Then why can't a mass escape from the E.H?

In this article (see the for this possibility at the end of the article) you can play around with various quantities related to a black hole. If you make the mass of the hole very large, the acceleration at the event horizon becomes very small compared to that of the earth. So why then is it so difficult to accelerate an object very near to the event horizon and make it move away from the hole? Is it because of the difference between the coordinate time $dt$ as measured by a far away observer, which makes the acceleration have such a small value, and the proper time $d\tau$ for the mass very near to the event horizon (which makes the time almost stop)?

• I naively would have assumed it was the curvature, once you are inside, all paths lead back down to the scary bit...... – user146020 Mar 21 '17 at 14:27
• Your linked article is in Dutch. It would be advisable on a primarily English language website to link to a translation in English. – StephenG Mar 21 '17 at 14:38
• Not exactly an answer, but a way of pointing out a core flaw in the reasoning here: even in Newtonian mechanics it is possible to find two bodies in which the surface gravity of one is larger than the other $g^\text{A} \gt g^\text{B}$, but the escape velocity of the other is larger than the one $v_e^\text{A} \lt v_e^\text{B}$. Surface gravity simply isn't the right metric for answering "How hard is it to get away from here?". – dmckee Mar 21 '17 at 17:00
• The only thing that bringing relativity into the frame does it make the nature of space near the event horizon asymptotic so that you don't have to look 'far' (an observer dependent notion, right?) from the horizon to know the answer to the escape question. – dmckee Mar 21 '17 at 17:03
• Consider to spell out acronyms. – Qmechanic Mar 23 '17 at 19:38

If you were to hover at a distance $r$ from a black hole of mass $M$ then the acceleration you would feel is given by:

$$a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}}$$

This is derived in twistor59's answer to What is the weight equation through general relativity?

At $r = r_s$ the acceleration goes to infinity regardless of the mass of the black hole, so for all black holes the "g forces" you feel at the horizon are infinite.

The article you link is talking about Hawking radiation, and Hawking radiation is proportional to a property called the surface gravity, $\kappa$, at the horizon. The surface gravity is the acceleration calculated above multiplied by the time dilation factor. This is discussed in my answer to Why do larger black holes emit less Hawking Radiation than smaller black holes? The surface gravity is given by:

$$\kappa = \frac{1}{2r_s}$$

so it does decrease as the mass of the black hole increases. The surface gravity is, in a rather handwaving way, the local acceleration at the horizon measured by an observer at infinity. It is relevant here because the Hawking radiation is measured at infinity as well.

So you are basically correct that it is the time dilation that causes the surface gravity to fall as the black hole size increases. However this is not the local acceleration experienced by an observer trying to escape the black hole so it does not mean objects can escape from a large black hole.

Only a small amount of momentum is exchanged between the very large black hole and the thing hanging near that black hole in one second of high altitude clock's time.

But as the thing hanging near the event horizon is very time dilated, in one second of the things proper time there are many seconds of high altitude clock's time. So if we ask the thing, there is a large amount of momentum exchanged in one second of its time. Another way to say that is that there is a large force.