Explain Escape Velocity

I understand the math but not the logic of escape velocity.

It seems to me that an object traveling upwards one mile an hour will leave Earth, albeit slowly, as long as it is moving forward, er, upwards.

That 12 km/sec speed everyone says is escape velocity will only shorten the trip to outer space.

But how does that object continue to slowly move away from the earth?

An object can’t just hang there over the ground. Gravity pulls the object down, so there has to be some force keeping the object up. Just to keep still in the air requires an enormous amount of energy.

Escape velocity is just a velocity that is so great that even with gravity constantly pulling the object back to the planet, it will never return. Note that objects in orbit are traveling at less than escape velocity.

It isn’t neccesary to get that escape velocity in one big push, but the faster you burn your fuel the more efficient that fuel use is.

You are correct. An object will excape Earth’s gravity as long as it continues to move forward. The problem is that it will have to continue burning fuel while it is “hovering”.

The logic is that all objects exert a gravitational force relative to it’s size and the distance of the object it is acting on. The furthur away from the Earth you are, the weaker the gravity.

Gravity acts as a force, causing objects to accelerate (F=ma). That means our rocket must continue to accelerate against gravity in order to maintain it’s constant upward velocity.

Or another way to think of it:
If I fire a cannon straight up, it leaves the barrel with a particular upward velocity. Gravity will cause the object to decelerate at 32 f/s/s (9.8m/s/s) until it eventually slows to a velocity of zero and begins accelerating back towards the earth.

If I fire that projectile at 12 km/s, it won’t decelerate to zero velocity until it has passed a point where gravity no longer has an effect

Escape velocity* is the speed you need to throw an object at, so it doesn’t need any additional push to escape the earth’s gravity pull.

If you throw a ball upwards at 1mph, it will fly up about 1/2 inch before falling down. If you shoot it at 100mph, it’ll go up maybe 300 ft before coming back down. If you shoot it at 12km/s, it’ll keep on going and never come back. (Ignoring atmospheric friction)

*Technically it’s incorrect to refer to it as “escape velocity” - it’s really "escape speed because the direction doesn’t matter, as long as you don’t aim at the ground.

Right. Escape velocity is a ballistic speed. It applies to objects whose propulsion ends after the beginning of its journey. All rockets are like this. They need 17,000 mph to go into orbit and not drop back to earth and 25,000 mph to leave earth orbit.

The space elevator is an example of the propelled object. One can travel up the elevator away from earth at any speed and just drop into an orbit because force is being applied to the object at every moment, not just at the beginning.

Wiki’s explanation covers this.

Also, escape velocity depends on where you are. The 12km/s number is escape velocity from the ground (or from low earth orbit, which really is very low.) Escape velocity from, say, geosynchronous orbit is much lower.

So if you have a rocket with a limitless fuel supply, you can set your autopilot to climb at a rate of 1mph. Eventually you’ll reach an altitude where escape velocity is less than 1mph, at which point you can cut off the engine and coast away.

Most significantly, I think - there is nothing magical about the escape velocity as a desired speed for rockets to leave earth. The ideal speed profile would depend on where you depart from and in what direction, and on your vehicle’s aerodynamic resistance and lift while moving through the atmosphere, and on the ability of your crew and payload to withstand force and vibration, and on the thrust you generate at different fuel consumption rates, and on how you want to be moving once you are fairly free of Earth (or of the Sun for that matter).

I do not believe this is correct. If a portion of your velocity is not directed away from the center of the earth, that portion will contribute to a resultant orbit, but I am not sure if it can be taken as a part of escape velocty. I will have to do the math, but I think that if you have the exact magnitude of escape velocity pointed at some angle away from the gravity vector, then you will just end up with an extremely eccentric orbit.

No, it’s just the speed. Think of conservation of energy, and balance the kenetic energy at the “escape speed” with the potential energy at the place where the object is located (ie, how much potential energy you’d lose if you move the object to infinity):

.5mv^2 = GMm/r, where G = gravitational constant, M = mass of earth (or other large body), m = mass of the object, and r = distance from center of m to center of M.

Close, but not quite.

If you got on your magic “Harry Potter Special Edition” flying broomstick, and aimed at the moon, and flew away at 50 kph, you would get there in (385,000 / 50 =) 7700 hours, or almost 321 days. (Eeeks! I better get the “Collectors Edition”, instead.) Your escape velocity was 50kph, applied over as much time as you needed.

You dont need to get to 11.2km/sec. That number is needed if you wanted to apply all of the thrust in one moment, like a cannon shot.

(I know Wiki is a little weak sometimes, but I think it’s safe enough to use, here.)

No. The term “escape velocity” is not defined that way. It’s defined assuming you achieve that velocity all at once. Note that your own cite says “without propulsion”. You can’t maintain a constant velocity going away from earth w/o propulsion (the broom, in your case).

…or that you achieve that velocity at some point in the trip. A rocket will be moving away from earth under propulsion until it reaches the escape velocity of whatever position it’s at when it achieves that velocity. So, if you propelled it to the moon, the escape velocity would be different, but that’s because “r” at the moon is different than “r” at the surface of the earth. The eqation is always valid, but “r” might change.

Ah. Ok. Sorry. I always get hung up when someone says you need to reach 11.2 k/sec to get to orbit. But that is from a perspective of starting at the surface, and applying the force all at once.

I don’t think that the US Space Shuttle gets up to 24,000 mph on launching, does it?(Because I think the acceleration time is several minutes, as opposed to an instant “boom”.)

See, my broom is magic. So it provides the propulsion as I need it, without fuel concerns. Top speed is 50 kph in any direction. 15 kph cruising.

But even when you move out to infinity you will have a component of your velocity that is not aligned with gravity (pointing at the center of the earth), thus all of your kinetic energy will not have been converted to potential energy.

And I just rederived that equation before checking back in here. Now I have to convert it to cylindrical coordinates and resolve to try and prove or disprove my point.

OK, damnit, resolving vectors in my head I see that it is likely that I am wrong due the oddities of taking something to infinity. I will check myself to be sure.

No, it doesn’t-- you can see it almost inch off the takeoff platform. The point is that “escape velocity” is going to be lower the father you are from earth. But if at some point you turn off your engines and you’re not travelling at or greater than the escape velocity for that point, you will get pullled back to earth. If you never turn off your engines, you’re not dealing with “escape velocity” at all.

You lost me. The equations assume idealized conditions, and you could never get “infinitely far away” anyway.

It turns out you are right due to my poor appreciation for infinity. Always have to be careful about that. A simple analysis of angular momentum shows that as object gets further away from the mass it is escaping, the component of velocity perpendicular to the pull of gravity drops such the scalar product of the distance and that speed is constant. Thus as we approach an infinitely large distance we have an infinitely small velocity and the only kinetic energy we have to consider is aligned with the force of gravity. I concede.

The space shuttle can’t get to 24000 mph at all, it can only go about 17000 mph, which is the speed required for Low Earth Orbit.

This is actually correct, if by “extremely eccentric” you mean “eccentricity equal to 1”. Of course, a conic section with eccentricity greater than 1 isn’t an ellipse any more, it’s a parabola, which, in fact, never returns to Earth. If your speed is greater than escape speed, meanwhile, your eccentricity will be greater than 1, and your orbit will be a hyperbola.

More on escape speed.

The Shuttle is not designed for deep-space travel. It only goes to low earth orbit, so it never reaches escape velocity.

The Shuttle does need to reach a speed of about 8km/s. It doesn’t matter how quickly it gets up to that speed, but it does need to get there. Otherwise, once the engines are shut off, the Shuttle in an elliptical orbit that intersects the earth’s atmosphere. If it achieves 8km/s, it’s in an almost circular orbit, and the entire orbit is above the atmosphere.