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I've heard that light can't escape from a black hole. Can it? If not, why?

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3 Answers 3

up vote 10 down vote accepted

A black hole has an event horizon which 'marks the point of no return'. So yes, light cannot escape from a black hole.

Why? Well, think of a 'spacetime fabric'. It's the easiest way to understand the physics at work here, in my opinion.

Usually, the fabric would look like this:


However, a black hole has so much gravity that one could say it 'rips' the spacetime fabric:

black hole fabric

When the light hits this area of amazingly intense gravity, it simply cannot get out - the light travels 'along' the fabric, and since there is a rip in the fabric, one could say it simply goes away - it becomes part of the singularity.

This is a simplification, of course, but it's enough to understand at least part of the physics behind this phenonenom.

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Is the time dilation effect of the event horizon itself enough to prevent light from escaping? Or is that only a small component of what keeps light trapped in a black hole? –  Mark Rogers Dec 13 '13 at 20:48
That's terrible. There is nothing there in the fabric-of-spacetime cartoon that addresses the horizon or why it's important. It's literally irrelevant--you're answering something like "what happens to stuff after it falls into a black hole?" instead. –  Stan Liou Dec 21 '13 at 15:47

I like to think of this in terms of escape velocity.

Escape velocity is the speed needed to escape the gravitational pull of a given object. For the Earth, that speed is 11.2 km/second (Mach 34!). When rockets blast off from Earth, they aren't trying to achieve a certain height or altitude, they're trying to reach a certain speed, the escape velocity.

Once a rocket reaches 11.2 kips*, it has attained the speed needed to leave the Earth completely. If a rocket fails to attain that speed, regardless of its height, it will fall back to the Earth. (You can imagine a magical balloon that slowly lifts you up into space, up past the ISS and most satellites, and then you let go: since you aren't going fast enough, you will fall back down, past all the satellites, and crash into the earth.)

Smaller gravitational bodies, like the moon, have smaller escape velocities. That's why the lunar landers were able to leave the moon with such a small ascent stage, compared to the massive Saturn V it took to leave Earth: they only had to go 2.4 km/second.

Saturn V vs Lunar Ascent Module Lunar Ascent Module

To escape the Sun, you'd have to go 617.5 km/second!

Fortunately for us, light goes faster than 617.5 kips, so we're able to see the light created on the Sun. However, as you increase the mass of an object, eventually the escape velocity would meet or exceed 299,792km/s, the speed of light. At that point not even light itself can go fast enough to escape the gravity well, and will always be pulled back down into the black hole.

*Short for "ki​lometers p​er s​econd"

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escape velocity for Earth is 11.2kps; however, this applies to thrown objects; you have to throw a rock at 11.2kps (ignoring atmospheric drag) in order for it to leave the Earth and not fall back; however, if your rock has an engine that can apply thrust, it can leave the Earth at a much lesser speed. The longer it is able to apply thrust, the slower it can go when leaving. –  jmarina Oct 30 '13 at 9:20
jmarina, that's interesting, I haven't heard that. Would you mind providing a link with more info or the name of the effect you're describing? –  brentonstrine Oct 30 '13 at 16:29
@jmarina is kinda right but the explanation is more interesting. Escape velocity actually decreases with distance from the body you are trying to escape from. For instance at 9,000 km up the escape velocity is about 7.1 km/s. The reason is that if you are going that speed aimed to just miss the Earth then you would pick up extra speed from falling towards it. And while the escape velocity from the sun at the sun's surface is 617.5 km/s, at Earth's orbit it is only 42.1 km/s. –  Jason Goemaat Oct 31 '13 at 5:37
Ah, I see. The escape velocity is factoring in the "free" velocity you get from gravitational pull if going back down (but angled enough to miss the planet). Is that right? –  brentonstrine Oct 31 '13 at 8:11
+1 for being the only answer so far that even attempts to address the question. –  Stan Liou Dec 21 '13 at 15:49

Don't forget that if a black hole has less than the current stable mass of a black hole (3 solar masses) then it evaporates - transforming its mass in radiation, in which case it would give off light, mostly X rays and gamma, at an increasing rate as its mass decreases, until the entire black hole is turned into a flash of hard radiation.


However, this light is the mass of the black hole escaping in the form of the most basic form of energy.

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OK would whoever is flagging answers here as "not an answer" please stop? There's a description under flags explaining when they should be used, and that isn't the case for any of the answers here in this thread. If that is not clear, please consult our help center pages or stop by in our Astronomy Chat to discuss what you think might be wrong with any of the answers provided here. Flags should be used for pressing issues with posts, not to indicate your disagreement with them. That's what voting is for! Thanks! –  TildalWave Dec 21 '13 at 18:58

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