I forgot about the spacetime distortion (heh, that sounds like Star Trek). Relativity is so weird.
Just to nitpick myself ( :smack: ), changing the value of π would require a rotating gravity source to cause the “frame dragging” effect.
Like you care. Well, anyway, now you know. :dubious:
Stranger
- Exapno Mapcase
No this is incorrect.
If during powered flight you are below the escape speed the object is still gravitationally bound to the earth (until it gets to infinity, which will never happen). So no escape will happen. You are doing little different then hovering in a helicopter.
Bold mine, there is no requirement for the object to return to it’s starting point if it does not equal or exceed escape speed, that helicopter, if we can hook it up with an infinite fuel supply and make it out of non-wearing parts (available in that theoretical physics catalog that Chronos mentioned) can stay hovering forever yet never escape.
Actually I think my nitpick was more right on then I thought. In order for any object to escape the earth it MUST equal or exceed the escape speed at some point (assuming it was gravitationally bound to start with)
Again, this depends on one’s definition of “escape”. The slow elevator will eventually reach any arbitrary distance from the Earth. I am quite content to say, therefore, that the slow elevator escapes, but it never reaches or exceeds escape speed. One might also, however, reasonably define “escape” to mean that even if you go ballistic, you’ll still approach infinity. If one defines “escape” in this way, then you do have to reach escape speed at some point.
On to black holes: One can treat a black hole as a Newtonian object whose escape speed is c, but this treatment is fundamentally flawed, and it’s only a coincidence that it predicts the correct value for the size of a black hole. A better description of a black hole is that, for any point inside the horizon, the horizon is in the past. One can no more climb slowly out of a black hole than one can climb to last Tuesday. In fact, any effort one makes to follow anything but a ballistic path, inside of a black hole, will only hasten one’s arrival at the singularity in the center.
Replacing the elevator with a rocket, I think it is more clear that what you are doing is just hovering over the earth, and slightly gaining altitude, I can’t accept an object hovering over the earth to have escaped from it. (But I’m willing to let you accept that :D)
This analogy simply doesn’t work. The elevator is not hovering about the earth in any sense. It is steadily moving away from the earth, just as the rocket that has exceeded escape velocity is. They are moving away at different rates, but neither one can be said to be hovering. And neither one will ever return to earth in our thought experiment. There is no conceptual difference between the two. You only create one by breaking the rules of the thought experiment, i.e. by turning off the power.
At what point do you consider the elevator to have escaped if it constantly remains below the relative escape speed? There is no difference between being one foot off the earth’s surface and being 1,000,000 light years away as far as I can see. Or is it merely that the thing is moving away and has the capability to do so for ever that defines it as having escaped?
Well, if we are positing the elevator to be rigid (and thus, the capsule to be rotating about the Earth at the same angular velocity as the Earth’s rotation) then escape occurs beyond unrestrained geosynchronous orbit, after which the capsule will be flung off into space, never to return. Where did the energy come from? From the Earth’s angular momentum, which decreased very slightly when the capsule detached. And yet, in terms of radial velocity, it need never have exceeded escape speed. If it is not moving away from or tangentially to the earth then it technically is not in an escape trajectory, but as a practical matter everything in space is in motion and anything that will not intersect the surface (or at least the upper atmosphere) has escaped from the planet.
Seriously, guys, you are trying to argue a point based upon a misapprehension. Escape speed is measured as the speed at which a ballistic object, fired from the surface (whether real or, in the case of gas giants, virtual) will escape in a hyperbolic orbit never to return. A similar, though more flexible definition is ejection velocity of an object in orbit (and we are all in orbit, powered by the interia of the ground and air around us), which is dependant upon altitude, but nobody calculates “escape velocity” as a varying function of altitude; we just figure that achieving escape velocity (or the kinetic energy level thereof) as measured from the surface is sufficient to launch something in a transplanetary trajectory. A vehicle capable of continuous thrust in excess of the gravitational attraction has “escaped” the earth even if it has a negligable upward velocity. (One can imagine, for instance, a vehicle made of Cavorite, which will automatically be in an escape trajectory no matter how weak the repulsion.)
Geezux, you guys are keeping me from my drinking. Give it a rest, will ya? 
Stranger
Correct me if I’m wrong here: A future light cone contains all the possible future positions of a particle in spacetime, given its current position. A black hole curves spacetime to such a degree that no particle inside the event horizon can have a future light cone that intersects the event horizon.
Two problems here. First, the dividing point is not at geosynchronous height. If you’re just a little above geosynch and let go of the elevator, then you’ll go into an elliptical orbit with perigee where you let go, and apogee somewhat further away. There is a point at which, if you let go, you’ll escape, but I’m too lazy at the moment to calculate exactly where it is.
Second, your radial velocity is irrelevant. As I said in the report, escape speed is a scalar, not a vector, and direction is irrelevant to it. If you go sufficiently high on a space elevator and let go, you are going at more than escape speed, even though most or all of your velocity is tangential, not radial.
sturmhauke, that is correct, if a little more technical than I put it.
I beg your pardon; you are correct. As I was trapped in a PowerPointless presentation review I took the liberty (anything to escape eight people arguing over nomenclature and semantics) of calculating out the escape velocity. I get 46,900 km, as compared to an unpowered geosynchronous orbit at 35,800 km. At this velocity you’ll travel along a parabolic path away from the planet. I think I did the calcs right, but I was going from memory on the constants, so hopefully I didn’t transpose any numbers or lose an order of magnitude.
Absolutely; my point was just that, with respect to ground (to which you can consider yourself tangentially fixed while attached to the elevator) you have no motion. Your only cost is the effort increase your radial distance from the surface, and on a fixed column you can go as slow as you like. You will, of course, pick up velocity with respect to a non-rotating reference at the center of gravity of the planet, and once you release from the beanstalk you’ll be flung off at tremendous speed (and the Earth loses a tiny bit in exchange.) But you need not achieve escape speed with respect to your rotating frame of reference (planet Earth).
Stranger
No then you misunderstood me. I replaced the (non roataing) space elevator with a rocket (due to budget cuts). This rocket will be powered in a way that it will never exceed the escape speed, just like the elevator. This makes it much easier to see that in our ‘earth only universe’ that it is really just hovering 5 lightyears above the earth, held up there by thrust. It has not escaped and it doesn’t matter that it is gaining another inch per year, it has not escaped.
I not totally sure this is correct, as both points are accelerating at different rates, due to rotation of the earth. You are taking a relative velocity on a rotating frame.
Wouldn’t that be when the elevator speed due to rotation = the escape speed for that elevation.
Yes, but you are still fixed to the Earth, even though not standing on the surface. You can stop at any time, lock your capsule to the column, and not expend any more energy to remain in place (unlike your rocket, which has to continue to fire.) You are under acceleration from rotation, of course, but that is true when you are on the ground, too. The Earth’s rotation is also calculated into orbital (and suborbital “ballistic”) launches, which is why orbital rockets are nearly always launched to the east, and preferably as close to the equator (for the continental US that’s California and Florida) as possible. You are still getting the required velocity magnitude to achieve escape speed, but you are getting it “for free” from the inertia of the Earth by moving in a tangential direction as liesurely as you like.
Escape can be very simply defined as having a trajectory that is parabolic or hyperbolic to (and non-intersecting with) the body in question. No unpowered object with a radial path through the center of mass of the body will ever technically escape as gravity will always be acting to accelerate it in the aft direction and will (taking the limit as t goes to infinity) return to the body, but for all practical purposes once it is to the point that gravity is measured in tiny fractions you may as well have escaped.
Stranger
And once again I ask, how in any way is this rocket - which is moving away from the earth forever, with no possibility, even theoretical, of its returning to the earth - different from the rocket that has exceeded a certain speed - and is moving away from the earth forever, with no possibility, even theoretical, of its returning to the earth? How do you give one the label of escaped and not the other?
Actually the error is that you can’t assign relative velocities to a accelerating frame. This is like 2 space ships accelerating side by side. One throws a ball to the other, but to the acceleration the ball appears to accelerate ‘backwards’ and misses the other spaceship. The rotating space elevator is no different in this respect.
The difference with the non-rotating elevator is it’s harder to see the forces, not that they are not present. Instead of the force from thrust you have a tensile force in the cable, the forces should be the same to balance out the force of gravity (and allow whatever acceleration you want).
Also for your claim that escape speed refers to ballistics, what do you think a rocket is. If you take any moment of time you have a rocket traveling at a certain speed being ‘shot’ ‘upward’ by a explosion behind it, giving it a acceleration, and a new velocity. This process is repeated over and over till escape speed is reached (if that was the goal). The ability of the rocket to thrust beyond the escape point is irrelevant to it escaping.
Replacing the rocket with the elevator just muddies the waters so to speak. Using an elevator really doesn’t change anything it just makes it harder to visualize the forces involved. A space elevator can be designed to launch a payload past escape speed, think railgun.
Yes the object will always be influenced by earths gravity on it’s trip to infinity, but the escape velocity, But you are the one who wanted to define escape speed in terms of balistics, which as I pointed out a rocket is at every moment of time and the one doing the same as the space elevator never exceeds the escape speed.
Every object in this earth only universe that escapes the earth has exceeded escape speed, your exception, while mathematically possible is physically impossible since infinity will never be reached. Your object will escape only when it reached infinity which will never happen.
Also another non-rotational elevator issue, how can you say that you have escaped the earth if you are on a elevator attached to the earth?
One more thing for now.
There are 2 types of space elevators being used here and they are not interchangeable:
Rotating elevator, attached to the equator that swings around the earth. If extended enough parts of it will be exceeding escape speed, further out still it will be exceeding the speed of light.
Non-rotating elevator, is on the north pole and on a frictionless device that allows rotation of the earth but not the elevator. This is a structure that is totally supported by the earth. It is basically standing on the earth and all it’s weight is supported by the ice shelf.
No, if you launch straight up at escape speed, you’ll escape just as truly and completely as if you had launched at any other angle. You will always be slowing down, but then, that would also be true if you launched on that (non-degenerate) parabola or hyperbola, and travelled a curved path. You will not, however, stop and return.
kanicbird, there is no meaningful definition of “ballistic” by which a firing rocket could be considered ballistic. “Ballistic” means that the only forces acting on an object are gravitational, and while a rocket is thrusting, there are other forces on it. Now, if the engine ever shuts off, then the rocket is ballistic after that point, and it doesn’t matter how it got up to speed. But if you allow it to be ballistic while firing, then everything’s ballistic, and the term becomes useless.
Oops. Again, mea culpa. Yes, any escape speed trajectory that doesn’t intersect the planet will escape. It’s a matter of energy; when the kinetic energy of the object (1/2 mv[sup]2[/sup]) is equal to or greater than the potential energy of the object in orbit (GMm/r) then it escapes. I’m morally certain Chronos knows this but for anybody else out there in TV land who wants to play with equations, there ya’ go.
Stranger
Why do they use missles which are basically rockets for ICB(allisatic)M’s then, and what does it mean when a fighter jet goes ballistic. my understanding is that you are opposing g with thrust (as opposed to lift in the case of the fighters). According to your statement those missles should be fired out of a cannon unfueled and jets should shut off their engines when going vertical