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Brady Yoon
2004-Apr-20, 04:32 AM
In the Astronomy Magazine I bought from the newsstand today, I saw in an article that time is slowed down when around a strong gravitational force. I knew of this for awhile, but I never thought about it. My paradox is, how do we know black holes exist when the collapse of a star appears "frozen" in time to an outside observer? I think I have brain damage... :)

The Bad Astronomer
2004-Apr-20, 04:52 AM
Try here (http://cosmology.berkeley.edu/Education/BHfaq.html#q4).

Brady Yoon
2004-Apr-20, 05:06 AM
I'm still a little confused. Am I right in saying that the solution to this "paradox" is Hawking radiation? Or am I way off?

AGN Fuel
2004-Apr-20, 05:59 AM
In the Astronomy Magazine I bought from the newsstand today, I saw in an article that time is slowed down when around a strong gravitational force. I knew of this for awhile, but I never thought about it. My paradox is, how do we know black holes exist when the collapse of a star appears "frozen" in time to an outside observer? I think I have brain damage... :)

There is no paradox, because we don't 'know' that BH's exist because we observe an object enter them as such. They are discovered in different ways.

Typically they are found from the emission of X-Rays from an accretion disc of material around them. This material is usually drawn from a companion star that has evolved to the Reg Giant stage (and so the outer envelope has filled the Roche lobe & encroached into the lobe of the BH).

Once the X-Ray emission has been detected, spectroscopic examination of the light from the companion star will reveal if there is a high-mass object orbiting about it. Calculation of the mass of the 'invisible' member of the pair will reveal if it is a BH.

Keep in mind also, that ~solar mass black holes are very small on an astronomical scale and the area of really high 'time-slowing' is quite localised. When you see animations of accretion material being drawn into a BH, the 'black spot' to demonstrate the hole itself is usually hugely overenlarged. So, we have never actually seen the area around a black hole to not see an object fall into it........ :-s

Tensor
2004-Apr-20, 12:13 PM
So, we have never actually seen the area around a black hole to not see an object fall into it........ :-s

I understood that, which for some reason worries me. 8-[

Disinfo Agent
2004-Apr-20, 07:55 PM
I'm still a little confused. Am I right in saying that the solution to this "paradox" is Hawking radiation? Or am I way off?

Black holes demonstrate some counter-intuitive properties of general relativity. Consider a hapless astronaut falling radially towards the center of a Schwarzschild black hole. The closer she comes to the event horizon, the longer the photons she emits take to escape to infinity. A distant observer will see her descent slowing as she approaches the event horizon, which she never appears to reach. However, in her own frame of reference, the astronaut crosses the event horizon and reaches the singularity in a finite amount of time.
I imagine that just as the astronaut reaches the singularity in finite time in her own frame of reference, so does the star collapse on itself in a finite amount of time in its own frame of reference…

Lunatik
2004-Apr-20, 08:24 PM
Black Hole (http://northdakota.asinah.net/en/wikipedia/b/bl/black_hole.html) says:

The Schwarzschild radius is given by

r = 2GM/c^2

where G is the gravitational constant, M is the mass of the object, and c is the speed of light. For an object with the mass of the Earth, the Schwarzschild radius is a mere 9 millimeters.
This is a variance on Newton's orbital velocity equation

r*v^2 = GM, whereby r = GM/v^2

Main difference is that v^2 is made to equal c^2 in the Schwarzschild.

The referenced illustration of a body falling into the 'black hole' is missing a very important component, namely, where any mass approaching such great gravity does not simply 'slip' into it, but is put into immense spin around it naturally. This spin is not counterintuitive. We know this from the vast number of stars crowding the black hole near the galaxy center, all at very high oribtal velocities. This very high velocity may account for X-ray radiation put out by the black hole, but am over my head again, so don't quote me. :oops:

Tensor
2004-Apr-20, 10:01 PM
I imagine that just as the astronaut reaches the singularity in finite time in her own frame of reference, so does the star collapse on itself in a finite amount of time in its own frame of reference…

Yes it does. While a frame far away will see it "freeze", a frame on the surface of the star will collapse through the horizon. The paradox of one frame collapsing, while the other frame seeing it freeze was solved by David Finklestein in 1958. He found a viable frame that included both the collapse and the apparent freeze. The apparent freeze from the far away frame was an illusion due to the gravitational red-shift of the light trying to escape from the collapsing star.

JohnOwens
2004-Apr-21, 12:12 AM
Black Hole (http://northdakota.asinah.net/en/wikipedia/b/bl/black_hole.html) says:

The Schwarzschild radius is given by

r = 2GM/c^2

where G is the gravitational constant, M is the mass of the object, and c is the speed of light. For an object with the mass of the Earth, the Schwarzschild radius is a mere 9 millimeters.
This is a variance on Newton's orbital velocity equation

r*v^2 = GM, whereby r = GM/v^2

Main difference is that v^2 is made to equal c^2 in the Schwarzschild.
It's an even closer and more appropriate variation on the potential energy/escape velocity equation
r*v^2 = 2GM, whereby r = 2GM/v^2
That's the usual Newtonian approach for oversimplifying the Schwarzschild radius. Apparently, that approach doesn't stand up under scrutiny, but it gives the right result anyway.

Brady Yoon
2004-Apr-21, 02:00 AM
I imagine that just as the astronaut reaches the singularity in finite time in her own frame of reference, so does the star collapse on itself in a finite amount of time in its own frame of reference…

I liked this explanation. Anything else to add? :)

Taibak
2004-Apr-21, 04:37 AM
Black Hole (http://northdakota.asinah.net/en/wikipedia/b/bl/black_hole.html) says:

The Schwarzschild radius is given by

r = 2GM/c^2

where G is the gravitational constant, M is the mass of the object, and c is the speed of light. For an object with the mass of the Earth, the Schwarzschild radius is a mere 9 millimeters.
This is a variance on Newton's orbital velocity equation

r*v^2 = GM, whereby r = GM/v^2

Main difference is that v^2 is made to equal c^2 in the Schwarzschild.
It's an even closer and more appropriate variation on the potential energy/escape velocity equation
r*v^2 = 2GM, whereby r = 2GM/v^2
That's the usual Newtonian approach for oversimplifying the Schwarzschild radius. Apparently, that approach doesn't stand up under scrutiny, but it gives the right result anyway.

Yeah, it comes out of the classical escape velocity equation. The problem is that the equation predicts that a photon launched from the event horizon will slow down, stop, and fall back to the horizon - classical ballistic motion. Since that requires the photon to change speeds, that can't happen.

The relativistic version is a bit more complicated, to say the least. For a non-rotating, electrically neutral, spherical object in otherwise empty space, the event horizon comes from the equation that describes the geometry of spacetime:

dT^2 = (1 - 2GM/rc^2)dt^2 - dr^2/(1 - 2GM/rc^2)

Basically, that tells you how space and time are warped near a spherical black hole. If r = 2GM/c^2, you're dividing by 0 in the second term - and that's what predicts the location of an event horizon.

JohnOwens
2004-Apr-21, 05:53 AM
That reminds me, speaking of these as "non-rotating, electrically neutral, spherical object[s]", is there an easy way to describe the shapes that result from the metric for Kerr and Reisser-Nordstrom and Kerr-Newman (rotating, charged, and rotating + charged, respectively) black holes? I'm particularly interested in the Kerr. Would it be an oblate spheroid, ellipsoid, or a similar shape up to a rate of spin where it gets weirder, maybe a torus or something? I don't know. I'd guess that the Reisser-Nordstroms ought to be spherical, perhaps with a different radius than an uncharged? Even then, I'm not having much luck trying to figure out whether it should be smaller or larger. And Kerr-Newmans just make my head hurt. #-o

Disinfo Agent
2004-Apr-21, 12:40 PM
Anything else to add? :)
I once had the same doubt as you, which is why I understood your question. ;)

Klausnh
2004-Apr-21, 02:03 PM
From physlink.com (http://www.physlink.com/Education/AskExperts/ae158.cfm)

"So if you can throw the object with enough initial upward velocity so that gravity's decreasing force can never quite slow it to a complete stop, its decreasing velocity can always be just high enough to overcome gravity's pull. The initial velocity needed to achieve that condition is called escape velocity."
Doesn't this mean that a rocket could escape from within the event horizon as long as it has fuel to accelerate?
and isn't this statement wrong from


Try here (http://cosmology.berkeley.edu/Education/BHfaq.html#q4).



"But if you find yourself inside the (event) horizon, then no matter how powerful your rockets are, you can't escape"

Taibak
2004-Apr-21, 02:38 PM
From physlink.com (http://www.physlink.com/Education/AskExperts/ae158.cfm)
[quote]"So if you can throw the object with enough initial upward velocity so that gravity's decreasing force can never quite slow it to a complete stop, its decreasing velocity can always be just high enough to overcome gravity's pull. The initial velocity needed to achieve that condition is called escape velocity."
Doesn't this mean that a rocket could escape from within the event horizon as long as it has fuel to accelerate?
and isn't this statement wrong from

No. The escape velocity within the event horizon is greater than the speed of light. It will be impossible to get out. Similarly, no matter how powerful the rockets are, any acceleration they can give will always be less than the downward acceleration due to gravity. Once you cross the horizon, it's a one way ticket to the singularity.

Lunatik
2004-Apr-21, 03:38 PM
Taibak: Yeah, it comes out of the classical escape velocity equation. The problem is that the equation predicts that a photon launched from the event horizon will slow down, stop, and fall back to the horizon - classical ballistic motion. Since that requires the photon to change speeds, that can't happen.
Since light is always at v=c, then the only other option is for it to redshift, where at the event horizon it redshifts out to nothing. I had seen (deleted "infrared") images of neutron stars (http://www.space.com/scienceastronomy/astronomy/death_spiral_sidebar_010111.html) where the light was redshifted around them. A black hole would be an extreme case where this light is stretched to invisibility. By thinking of it this way, should this prove true, makes light less a particle and more a wave, and it makes the black hole a special case. That special case may also mean that the gravity there is maximum G, unlike the very weak dilluted G we experience here in our solar system. If so, that leaves the door open for G to be a variable between its mximum in black holes, and minimums in hot solar systems, with lots of space variations in between... just a possibility to watch for in all our astrophysical observations.

JohnOwens
2004-Apr-21, 06:34 PM
From physlink.com (http://www.physlink.com/Education/AskExperts/ae158.cfm)

"So if you can throw the object with enough initial upward velocity so that gravity's decreasing force can never quite slow it to a complete stop, its decreasing velocity can always be just high enough to overcome gravity's pull. The initial velocity needed to achieve that condition is called escape velocity."
Doesn't this mean that a rocket could escape from within the event horizon as long as it has fuel to accelerate?
For one thing, like I said above, treating the Schwarzschild radius as the Newtonian solution for the escape velocity being the speed of light isn't accurately correct; it just happens to give the right answer. Also, in addition to the Taibak's explanation, remember that according to GR, your time (as viewed by an outside observer) slows down approaching infinitely as you approach the event horizon (Does it even make sense to consider whether this is true from the inside as well? I know we can never know, as far as we know, because no information will ever get out about what's in there, but is there supposition of a symmetry which would have time pass more slowly towards the EH than at the singularity?), so you'll never have time to use enough of that thrust.


That special case may also mean that the gravity there is maximum G, unlike the very weak dilluted G we experience here in our solar system. If so, that leaves the door open for G to be a variable between its mximum in black holes, and minimums in hot solar systems, with lots of space variations in between... just a possibility to watch for in all our astrophysical observations.
Remember, with a big enough black hole, the acceleration of gravity can be arbitrarily weak at the event horizon. There's no direct correspondence between being at the EH and being subjected to high gravitational acceleration. (Hmm, I assumed by G you meant the acceleration of gravity, but I'm starting to think you did mean the gravitational constant after all, remembering what I think you've posted elsewhere.)
P.S. You do know that the neutron star/black hole pictures in that article are artist's representations or computer simulations, not actual astronomical images, right?

Lunatik
2004-Apr-21, 08:02 PM
JohnOwens: P.S. You do know that the neutron star/black hole pictures in that article are artist's representations or computer simulations, not actual astronomical images, right?
Didn't know, but wondered if so. Thanks for clearing that up, just an artist's rendition, not the real thing.

And yes, I meant the G constant as being a possible variable from maximum strength inside EH, and minimum in our region. However, shelve this as 'extremely speculative' for now, since we have no direct observational evidence this is true, merely conjecture. Interesting that G es expected to be the same as here on the event horizon itself, hadn't thought of that.

darkdev
2004-Apr-21, 11:22 PM
So if from the outside, an observer sees the object at (just above) the EH for infinity, shouldn't black holes be very very bright, emitting all the radiation of everything thing that has ever passed the EH, for all eternity?

Does that make sense?

Ricimer
2004-Apr-21, 11:33 PM
So if from the outside, an observer sees the object at (just above) the EH for infinity, shouldn't black holes be very very bright, emitting all the radiation of everything thing that has ever passed the EH, for all eternity?

Does that make sense?


That would be true, except the light is also redshifted. And the light that gets emitted at the horizon, is redshifted infinitely, on top of not going anywhere.

Even the stuff just this side has an immense redshift, sending the emitted ligh t way off any scale.

Andreas
2004-Apr-21, 11:41 PM
So if from the outside, an observer sees the object at (just above) the EH for infinity, shouldn't black holes be very very bright, emitting all the radiation of everything thing that has ever passed the EH, for all eternity?

Does that make sense?

It's also redshifting, the more the close it gets to the event horizon. I don't think we could see something falling in freeze at the event horizon, since that requires almost infinite time dilation from which almost infinite redshift follows.

The total amount of light emitted is exactly what the infalling object emits before crossing the event horizon from its own frame of reference, only to the outside observer the emission of that amount will stretch over an infinite amount of time. Well, not infinite if we consider the quantisation of light. Am I correct?

JohnOwens
2004-Apr-21, 11:44 PM
So if from the outside, an observer sees the object at (just above) the EH for infinity, shouldn't black holes be very very bright, emitting all the radiation of everything thing that has ever passed the EH, for all eternity?

Does that make sense?
Yes, they do, but, very, very slowly, i.e. not "very very bright". For instance, a black hole the mass of our Sun would have an effective temperature... let me check my notes... 6.17*10^-8 K, a little bit chilly, total power of 9.01*10^-29 W, lose mass at about 1.00*10^-45 kg/s, and have a lifetime of about 2.1*10^67 years. So like I said, yes, it radiates away everything that ever passed the EH, but not at all that quickly, though not quite for all eternity.

Added: I see other people took you to mean radiation from the actual matter approaching the EH with infinite slowness, and this time I'm not sure my interpretation of it as referring to the Hawking radiation is correct. :-?

darkdev
2004-Apr-22, 12:12 AM
Between the three of you ;), I think I get it, it [the radiation] is there, but spread out (red-shift and in time) such that it appears very very weak, if it does indeed escape.

JohnOwens, I did mean matter radiation from just above the EH, and not Hawking radiation.

:D :D :D

Disinfo Agent
2004-Apr-22, 10:46 AM
Well, correct me if I'm wrong, but there should be a lot of X-rays emitted from just above a black hole. In that sense, it is "bright". So bright we can't see it. ;)

Edited.

Richard of Chelmsford
2004-Apr-22, 11:19 AM
I'm sure this has already been covered somehow, but I saw a prog on telly called 'Space' which was fronted by actor Sam Neill of Jurassic Park and other stuff.

At 8.30pm he said "A black hole has such strong gravity that nothing, not even light can escape from it."

At 8.45pm he said "The black hole will sometimes eject excess material."

#-o

Disinfo Agent
2004-Apr-22, 11:30 AM
At 8.45pm he said "The black hole will sometimes eject excess material."

#-o
He could be talking about material outside the event horizon. Like in an accretion disk.

Klausnh
2004-Apr-22, 04:05 PM
For one thing, like I said above, treating the Schwarzschild radius as the Newtonian solution for the escape velocity being the speed of light isn't accurately correct; it just happens to give the right answer. Also, in addition to the Taibak's explanation, remember that according to GR, your time (as viewed by an outside observer) slows down approaching infinitely as you approach the event horizon (Does it even make sense to consider whether this is true from the inside as well? I know we can never know, as far as we know, because no information will ever get out about what's in there, but is there supposition of a symmetry which would have time pass more slowly towards the EH than at the singularity?), so you'll never have time to use enough of that thrust.
Thanks for the explanation. So it's the time component of spacetime that prevents anything from escaping a black hole?
Many of the articles I've read, use the Newton explanation (using escape velocity) of why light cannot escape the black hole and that never made any sense to me.

You can think of the horizon as the place where the escape velocity equals the velocity of light. Outside of the horizon, the escape velocity is less than the speed of light, so if you fire your rockets hard enough, you can give yourself enough energy to get away. But if you find yourself inside the horizon, then no matter how powerful your rockets are, you can't escape.
Am I right in interpreting the above quote as a Newtonian explanation?

Taibak
2004-Apr-22, 05:16 PM
For one thing, like I said above, treating the Schwarzschild radius as the Newtonian solution for the escape velocity being the speed of light isn't accurately correct; it just happens to give the right answer. Also, in addition to the Taibak's explanation, remember that according to GR, your time (as viewed by an outside observer) slows down approaching infinitely as you approach the event horizon (Does it even make sense to consider whether this is true from the inside as well? I know we can never know, as far as we know, because no information will ever get out about what's in there, but is there supposition of a symmetry which would have time pass more slowly towards the EH than at the singularity?), so you'll never have time to use enough of that thrust.
Thanks for the explanation. So it's the time component of spacetime that prevents anything from escaping a black hole?

It's a little bit of both, actually. One of the things that the Schwarzschield equation predicts is that within the event horizon, space becomes timelike. Jargon aside, that means that within the horizon, space lets you travel in only one direction - in.


Many of the articles I've read, use the Newton explanation (using escape velocity) of why light cannot escape the black hole and that never made any sense to me.

You can think of the horizon as the place where the escape velocity equals the velocity of light. Outside of the horizon, the escape velocity is less than the speed of light, so if you fire your rockets hard enough, you can give yourself enough energy to get away. But if you find yourself inside the horizon, then no matter how powerful your rockets are, you can't escape.
Am I right in interpreting the above quote as a Newtonian explanation?

No, I think it works under relativity, to a certain extent. It's perfectly fine to say that inside a black hole your escape velocity is greater than the speed of light. You just need to keep in mind that acheiving that velocity is impossible.

Where this stops working is when you're talking about light. A photon launched from within the event horizon will, according to Newton, slow down, stop, and then turn around and fall back in. Under relativity, that doesn't happen. What does happen to the light is that as it tries to move out of the black hole, it gets severely redshifted. In fact, it gets redshifted so that it's wavelength becomes infinite and its energy gets transferred to the black hole. An infinite wavelength means, essentially, that the wave is gone (try to visualise this - it's pretty easy). For all intents and purposes, the light gets redshifted out of existence.

Ricimer
2004-Apr-22, 07:21 PM
richard, he, and the writers, made a typical slip.

He was talking of

a: Accretion disk phenomena (jets and the like)
b: Hawking radiation, an exception to the rule.

Taibak: I've always viewed it as all the space-time paths are curved back to the singularity. Is that valid, or should I think of it as the redshift method? Both seem to work in my mind.

Klausnh
2004-Apr-22, 09:24 PM
No, I think it works under relativity, to a certain extent. It's perfectly fine to say that inside a black hole your escape velocity is greater than the speed of light. You just need to keep in mind that acheiving that velocity is impossible.

Where this stops working is when you're talking about light. A photon launched from within the event horizon will, according to Newton, slow down, stop, and then turn around and fall back in. Under relativity, that doesn't happen. What does happen to the light is that as it tries to move out of the black hole, it gets severely redshifted. In fact, it gets redshifted so that it's wavelength becomes infinite and its energy gets transferred to the black hole. An infinite wavelength means, essentially, that the wave is gone (try to visualise this - it's pretty easy). For all intents and purposes, the light gets redshifted out of existence.Let me break down my reasoning.
To get away from earth's gravitational effects, a rocket does not need to have an escape velocity, it only needs to maintain an acceleration that is greater than the gravitational acceleration of the earth, ie about 9.8 m/s/s on the surface and decreasing as it rises. It's velocity can be as little as 5kps. Translate that to a "newtonian" black hole, a rocket should not need to reach the black hole's escape velocity. It only needs to maintain an acceleration that is greater than the black hole's gravitational acceleration. As long as that gravitional acceleration is not infinite, a rocket should be able to escape the black hole's gravitational effects. However, as JohnOwens points out, the relavistic effects (curvature of spacetime) has to be taken into account in a "relativistic" (non-newtonian) black hole. In the "relativistic" black hole, gravitational acceleration is not enough to prevent anything from getting out. It's the time component which keeps matter inside the black hole.
Are my assumptions and reasoning right?

Ricimer
2004-Apr-22, 10:14 PM
escape velocity is actually an interpretation of energy. The escape velocity is velocity required to have enough kinetic energy (imparted in one swift kick) to escape a gravity source.

So, while you don't have to ever achieve that velocity to get away, you do have to expend the corresponding amount of energy.

This can easily be transfered to a BH situation (unlike the acceleration where it requires an infinite acceleration scenario). To leave a black hole, you merely need an infinite amount of energy. Actually, more than that. This is becuase the escape velocity is = or > C. And that corresponds to an infinite, or greater than infinite amount of energy.

Richard of Chelmsford
2004-Apr-22, 10:33 PM
At 8.45pm he said "The black hole will sometimes eject excess material."

#-o
He could be talking about material outside the event horizon. Like in an accretion disk.

So material CAN get away from a black hole?

Taibak
2004-Apr-22, 11:01 PM
Taibak: I've always viewed it as all the space-time paths are curved back to the singularity. Is that valid, or should I think of it as the redshift method? Both seem to work in my mind.

As far as I'm aware, both work. I think I've seen it modeled both ways.

AGN Fuel
2004-Apr-22, 11:03 PM
At 8.45pm he said "The black hole will sometimes eject excess material."

#-o
He could be talking about material outside the event horizon. Like in an accretion disk.

So material CAN get away from a black hole?

From the region that surrounds the Black Hole. Black holes are frequently accompanied by bi-polar jets of material that are accelerated away from the region around the hole to very high velocities. Black Holes appear to be messy eaters!

For an example, have a look at images of the huge galaxy M87, which appears to have a 3 billion solar mass black hole at it's nucleus. There is an amazingly tightly collimated jet that extends for thousands of light years away from the central BH.

Ricimer
2004-Apr-22, 11:08 PM
At 8.45pm he said "The black hole will sometimes eject excess material."

#-o
He could be talking about material outside the event horizon. Like in an accretion disk.

So material CAN get away from a black hole?

Yeah, its the stuff that "almost" falls in. It gets sped up enough to escape before (sometimes just before) it crosses that point of no return.

Klausnh
2004-Apr-27, 12:09 AM
escape velocity is actually an interpretation of energy. The escape velocity is velocity required to have enough kinetic energy (imparted in one swift kick) to escape a gravity source.

So, while you don't have to ever achieve that velocity to get away, you do have to expend the corresponding amount of energy.

This can easily be transfered to a BH situation (unlike the acceleration where it requires an infinite acceleration scenario). To leave a black hole, you merely need an infinite amount of energy. Actually, more than that. This is becuase the escape velocity is = or > C. And that corresponds to an infinite, or greater than infinite amount of energy.
Using energy instead of velocity:
Energy required to leave earth's gravitational influence is equal to the kinetic energy required to produce escape velocity. For the earth, I believe escape velocity is 11m/s, so energy required is 60.5*m of the rocket. If a rocket has an initial velocity of 10m/s (kinetic energy = 50*m of the rocket) and has no energy added to it, it will not have enough kinetic energy to leave earth's gravitational influence. However, if the rocket has a constant velocity of 10m/s and add additional energy (10.5*m), it will leave earth's gravitational influence. (kinetic energy (50*m) + 10.5*m = 60.5m)
Substitute a "Newtonian" black hole for the earth: Kinetic energy required is m*c*c/2. A rocket has an initial velocity of .9c (kinetic energy = .405c*c*m) of the rocket, It will not have enough kinetic energy to brake away from a "Newtonian's" black holes influence. Now give the rocket a constant velocity of .9c and add additional energy (.095c*c*m), it will escape the "Newtonian" black hole. (kinetic energy (.405c*c*m) + extra energy (.095c*c*m) = m*c*c/2).
Is this correct?

JohnOwens
2004-Apr-27, 02:02 AM
Using energy instead of velocity:
Energy required to leave earth's gravitational influence is equal to the kinetic energy required to produce escape velocity. For the earth, I believe escape velocity is 11m/s, so energy required is 60.5*m of the rocket....
One correction here, although it doesn't really affect any of your reasoning here: It's 11 km/s, so 60,500,500*m, and apply the same factors to the rest of your m/s & (m/s)^2 figures. Doesn't affect the ones using c at all, obviously.

Klausnh
2004-Apr-27, 02:25 AM
One correction here, although it doesn't really affect any of your reasoning here: It's 11 km/s, so 60,500,500*m, and apply the same factors to the rest of your m/s & (m/s)^2 figures. Doesn't affect the ones using c at all, obviously.
Thank you. I should have looked it up instead of relying on memory.

Ricimer
2004-Apr-27, 04:04 PM
Klausnh: Yeah, thats basically how it works. Viewing it that way is one of the easiest ways to explain how something can escape under constant acceleration, but never hitting the escape velocity.


Another way is to realize the escape velocity drops, the further out you are (you know, gravity is weaker) and so you only need to exceed the escape velocity of the object, at some point in the trip. And since the requirements drop, and if you accelerate throughout, you'll pass that barrier. You still need to accelerate. If you don't and go slower than 11km/s for the earth, gravity will turn you around before you get far enough.

But yeah, you've got it.