ICBMs are ballistic because they only fire part way along their trajectory in what is called the boost phase. While they are still on the way up, they expend all motors and coast on a ballistic track (with minor guidance adjustments). Strictly speaking, it is actually the payload (the individual RVs and the RV bus) that are ballistic. They don’t maintain acceleration along their entire track, like a SAM (surface to air) or AMRAM air to air missile does.
When a fighter jet goes “ballistic” (such as the MiG-25 attempted intercepts of the U-2 and TR-1 spy planes) it basically shoots up at maximum ascent angle and rides its afterburners up until the atmosphere is so thin that the control surfaces don’t offer any control. It is then (until it returns to thicker atmosphere) on a pure balllistic track (the afterburner is turned off so as not to destabilize the plane). U-2 pilots on overflights of Cuba durning and after the Cuban Missile Crisis reported MiG-25s doing this; they’d just slide over a few hundred meters and watch the fighters fly right on past, unable to direct fire or intercept.
kanicbird, you need to stop anthropomorphizing what is actually an abstract physical problem. Usually such a substitution aids in visualizing the process but this time it seems to have hindered understanding.
All we are talking about is a continually powered object. It does not have to have any earthly analogy, whether of an elevator, rocket, or large guy blowing smoke out of his ass. The contrast is to a ballistic object, i.e. one whose propulsive mechanism stops after a given time. As said, this is what makes an ICBM ballistic and not conceputally different from a cannonball. You’ve been implicitly slipping in the condition that the power will stop at some point in the journey, even after being corrected.
But the key word is “continually.” In a thought experiment physical constraints do not apply. This continually powered object will not turn off its power at any time. This makes it on the most fundamental level a different type of physical object than one that has been launched ballistically.
To go back to the beginning. Escape velocity is a concept that applies only to ballistic objects. A continually powered object will duplicate any journey that an object that has exceeded the escape velocity will make. In that sense, both objects can be said to have escaped the earth no matter that their speeds differ from one another.
Anyway let’s take this to another tack, and at lease clear up a few items:
Using the ‘more flexable’ definitions, and using one of those advanced cannonballs that can fire though solid (liq,gas) objects w/o drag. every point in the earth only universe except one is a potential orbit point (point which an orbit could pass through. The orbit could be 2 inches from the center or a google lightyear. The only point that is not included it the center itself. So escape speed applies to every point at every altitude (except for altitude = 0 from center of mass)
So I think it’s clear that escape speed applies everywhere.
I don’t see how the fact that a object can provide it’s own motion in the future has anything to do with it not escaping the earth at this moment of time.
Also a powered object that has exceeded the escape speed can be recaptured by use of it’s own power.
The difference between powered and non-powered is that a powered object CAN change from escaped to captured by itself.
OK, how about this? Suppose we put a whopping big electric charge on the Earth, and then put a charge of the same sign on our cannonball. Our cannonball now has an electrostatic force on it, which will partly counteract the gravitational force. If we now fire our cannon at just the right speed (which will be slower than escape speed), our cannonball will never return to the Earth, but the electrostatic force will die off at just the right rate that it will never exceed escape speed, either. There’s no question here of the cannonball using any power, and there’s no way to turn it off. Can one say that this cannonball has “escaped”?
more on the charged cannonball, if the forced can be balanced between g and electric charge the ball will not decelerate at the same rate as a non-charged ball. It will eventually exceed escape speed.
OTOH a neg charged ball can be launched at greater then escape speed and eventually slow down and be recaptured.
Of course the charged cannonball is pushing off of the Earth. Why is this a problem? The only reason I chose an electrostatic force is that it requires no power and cannot be turned off.
And so long as the electrostatic force is less than the gravitational force, there will exist a speed at which you can launch your ball so that it will not ever exceed escape speed. The adjustment of the forces to keep going but still stay slow will be automatic, from the nature of the electrostatic force. It will not decelerate at the same rate as the non-charged ball, but it will still decelerate.
Escape velocity is really just a formalism which stems from the following.
At any distance from the Earth, you have an amount of potential energy. This potential energy is decided to be zero when you are at an infinite distance from Earth, and decreases as you get closer, implying that a positive amount of work must be done to push you back up to zero potential energy i.e. you have to do work against the gravitational field. The amount of work you need to do i.e. energy you need to get back to zero is equal to the value of your potential energy.
Now, if you were to give an object an instantaneous velocity on the surface of the Earth, it will instantaneously acquire a certain amount of kinetic energy. If the kinetic energy associated with this velocity is equal to the object’s potential energy, it is said to have “escape velocity”.
The reason this is a defined velocity (something like 25,000mph - I can’t remember!) is because the potential energy of everything on the Earth’s surface is proportional to its mass AND in general the kinetic energy of anything is also proportional to its mass, hence the mass cancels in the equations.
Your “escape velocity” only ever depends on your distance from the Earth’s centre, and all this talk of “changing” escape velocity is rather useless. The point is, all you ever need to reach infinity is to have more kinetic energy than your potential energy.
1st I want to point out that when I started this thread, I did so as a nitpick. But now it seems like my OP was the case most of the time and you (plural) are scrambling for the nitpicks, coming up with unrealistic or even impossible examples.
OK enough crowing.
We have the same problem with your charged cannonball as the elevator, it is pushing (and thereby supported by) the earth. And object that needs the support of the earth can’t be said to escape from it.
Also lets put another nail in the rocket coffin. If you have a powered rocket that is going to infinity under power at a speed that is less then escape speed, the rocket is loosing mass, I can ignore it running out of propellant, but not that it is now a different object, containing less mass then before the burn.
As I’m reading this, it occurs to me that the language itself is hindering understanding.
When you’re talking about “escape velocity”, infinite distance cannot be reached, so an object will always be within Earth’s gravitational field, to some minor extent. This means there is no “escaped” condition. There is only “escaping” or “not escaping”. Yet obviously there have been a number of orbital spacecraft and interplanetary spacecraft, so what’s the right way to describe their ascent?
Here’s how I figure it:
Objects may be in 1 of 2 states:
** Escaping ** - object has sufficient velocity that it will forever continue increasing distance from “Earth” without adding thrust (unless acted on by another force).
Not escaping - object may continue leaving earth, but must have continuous thrust added (either intrinsically or extrinsically). If thrust is removed, object will either enter orbit or eventually fall back to earth. Objects in the “not escaping” category may fall into sub-categories of “powered climbing”, “orbiting”, or “falling.”
I think interplanetary or interstellar vehicles must reach escape velocity at some point if they are intended to reach their destinations. ICBM’s need not reach escape velocity since they are intended to return to earth. Orbital probes need not reach escape velocity because they intended to reach a certain distance from Earth and stay there. Lunar probes need not reach escape velocity because they will be picked up by the gravitational field of their target.
I rented the earth only universe (on pg 1006 of the theoretical physics supply catalog), allowed the construction of a space elevator to infinity (and most likely lost my security deposit on that one), allowed infinite power supplies, charged planets and cannonballs.
What part of this thought experiment am I not understanding?
Perhaps you object to me wanting to stay with the 4 known forces, do you want to consider space-time warp drives, trans-dimentionnial displacement devices (so called TDDD’s), transporters or stargates?
Perhaps you don’t like me saying that the ship is a different object, but I see that no different then using a exploding frictionless cannonball fired at less then escape speed then after exploding a fragment exceeds escape speed. The cannonball never did, the fragment does exceed E.S…
When you introduce another gravitational body, the object can become captured by that object which may or may not excede E.S. of the earth. Objects on the moon have not escaped from the earth as the moon is captured by the earth. Objects on Mars have escaped the earth, but have not escaped the sun, nor the solar system which can also be considered a gravitational body.
As somebody pointed out, I am indeed a stranger. Though not on a train. What do all you gentlemen do for a living? And what exactly is the nature of your discussions? I am intrigued.
Just as Einstein conceptualized relativity by a thought experiment of what he would experience traveling on a train approaching the speed of light, we can discuss an ideal self-powered object without worrying about whether it has enough fuel or is losing mass. As I said earlier, thought experiments are not constrained by physical reality. They reduce the universe to ideal systems to illustrate the basic physical principles that govern it.
So we can create a thought experiment involving a universe with three objects: the earth, an ideal ballistically traveling object and an ideal powered object. In such a universe both objects have the same property of traveling away from the earth forever, never to return. Both can therefore be said to have escaped the earth.
This is the same principle that allows us to conceptualize gravity without worrying about terminal velocity in an atmosphere. We need to get rid of the factors that confuse the underlying physics and confound the observer to get at the basic truth.
So, no cannonballs, no rockets, no elevators, nothing that complicates the physics. Just ideal objects illustrating the difference between ballistic flight and endlessly powered flight.
Worzo, I’m a physics graduate student, currently (hopefully) working towards my PhD, with a side business of writing Staff Reports, for which I have thus far been paid the grand total of three calendars and a coffee mug. The nature of our discussions is as you see here; we like to nitpick, hyperanalze, and take apart into many pieces, ideas ranging from physics to philosophy to literature, among others (though not all in this forum; we have different places for different types of discussion). Your comment was both correct and in tune with the tenor of the discussion; Stranger on a Train was only correcting your statement that you were the first to make that comment.
kanicbird, the fuel problem was one of the reasons I switched to the charged cannonball. It may be impractical, but it is absolutely physically possible, and one need never worry about it running out of fuel. Yes, there will always be an electrostatic force between the cannonball and the Earth, but then, there will also always be a larger gravitational force between the cannonball and the Earth. Why does the electrostatic force disqualify an object from being “escaped”, while the gravitational force does not?
Wow, I didn’t know it paid so well! Where do I apply? Oh, Una Persson already has the mechanical engineering gig locked up.
Instead of a cannonball with an electrostatic repulsion, how about a ball made out of Cavorite (or a solar-powered flubber engine), which has a negative gravitational attraction. It will automatically be in an escape trajectory even if its initial speed is zero, yes? Ditto for Chrono’s electrorepulsive cannonball, so long as the repulsive force of the EM field overcomes the attractive force of gravity.
As Exapno Mapcase points out, “escape speed” is really only applicable to a purely ballistic (unpowered) object; it assumes that additional delta-V (change in velocity) isn’t available except by the gravitational field it is in. A powered vessel that can lift itself under gravity indefinitely is already “escaping”, regardless of speed. We use powered vessels (rockets) to launch stuff into an escape trajectory, of course, because a catapult or gun applying the kind of acceleration required to (almost) instantaneously achieve escape speed would destroy whatever it was launching, especially if it was manned, but even rockets only fire for a short part of their trajectory, after which they become ballistic objects. With the Apollo moon launches, for instance, the Saturn V launched the Command Service Module into a (slightly) sub-escape orbit, but only fired for about 900 seconds (922 seconds if you add up all the stage design burn times) out of a 7-10 day mission. For the vast majority of its transit it is a purely ballistic object with Sir Issac Newton in the driver’s seat (nods to James Lovell), and whether it escapes or not is purely dependant upon how much kinetic energy it has relatative to its gravitational potential.
No, there is no “escaped” or “captured” state. The gravitational field of another body adds force to the object. Thus, Apollo 13 could reach the moon without ever having reached Earth’s escape velocity.
Again… there is no “escaped.” Why would objects on Mars be subject do different physics from object on Luna? They’re both in the same universe. To be specific, objects on both Mars and Luna rest on bodies that are large enough so that at their surfaces, the effects of other bodies are captured.
There is no “escaped” or “not escaped.” There is only escaping and not escaping. Objects that are not escaping are either powered climbing, orbiting, falling, or at rest (Newtonically speaking).
Gar. In my old age, the fingers are outrunning me brain, and they’re dyslexic phoenetic spellers.
In previous post, “the effects of other bodies are captured” should read “the effects of other bodies are canceled.” Even though they’re not completely canceled, just masked to the point of becoming trivial.
