From http://www.straightdope.com/mailbag/mescapevelocity.html
Though it is a nitpick, you will need to exceed escape velocity at some point to escape the earths gravity. This is because the escape velocity is not a constant but changes with distance from the earths center of gravity, and at one point will approach zero (very far away from earth), a velocity you must exceed to break free.
I think this is wrong. By definition, escape velocity is the speed needed to escape. So all one need do is equal it, not exceed it.
And your speed will decrease as well with distance, as the article states. Your speed and escape velocity are always equal. They will approach zero but that’s irrelevant.
If you are changing your speed, or continuing to fire rockets or use any other propulsion, then again, by definition, escape velocity doesn’t apply.
You nitpicked my nitpick.
But for clarity what I was nitpicking was
, which is incorrect, that rocket will have to get to escape speed at some point in it’s travel.
Also another escape speed Q, what does it actually mean? What I assume it means is if there is only the earth and the totally no-drag cannonball (boy are those special order cannonballs expensive) in the universe, that at the escape speed the cannonball will reach a zero speed (relative to the earth) when it’s distance it infinity?
No it won’t. Think of a space elevator of indefinite size. As long as it is powered it never has to exceed escape velocity to travel as far as it likes.
It will approach zero, but can never reach zero in a finite universe. It’s speed can become unmeasurable by any conceivable instrument, but not zero. For that matter no material object can ever reach infinity or zero even in an infinite universe.
.
I disagree. It never has to get to that X thousand MPH speed, but as that elevator continues ‘up’ at 5 mph, it will eventually get to a point (way way up high) that the escape velocity = 5 mph.
Think of it this way. The escape velocity never goes to zero. So there is always a speed lower than the escape velocity but above zero. The elevator does not have to travel at a constant speed. It can travel indefinitely far slower and slower but always under the escape velocity.
IOW, No matter what the escape velocity is, the elevator can continue to go farther and farther out into space without ever exceeding or equaling it.
But you’re assuming that the elevator travels at a constant speed. It would be quite possible (though admittedly impractical) to make the elevator gradually slow down, such that at any given height, it’s always travelling at (say) exactly half of escape speed at that height. An elevator travelling in such a manner would eventually reach any arbitrary distance from the Earth (that is to say, it would escape), but it would never reach or exceed the escape speed for its height.
Assuming we ignore mathematical quibbles to the effect that it never actually reaches infinity (hey, we’re physicists, not mathematicians; we’re allowed to do that), yes, this is correct. And the Frictionless Cannonballs aren’t too expensive; the Theoretical Physics Supply Catalog sells them for about the same price as the Massless Pulleys, and only about twice as much as a ten-meter length of Unstretchable String ;).
Thank you for that report, Chronos. It’s intuitively obvious that “escape velocity” has more to do with cannon than slow rockets with plenty of fuel but people have always told me I was wrong to think that, especially my wife. I’d show her your article but she’d just deny she ever said such a thing.
What about Infinitely Inelastic Rods in ly lengths (so you can transmit information instantaneously across the reaches of space) and Relativistly Objective Stopwatches?
I was looking for some sheets of .050 unobtainium sheet stock the other day, but Ryerson said they could only order it in mill runs of 20 tons minimum. Does TPSC offer it in small quantities?
Stranger
Then again, at the height that, “for all practical purposes” the object does escape, the escape speed would be “for all practical purposes” zero, so the object’s speed rel. to earth must be greater then or equal to the escape speed since we can’t have a negative speed. So for an object that is escaping MUST at some point equal or exceed the escape speed.
I think the only exception might be if we allow infinity, as the object may be able to approach it slower then the escape speed will allow, but since it will never reach it, it would never really have escaped, and will crash back to earth (or perhaps orbit if some angle was used) at a greater speed and quicker time then it took the elevator to get the object to that point at the 1st place.
IANA(n) engineer, but for practical purposes, can’t escape velocity be established relative to orbital distance? It’s true that an object can never escape the pull of earth’s gravity, but objects can and do reach orbit on quite a regular basis.
OK, so at that height the escape speed is really really small. Then make the elevator’s speed at that height really really smaller. It’s not true to say that in order to be less than something close to zero, you have to be negative. You just have to be closer to zero.
Or, to put it another way: If I have my elevator that’s always moving up at half of escape speed, at what point would you say it exceeds escape speed? You say there is such a point; where is it?
And NattoGuy, the whole point of escape speed is that an object can escape the pull of Earth’s gravity. It never reaches infinite distance, but it’s going fast enough that there’s nothing to stop it from doing so. Escape speed is related to orbital speed, though: It’s always exactly sqrt(2) times the circular orbital speed at that height. In other words, if you took a satellite in a circular orbit (at any distance from the Earth), and increased its speed by a factor of 1.41…, it would fly off into the depths of space never to return.
Hmm Chronos, I think you’re wrong on this one. In order to maintain a speed at say half escape speed, you’d need to continually apply thrust yes? The earth’s gravity will always be exerting a pull such that you need to maintain slight thrust even though you are constantly slowing down. How can you ever consider yourself to have “escaped” the effects of earth’s gravity (lets ignore all the other gravity sources for now)? If you cease the thrust you will eventually head back to earth. However if you maintain constant speed such that it at some point equals the escape speed for earth at that distance, then you can remove all thrust and continue for ever without ever falling back to earth. Only in this way will you have escaped and as, Kanicbird says, you will have equalled escape speed at some point, which validates his nitpick.
Obviously in the real world, all you need to do is maintain your speed until you come within the grips of some other gravity field stronger than the earth’s, escape speed need not be met.
Touche. I suppose it depends on how one defines “escape”. I would say that if, as t -> infinity, the object’s position -> infinity, then the object has escaped, regardless of how it does it, but I can see the argument for only calling it escaped if it can turn off the thrusters and still approach infinity. But there’s no point in arguing over definitions.
Switching gears here, The moon will eventually leave us and fly off to infinity. Does this mean it’s current speed is equal to or greater then ‘about’ 1.41x it’s stable orbit speed? Does the moon already exceeded escape speed? I say about because of the non-circular orbit of the moon.
Yes, there is a point in arguing over definitions, because a definition is at the heart of this argument.
There is a basic distinction between ballistic flight and continually powered flight. What 1920s Style “Death Ray” does, in his otherwise correct post, is merge the two. If you turn off a rocket then you are changing a continually powered flight into a ballistic flight. Then the restrictions on escape velocity come into play. But we’re dealing with a thought exercise here in which the power is never turned off. That means that the craft can travel arbitrarily far and never return.
If you never enter ballistic flight then you can travel as slowly as you want, never exceed escape velocity and still never return. That’s a necessary conclusion from the definition of continually powered flight.
kanicbird’s problem was that he kept confusing the two modes. I think he is still doing so in his last post about the moon, because the recession of the moon from earth is not ballistic, but due to tidal frictions, which is a source of power.
Er, no. As discussed in this thread, the moon will not fly off to infinity. While it’s average orbital distance is increasing (very gradually), it will eventually come to an equilibrium position where the moon and the earth are tidally locked, and absent of any other gravitational effects will stay in that configuration indefinately. If the moon had exceeded escape speed it wouldn’t continue to orbit. I would instead take of on a hyperbolic trajectory, never to return.
As for the OP and Chronon’s Staff Report that proceeded it, I think the confusion rises from an overextension of the term “escape speed”. Escape speed (or velocity, if you don’t mind being a little sloppy with the terminology) is the instantaneous speed you need require in order to achieve a hyperbolic (escape) orbit for a given altitude. (Technically, the path you’ll get exactly at escape speed is a parabolic, but we’re assuming you want a little lattitude in case Engine 5 fails to ignite.) On earth 36.7 kfps is sufficient (at the surface and neglecting air resistance) to achieve orbital escape. At higher altitudes, the speed decreases by a squared factor.
Certainly, at some point the magnitude of your ballistic velocity will exceed escape speed, if only because it dwindles to an infinitesimal value as your distance from the surface increases. But generally when we talk of escape speed we are speaking in reference either to the ground or to an orbital injection/ejection mode because we use high thrust but (relatively) low specific impulse, thus the altitude gain during thrust is less important, in terms of reduction of escape speed, than the total momentum increase. For a high specific impulse engine, like an ion engine, the velocity is less important than the total kinetic energy gain over the span of time, assuming your “orbit” doesn’t take you back through the surface of the planet you are attempting to escape. But instead of calculating escape speed as a function of altitude, it is more useful (and an easier function to integrate) to look at energy levels with respect to the gravitational frame of reference.
At any rate, Chronos is correct in noting that one not need achieve escape speed at the surface in order to achive orbital ejection velocity. The sky elevator concept to which he refers could raise a payload up above geosynchronous orbit at any speed whatsoever and then release it; the momentum imparted by centrifugal force (or the reaction to the centripetal constraint, if you are one of those blokes who thinks centrifugal force is “imaginary”) will be sufficient to eject the object from orbit. Don’t forget to bring your towel.
Now, all we need are incredibly strong self-damping carbon nanotube molecules to build our beanstock ribbon from. Get cracking, you material science geniuses. 
Stranger
Here’s a tougher question then: Is it possible to slowly climb out of a black hole? Ballistic flight won’t work obviously, but can powered flight work? Pretend the ship is made of unobtanium so we can ignore gravity shearing effects.
If you’re within the event horizon of a black hole you have to exceed light speed to escape. This is a part of the basic definition of a black hole.
Here’s a rather good link on ballistic trajectories and orbits that I totally neglected to insert into my previous post. :o
If your ship is within the event horizon of a singularity there is no way to climb out of the hole. The time-distance you have to travel to reach the limit is infinite; even light, “which travels so fast it takes most civilizations thousands of years to realize it travels at all” redshifts to zero as it reaches the horizon. (Technically speaking, from the outside, you’ll never reach the actual surface of the hole–time contraction will prevent you from actually touching the surface–but you’ll fall through the event horizon from which you can’t transorbit.) It’s helpful to think of a black hole and the surrounding ergosphere not just as an area where gravity is larger but where space is stretched and twisted, causing it to take much longer to get from point A to point B than the path described by a Cartesian line or arc. (You want to change the value of π to 3? There’s the place for it.)
So no, even neglecting all the other tidal effects a severely graduated gravity field would have on your ship, you couldn’t escape from a black hole, no matter how much velocity or energy you have. You can’t even send a message home, unless you can figure out some way to control the reactions that create Hawking radiation at the horizon boundary. This is theoretically impossible, I think…but so is a baby, according to some, so perhaps one day some bright social reject with nothing better to do than play with numbers will figure out a workaround. I wouldn’t go banking the retirement account on it, though.
Stranger