If the speed of light,is the fastest thing we know, Why can't light escape the pull of a Black Hole.
2007-06-29 13:22:41 · 10 answers · asked by raybbies 5 in Science & Mathematics ➔ Astronomy & Space
The gravity of a black hole at its event horizon is so hight that even the speed of light isn't fast enough to escape a black hole - the escape velocity of a black hole at its event horizon is greater than the speed of light.
The more mass of an object, the faster you have to move to escape the surface (called the escape velocity). For example, the escape velocity for the Earth is 11.2 km/s, and the escape velocity for the sun is 617.5 km/s.
But the escape velocity of a black hole is more than the speed of light.
2007-06-29 17:17:59 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Hi. Light travels in a straight line, ALWAYS. The line is the point between two positions in space. A black hole causes the space around it to bend. Light simply follows the bend. It's not that it can not escape so much as that the line's curve in folded back on itself. Make any sense?
2007-06-29 13:29:51 · answer #2 · answered by Cirric 7 · 1⤊ 0⤋
Black holes actually bend space, not just pull things into them. So, when light gets sucked into a black hole, the light is actually still traveling in a "staright" line (according to the light), but, because of the bending of space by the black hole, the light is bent, or somewhat turned into it.
I hope that makes sense. If you have any other questions feel free to e-mail me.
Good luck!
2007-06-29 13:28:56 · answer #3 · answered by Nick Name 3 · 1⤊ 0⤋
Think it does. If the grav field is large enough it can create a complimentary particle pair for as long as is allowed by E*t=h/2*pi.(like when a gamma ray gets up close and personal with a massive nucleus) one particle gets absorbed, the other has nothing now to annihilate with an fires off into real space, so effectively your 'black hole' is emitting energy. incidentally where do you guys get r=G*M/2C^2 for the radius..where does the factor of two come from? Also if we were to create a micro black hole, would it necessarily evaporate very quickly given the restrictions involved in quantum emission of energy??
2007-06-29 18:09:46 · answer #4 · answered by RTF 3 · 0⤊ 0⤋
Very simple. The black hole is so massive that its escape velocity is higher than the speed of light.
2007-06-29 13:46:04 · answer #5 · answered by Owl Eye 5 · 1⤊ 0⤋
Kinetic energy is one half the product of mass and velocity squared, okay? E=(M*V^2)/2.
It is energy that overcomes gravity. A small particle with little mass, like a photon, may move very fast, but speed is only one aspect of energy.
An anvil moving at the speed of light might go right through a strong gravitational field and out the other side before its momentum was exhausted. But a photon, which is "all speed and no mass" would not have the same kind of momentum.
2007-06-29 13:35:43 · answer #6 · answered by aviophage 7 · 0⤊ 1⤋
1) By definition, escape velocity from within the event horizon exceeds lightspeed. Gravitation has no upper limit.
2) All paths within the event horizon point at the singularity. There is no way out.
(2) is not so unusual. Look up, sideways, down. All 4(pi)steradians of direction point exactly at the Big Bang. The Big Bang is the same distance away in all directions. Every point in this universe is at its exact center.
Are we within a black hole?
2007-06-29 13:31:49 · answer #7 · answered by Uncle Al 5 · 2⤊ 1⤋
Because black holes warp the space around them, and light follows the curvature of space - right into the black hole.
2007-06-29 13:34:59 · answer #8 · answered by eri 7 · 1⤊ 0⤋
because light is still affected by matter and mass.....a black hole simply has so much mass, its gravity prevents photons from escaping
2007-06-29 13:28:36 · answer #9 · answered by Anonymous · 0⤊ 0⤋
the gravity is too strong
2007-06-29 14:22:02 · answer #10 · answered by futureastronaut1 3 · 1⤊ 0⤋