# How gravity works beyond the Event Horizon

1. Jun 9, 2015

### Gaz1982

Forgive my ignorance here, I'm not a physicist, but the exact nature of gravity beyond the Event Horizon has me a bit stumped.

Say I cross the Horizon of a Supermassive Black Hole, meaning the pull on me isn't really that much stronger than Earth's pull initially - why can't I then escape?

What exactly changes the escape velocity when I'm still only subjected to Earth-level gravitational pull?

2. Jun 9, 2015

### phinds

That is incorrect. The TIDAL FORCES on you are very small as you cross the EH of a supermassive BH, but the gravity is so strong that not even light can escape to back outside the EH.

3. Jun 9, 2015

### Gaz1982

That's my question, as I should have worded it.

What's the difference between tidal force and gravity in this situation?

4. Jun 9, 2015

### phinds

The tidal force is the difference in gravitational attraction between two points not far from each other ... say from your head to your feet. As you enter the EH of a supermassive BH, the tidal force on you is trivial. Long before you enter the EH of a very small BH, the tidal forces will have turned you into a long thin strand of bloody mess. This is called sphagetification.

5. Jun 9, 2015

### Gaz1982

Sure.

But tidal force weak but gravity inescapable. I'm struggling to compute

6. Jun 9, 2015

### phinds

Yes, that's because you are a member of a species that evolved in a place (the surface of the Earth) where tidal forces are trivial. There is no survival value in having any "intuition" about what happens near a small BH.

Just do the math.

EDIT: when I say "do the math", I mean

- look up the mass of a small BH
- figure out the "radius" to the EH
- calculate the force of gravity at the EH
- calculate the force of gravity 6' outside the EH
- compare the two

7. Jun 9, 2015

### Gaz1982

Right, sorry to sound thick. Bear with me.

To me the tidal forces on earth aren't trivial. We're being pulled towards the centre of the earth quite strongly. A 15 feet fall can kill someone.

The escape velocity of Earth is about 25,000 miles an hour

As a non-physicist I'm struggling to reconcile tidal force with gravity.

8. Jun 9, 2015

### Gaz1982

Yes I know the moon exercises a tidal force on the Earth for example, but doesn't drag people upwards. I'm struggling to define tidal force.

I'll check Wiki

Last edited by a moderator: Jun 10, 2015
9. Jun 9, 2015

### Quds Akbar

That is not true, why would a supermassive black hole subject you to the same amount of gravitational pull as Earth? A black hole definitely has stronger gravitational pull than Earth's gravity, that is why it is a black hole, because light cannot escape at a certain point, the gravity is so strong that you reach a point where the escape velocity and the speed required to escape the pull is far greater than the speed of light, which violates relativity.
So I think what you missed here is that a black hole has much, much stronger gravitational force than our Earth that is nothing compared to a supermassive black hole's gravity.

10. Jun 9, 2015

### Noctisdark

The force of gravity changes when varying the distance, closer objects will feel more attracted than further objects, In classical mechanics it varies with the inverse square of distance, [The math of general relativity is different, some math I totally don't understand], anyway when you calculate the rate of change of that force when you change the distance, you'll find that it changes very quicky when you are very close to the center of mass, and the same "should" happen near a black hole, when you are close the force on your feet is very different and larger than the force on your head, these are tides waves that tear you apart, so keep away from a black hole and just one thing, crossing an event horison is the END, the pull is way stronger that the pull of earth, it so strong that it made me say this sentence, even light cannot escape from it, and don't worry, no one will shout on you, good luck

Last edited: Jun 9, 2015
11. Jun 9, 2015

### phinds

Which has absolutely NOTHING to do with tidal forces. That's just gravity.

I'll say it again. DO THE MATH.

EDIT: and if you can't be bothered to do the math at least look up sphagetification where it will all be explained.

Last edited: Jun 9, 2015
12. Jun 9, 2015

### cosmik debris

I think you are confusing the pull of gravity with the tidal force which is the difference in gravity between two points. If you imagine Gravity to be a field with values at each point then each point can have a high value which will cause an acceleration, but it is only the difference between two points that you will "feel". If this difference is large and your head is at one point and your feet another you will feel unwell.

13. Jun 9, 2015

### phinds

Yes, that's what I told him in post #4. It doesn't seem to be clear to him yet. It SEEMS like a trivially clear distinction, but I can recall not "getting" stuff that later became crystal clear, so I sympathize.

14. Jun 9, 2015

### DaveC426913

That is not tidal force; that is gravitational force.
Tidal force on Earth is the difference between the force at your head and the very slightly stronger force at your feet (which are closer to Earth's centre of mass). If you were in free-fall, you would find your feet being pulled toward Earth just a little stronger than your head. You'd get taller!

Satellites in orbit around Earth are sometimes large enough to feel this very gentle tug along their lengths. Unless corrected, they will have a tendency to orient themselves with their long axis pointed toward the Earth, as their near end is pulled down from the centre of mass and their far end is pulled up from the centre of mass. (The sat's centre of mass is in a stable orbit, but the near end of the satellite is actually going too low for its slightly lower orbit, so it tends to drop. The far end of the satellite is actually going too fast for its slightly higher orbit and this pulls it upward).

This is also why the Moon is tidally locked. It is not a perfect sphere, and a tiny tidal force over billions of years has oriented the Moon with its larger diameter pointed along the Earth-Moon axis.

Last edited: Jun 9, 2015
15. Jun 9, 2015

### phinds

Well, we've explained it pretty much exactly the same way at least 3 times now. @Gaz1982 did you get it?

Last edited: Jun 9, 2015
16. Jun 9, 2015

### Staff: Mentor

Because spacetime is curved at the horizon to the point where radially outgoing light no longer moves outward. In other words, outside the horizon, an object that "stays in the same place" is moving slower than light. But at the horizon, an object that "stays in the same place" must be moving at the speed of light, so only light can do it. And inside the horizon, nothing, not even light, can "stay in the same place"; everything falls inward, including radially "outgoing" light.

Escape velocity isn't the same as "gravitational pull". This is true even in the Newtonian regime, i.e., for objects much less compact than black holes. Just look at the Newtonian formulas:

Escape velocity: $v_e = \sqrt{2GM/r}$

Acceleration due to gravity: $a = GM / r^2$.

So it's perfectly possible for two different planets, say, to have the same $v_e$ but different $a$ at their surface, or the same $a$ but different $v_e$.

17. Jun 10, 2015

### Gaz1982

Yes,

No need to be rude

18. Jun 10, 2015

### PWiz

@OP I would just like to add that gravity is not a force (in GR). You would feel no force at all in free fall, because your body would be following a geodesic path (objects always follow straight path in spacetime, and when spacetime becomes curved, objects continue to follow straight paths along the curved spacetime). When you are near a black hole, what really tears you up is not this gravitational "pull," (curvature) but rather the tidal effects - you're head is trying to follow a slightly different geodesic compared to your feet (if you're falling feet/head first), and these differences become really noticeable the closer you get to a black hole where curvature (measured by the Riemann tensor) varies much more over shorter spatial distances.

The escape velocity (which becomes $c$ at the EH) at a point is the minimum velocity at which an object must be moving to move "against" the geodesic path at that point. These are different concepts.

19. Jun 10, 2015

### Gaz1982

That makes sense

Thank you

20. Jun 10, 2015

### phinds

I was kidding US more than you (and I WAS kidding, not being rude). We have a knack for saying the same thing over and over with each person thinking that his particular way of saying it is better than the previous n versions.