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Earth's "escape velocity" is roughly 11.2 km/s. This means that if an object is thrown upward at, say, 5 km/s it will fall back to the surface of the Earth. If the object is thrown upward at 7 km/s it will go higher, but still eventually fall back to the surface. If the object is thrown upward at 15 km/s, Earth's gravity will not be able to stop the object and it will never fall back down (it has exceeded Earth's "escape velocity"😉.
Since stars have a lot more mass than Earth, they have higher escape velocities. A black hole is an object whose escape velocity is higher than the speed of light, 300 000 km/s.
Suppose on object has an escape velocity of 330 000 km/s. If someone, somehow, were standing on this objectand were to point a flashlght straight up, the beam of light would go up, slow down, turn around and return to the surface.
Question: Would this be a black hole? If someone tried to observe this from far away, the light from the flashlight would have never made it to the observer, so the object would seem like a black hole, but if they moved closer, they may be able to see the beam of light before it turned around and fell back to the object, so it would no longer be a "black hole".
Does this seem right? Does it matter how far away an object is for it to be a black hole or not?
Are there any astronomers in the house?
Originally posted by smomofoMy first thought: Read Hawking, then think again about this.
Earth's "escape velocity" is roughly 11.2 km/s. This means that if an object is thrown upward at, say, 5 km/s it will fall back to the surface of the Earth. If the object is thrown upward at 7 km/s it will go higher, but still eventually fall back to the surface. If the object is thrown upward at 15 km/s, Earth's gravity will not be able to stop the objec ...[text shortened]... object is for it to be a black hole or not?
Are there any astronomers in the house?
I'll give my thoughts later, am in a hurry right now. 🙂
Originally posted by smomofoIf 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.
Suppose on object has an escape velocity of 330 000 km/s. If someone, somehow, were standing on this objectand were to point a flashlght straight up, the beam of light would go up, slow down, turn around and return to the surface.
Question: Would this be a black hole? If someone tried to observe this from far away, the light from the flashlight would h ...[text shortened]... away an object is for it to be a black hole or not?
Are there any astronomers in the house?
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 horizon.
But if the observer traverse the event horizon en go nearer the center of the black hole (and survives) then possibly he will se the light. But that is mere speculation.
A black hole is a black hole, no matter the distance of the observer.
Originally posted by smomofoYou use the character combination of " and ) which generate this kind of MrSmiley automatically.
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?
Put a space in between when you get a MrSmiley where you don't want it.
actually escape velocity of a black hole is not higher than speed of light. just because it depends on the square root of a ratio of a mass of an object you are escaping from (black hole)and in denominator is your distance from the center of an object.
and speed of light is maximum possible speed in our universe.
Originally posted by iambilkoActually no. There exist velocities exceeding that of light. But this is technicalities...
and speed of light is maximum possible speed in our universe.
If you in a equation yield a v > c means that you will have problems. And the problem for the experimentator in the bottom of a black hole trying to throw something out of the black hole is theat the object must traverse the barrier v=c to excape the gravitation. That is theoretically not possible
Take a paper and draw a straight line on it.
Then take the paper and bend it a little.
Is the line still straight?
Take a space and let a photon pass it in a straight line. Record the line.
Apply a (strong) gravitational field (with a black hole or something).
Is the recorded photon line still straight?
The answer is no of the both questions.
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Originally posted by FabianFnasOkay, so forget about my carburetor for a minute.
Take a paper and draw a straight line on it.
Then take the paper and bend it a little.
Is the line still straight?
Take a space and let a photon pass it in a straight line. Record the line.
Apply a (strong) gravitational field (with a black hole or something).
Is the recorded photon line still straight?
The answer is no of the both questions.
Considering my first post with an object, like a baseball, being thrown upward and falling back down, and assuming that light travels only in straight lines, does this mean that a black hole bends space right back around on itself so that my baseball analogy works, or is the analogy not a good one?
Would it be possible for an object's gravity to bend space just right so that light would orbit the object like a satellite?