Escape velocity and black holes

according to fabian it should

If you near (or at?) the event horizon emit a laser paralell to the surface of the event horizon - then you get the photons in a orbit around the black hole. Theoretically.

As you can launch a baseball so that go in an orbit around the earth.

But we have to remember one very important thing:
When we compare something hard imaginable to somthing easy imaginable in order to get a similar result - we have to be very careful not to beleive it to much.

Like - if the black hole rotates (as black holes do), does the comparison still work?
Not likely.

Is my comparison between a straight on a bent paper and a photon movement in a bent space valid?
Yes, on low level but not in details.

Be careful here...

but light still has no mass and there is nothing in space to bend, so your comparisons dont make sense to me

Originally posted by Mephisto 666
but light still has no mass and there is nothing in space to bend, so your comparisons dont make sense to me

That was what everyone thought before Einstein. But Einstein showed otherwise.

Astronomers has detected stars behind the eclipsed sun where no stars should be seen. Explanation is that the light of the star has been bent by the gravitational field of the sun.
This is a fact. There is no other valid explanation. Einstein was right.

Originally posted by Mephisto 666
but light still has no mass and there is nothing in space to bend, so your comparisons dont make sense to me

You have to think of space as a matrix, kind of like a sheet of rubber, that would be a two dimensional representation, then when this sheet is stretched out in both directions and you put a golf ball on it you see there is a little dimple in the sheet, that represents the bending of space. Now imagine replacing the golf ball with a bowling ball, of course the sheet will have a significantly deeper dip in it. Now if you leave the bowling ball there, you can imagine rolling the golf ball on the sheet around the bowling ball and if there is low friction, it can make several 'orbits' around the bowling ball but the friction will eventually make it crash into said bowling ball but while there is some velocity to the golf ball you can see it simulates a planet orbiting around a sun or a moon orbiting around a planet. The thing to notice here, if you had drawn a line on the sheet before you put the balls there you would have a straight line because you used a ruler but if the bowling ball in placed near the line you will see the line stretch into a curve. That is exactly the path that light will take when it goes nearby a mass in space, like the sun. That was one of the key predictions of Einstein, that light would bend by about 1.7 arcseconds from the path it would have taken had the sun not been there and was measured a few years later which is what propelled Big Al to the world scene. It was verified about 30 years after that by radio astronomers who did the same experiment but on a radio frequency source, they measured that bend even more accurately than the original optical measurements.

This is one of the dumbest questions I've ever heard.

Originally posted by Bowmann
This is one of the dumbest questions I've ever heard.

Well a beginners question for sure, but I would not say dumb. Dumb answers maybe but not dumb questions.

Originally posted by smomofo
Also, why is there a smiley in my original post? I meant to close my brackets there, but a smiley appeared instead. How did that happen?

The winking emoticon is, itself, a black hole, gobbling up all brackets within its periphery. As said periphery is described as one space, anything which finds itself within the gravity pull range of .00001 through .99999 (and including the next integer, namely, 1) will necessarily yield to the winking emoticon.

Exceptions, as usual, abound.🙄

Originally posted by FabianFnas
If you manage somehow to stand on the 'surface' of an object with an escape velocity exceeding the speed of light - yes it would be a black hole.

A light beam could not escape the gravitational influence from the black hole.
Therefore an outside observer cannot ever see the light coming out of the black hole, no matter how near he is from the event ho ...[text shortened]... is mere speculation.

A black hole is a black hole, no matter the distance of the observer.

Almost sounds as though black holes have no boundary.

Originally posted by FreakyKBH
Almost sounds as though black holes have no boundary.

A black hole is defined to be inside its event horizon. But the influence of its presens is of course also recognizable outside this boundary.

Even on the Moon you are influenced by the Earth. This doesn't say that Earth is on the Moon.

The influence of a black gole has no gravitational boundary, just diminishing by the inverse square of the distance to it. As every body in the space.

Originally posted by FabianFnas
A black hole is defined to be inside its event horizon. But the influence of its presens is of course also recognizable outside this boundary.

Even on the Moon you are influenced by the Earth. This doesn't say that Earth is on the Moon.

The influence of a black gole has no gravitational boundary, just diminishing by the inverse square of the distance to it. As every body in the space.

No gravitational boundary, and no detectable boundary as well?

Originally posted by FreakyKBH
No gravitational boundary, and no detectable boundary as well?

The only abrupt change of a black hole in ralation of the distance from the center (aside for the center itself of which no one really knows anything anyway) is the event horizon.

If you pass the event horizon then you have no way getting back. Stay away from that.

Originally posted by sonhouse
You have to think of space as a matrix, kind of like a sheet of rubber, that would be a two dimensional representation, then when this sheet is stretched out in both directions and you put a golf ball on it you see there is a little dimple in the sheet, that represents the bending of space. Now imagine replacing the golf ball with a bowling ball, of course ...[text shortened]... y source, they measured that bend even more accurately than the original optical measurements.

thanx 😀 that helps a lot

So if you were within the event horizon of a black hole, would you be able to see light emitted from the black hole? Would the light seem to be approaching from all directions?

Originally posted by smomofo
So if you were within the event horizon of a black hole, would you be able to see light emitted from the black hole? Would the light seem to be approaching from all directions?

I may be off by a lot here (thus is the peril of a passing intrest in physics), but because, A, a black hole is commonly considered to be a singularity; and B, a black hole is the maximum expression of entropy, there can be no set direction for the light to come from. The light would exist as part of the theoretical observer and would exist as something like a constant field (relax, I'm not talking about aether).

To add a little extra, light can never aproach from all directions as photons exist only in the three dimensions we see (as far as popular theory goes).

I hope I have not made a greavious error in my explaination.