# Does the definition of escape velocity apply strictly to an initial velocity of a ballistic projectile?

In Shivam Sarodia's answer to Escape Velocity question he states:

"For a rocket launched exactly at escape velocity"

This seems to be a repeated, and unclear way of discussing the matter of escape velocity since it doesn't say when or how long escape velocity is occurring for the rocket.

1. Does the escape velocity only need to occur momentarily and then thrust can be removed ( so therefore an initial velocity to a ballistic trajectory)

2. Or does the velocity need to be maintained for some period of time before thrust is cut?

3. Or are there other details/conditions that need to be considered?

I suspect 1. above is the correct interpretation, and that if I write the equations of motion I can demonstrate that's so in simulation. But it seems so many text books and discussions on the subject do not really make it clear the circumstances surrounding the definition of escape velocity. Even Feynman's lectures are not clear on this point - but they do have other interesting aspects.

• It's the Newton cannonball idea. Fire a cannonball and it falls to the earth after some distance. Fire a cannonball fast enough and it can escape the Earth and enter into orbit. Granted, air resistance changes the equation somewhat. Also, a sudden cannonball style launch at 11.2 KM per second would tend to break any equipment and the temperature would grow white hot flying through the air at that velocity, so it's much more practical to launch rockets over a few minutes not all at once. (that's probably kinda obvious). – userLTK Aug 15 '15 at 9:34
• Physics instructors/teachers and textbooks really don't make this point clear. The 'escape' velocity commonly taught applies to an initial velocity of a ballistic trajectory. Perhaps they should equally illustrate the fact that one can escape the Earth's gravity at 3 mph if they wish using continuous thrust - and assuming enough fuel to complete. – docscience Aug 15 '15 at 14:19

Yes, it's an initial velocity in a ballistic trajectory in a vacuum. The escape velocity is derived using conservation of energy at two different points in time, as $\frac {1}{2} mv_0^2-\frac {G M m}{r_0}>0$. $v_0$ means initial velocity.

In practice you haveto worry about the atmosphere and all that.

From the Wikipedia article "Escape velocity":

Defined a little more formally, "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity with a residual velocity of zero, with all speeds and velocities measured with respect to the field.

Additionally, the escape velocity at a point in space is equal to the speed that an object would have if it started at rest from an infinite distance and was pulled by gravity to that point.

Now, think a little bit about that last sentence.

By symmetry, a (freely falling) object with escape velocity at one point has escape velocity at every other point. So, if a rocket thrusts 'upward' until it has escape velocity and then terminates thrust, the rocket's (decreasing) velocity afterwards is, at any point, the escape velocity at that point.

In this sense, the rocket 'maintains' escape velocity from the moment the thrust is terminated.

(The above assumes the typical ideal context).

The earth is at the bottom of a gravity well. The amount of kinetic energy you need to escape that well depends on how far down you are in it.

At the surface of the earth, or rather in low earth orbit above the atmosphere, escape velocity is about 25,000 mph. That means if the rocket accelerates to that speed, it will go up out of the well, and eventually escape the well altogether. However, when it is partway up, it will have slowed down, but that is still escape velocity because it is no longer so close to the earth, it is farther away.

When it finally climbs out of the well, it can actually be moving very slowly, but if it is far enough away from the earth that the earth's weaker gravity cannot stop it, then it is still moving at escape velocity.

If it is launched with any speed in excess of escape velocity, then when it leaves the gravity well, it will have that excess velocity.

To put it another way, do it in reverse: place a ball motionless in space with respect to the earth at some good distance away, like 10 million miles. It will start being slowly attracted toward the earth, and as it gets closer it will pick up speed, until when it drops into the atmosphere it is going about 25,000 mph. So it is going at escape velocity (but down, not up), not only at the bottom, but at every step along the way, because if it hit a perfect trampoline anywhere along the way, it would bounce right back out of the gravity well.

It would be perfectly possible to reach a distance from the Earth where the sum of an object's (space ship) kinetic energy and its gravitational potential energy is equal or larger to zero, with a continuously powered (thrusted) vehicle, as long as during flight the vehicle’s thrust always overcomes the gravitational attraction exerted by the Earth.

But for a ballistic object that has no thrust of its own and is launched at a certain initial speed from the Earth, that initial speed needs to exceed the escape velocity (with the caveats NeuroFuzzy mentioned re. air drag)