Can the force of gravity from a Black Hole be escaped?

[rusher38]

I'm not exactly sure if anyone can answer this because no one (that we know of) has encountered one close enough to be "sucked" in (if they actually do that)

Also, how do they "spaghetti" things?

Also, I have heard they produce massive forces of gravity, though only one of 3 types of black holes can "suck". Correct me if I am wrong.

 

[Char. Limit]

I'm not exactly sure if anyone can answer this because no one (that we know of) has encountered one close enough to be "sucked" in (if they actually do that)

Also, how do they "spaghetti" things?

Also, I have heard they produce massive forces of gravity, though only one of 3 types of black holes can "suck". Correct me if I am wrong.

A black hole works in exactly the same way that Earth works... just much, much stronger.

And yes, I've gotten close enough to be sucked in... I'm standing on the surface of a black hole right now, typing this message to you. Luckily, the modem is outside of the Schwarzschild radius. :lol:

Firstly, how does a black hole work?

Well, to help the answer (I'm going to simply use classical mechanics here, as quantum mechanics is far beyond me and quite incomprehensible), I'll present the gravitational acceleration formula!

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

For Earth, the equation becomes this:

$$9.8 \frac{m}{s^2}=\frac{Gm_e}{r^2}$$

For a black hole, g is such that light cannot escape, and is actually quite a large number.

So, a black hole works exactly like a planet, except that the value of g is much stronger. Thus, everything, including the mass of the black hole itself, gets sucked in, and the gravitational force sucking even the black hole itself reduces the mass of the hole to a singularity with zero volume, and thus, infinite density.

Now, spaghettification is when the difference between r^2 at, say, your feet, and r^2 at your head, is different enough that the g-force from the black hole at those two points is different. One part gets pulled faster than the other, and you get stretched out.

 

[rusher38]

So could a black hole be formed by a planet's gravity becoming so great it just collapsed on itself?
And could a black hole actually be a literal hole to the 'end' of space?

 

[diazona]

@Char. Limit: that's the general idea, but you can't really talk about gravity and light quite that way. Light always travels at the same speed in a vacuum, it doesn't accelerate or decelerate... so it doesn't make much sense to say that the acceleration is so large that light can't escape.

The better argument has to do with escape velocity, which is the minimum speed needed for a projectile to escape the gravitational pull of some object (Earth, Sun, black hole, whatever). Any slower than escape velocity and the projectile will eventually come falling back down. The formula can be gotten from conservation of energy:

$$\frac{1}{2}m v^2 - \frac{GMm}{R^2} = 0$$

(on the left: energy at the surface of the planet/star/hole, on the right: energy at an infinite distance away) That works out to

$$v = \sqrt{\frac{2GM}{R^2}}$$

So the more massive the planet/star/hole, and/or the smaller it is, the faster you have to be moving to get away from it. A black hole is just an object that is so massive and so small that its escape velocity is faster than the speed of light. Since nothing can travel faster than light, anything that is unfortunate enough to wind up on the black hole's "surface" (inside its event horizon, in the lingo) is stuck there.

One caveat worth mentioning is that black holes are actually a relativistic phenomenon. You need the theory of general relativity (not so much quantum mechanics) to describe their structure and effects in detail. But luckily, the classical calculation I did above just happens to get the escape velocity right.

Also, I have heard they produce massive forces of gravity, though only one of 3 types of black holes can "suck". Correct me if I am wrong.

Well... there are different types of black holes, i.e. rotating or non-rotating, charged or uncharged... but those are tiny differences. All black holes have an incredibly strong gravitational pull when you get too close.

 

[Char. Limit]

So could a black hole be formed by a planet's gravity becoming so great it just collapsed on itself?

Well, not really. If the planet were large enough to collapse in on itself, it would first collapse into a star. Then if the resulting star burned out its life, it would then become a black hole, assuming that the star was large enough.

And could a black hole actually be a literal hole to the 'end' of space?

A black hole is not a hole.

 

[diazona]

So could a black hole be formed by a planet's gravity becoming so great it just collapsed on itself?

No, because there are forces that act between atoms and molecules that keep them from getting too close to each other. In order to form a black hole, you would need gravity to get strong enough to overcome those forces, and that in turn requires a lot of mass. The mass of a very large star, to be precise.

And could a black hole actually be a literal hole to the 'end' of space?

As far as we know, space doesn't have an end.

There's a lot of speculation about the true nature of black holes (like whether they could somehow be connected to other regions of space) but not much certainty.

 

[Char. Limit]

@Char. Limit: that's the general idea, but you can't really talk about gravity and light quite that way. Light always travels at the same speed in a vacuum, it doesn't accelerate or decelerate... so it doesn't make much sense to say that the acceleration is so large that light can't escape.

The better argument has to do with escape velocity, which is the minimum speed needed for a projectile to escape the gravitational pull of some object (Earth, Sun, black hole, whatever). Any slower than escape velocity and the projectile will eventually come falling back down. The formula can be gotten from conservation of energy:

$$\frac{1}{2}m v^2 - \frac{GMm}{R^2} = 0$$

(on the left: energy at the surface of the planet/star/hole, on the right: energy at an infinite distance away) That works out to

$$v = \sqrt{\frac{2GM}{R^2}}$$

So the more massive the planet/star/hole, and/or the smaller it is, the faster you have to be moving to get away from it. A black hole is just an object that is so massive and so small that its escape velocity is faster than the speed of light. Since nothing can travel faster than light, anything that is unfortunate enough to wind up on the black hole's "surface" (inside its event horizon, in the lingo) is stuck there.

One caveat worth mentioning is that black holes are actually a relativistic phenomenon. You need the theory of general relativity (not so much quantum mechanics) to describe their structure and effects in detail. But luckily, the classical calculation I did above just happens to get the escape velocity right.


Well... there are different types of black holes, i.e. rotating or non-rotating, charged or uncharged... but those are tiny differences. All black holes have an incredibly strong gravitational pull when you get too close.

Yes, I knew that it wasn't quite working when I was trying to find a gravitational acceleration to match a velocity, but I couldn't remember the exactly right way, so I went with this general idea. I was thinking that you had to use this equation somehow:

$$v_f^2-v_i^2=2ax$$

Who knows? Maybe it would have worked if I'd have worked it out.

 

[diazona]

I don't know about that, because that equation is for constant acceleration, and when you're escaping from a gravity well, your acceleration decreases with height. I've only ever seen this done with energy conservation. (And then of course there's the proper calculation in general relativity, but that's another story entirely ;-)

 

[Char. Limit]

I don't know about that, because that equation is for constant acceleration, and when you're escaping from a gravity well, your acceleration decreases with height. I've only ever seen this done with energy conservation. (And then of course there's the proper calculation in general relativity, but that's another story entirely ;-)

It would work if you could set initial velocity equal to zero (entirely reasonable) and position equal to 1 (I don't know why that would be). Then you get $v^2=2a$, and I showed before that a=Gm/R^2, and you quickly get your equation for escape velocity.