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It's been a while since I studied physics so I was wondering if anyone could clear up my understanding of escape velocity.

I'm being told that the escape velocity is a the same for powered craft and non powered craft. This makes no sense to me. I'm also being told its a constant for any given planet.

The only thing that I'm starting to think is that somewhere there must be a formal definition of what escape velocity is and it applies only within the domain of the definition.

For example my hyperthetical craft has a power source that lasts for years and more than enough thrust to lift its own weight 10 times over. Given that senario I can run my engines and maintain 20km/hr velocity straight up. After x amount of time I would have passed the point where my 20km/hr velocity is enough to keep me from falling back even if I shut the engines down.

So whats my escape velocity?

By my reasoning it's the velocity at any given altitude that an un powered object must have in order to not fall back.

My friends disagree and state it's the velocity an object must atain to not fall back which is 11km/sec for Earth no ifs or buts. I think they said 11km/sec but can't be quite sure of that figure. Again this just makes no sense to me.



Anyone have the formal definition or at least a better explanation?

Regards,

The formal physics definition of escape velocity is:

V = SQRT(2GM/R)

Where G is 6.672E-11 (the Gravitational constant), M is the mass of the planet in kg, and R is the radius of the planet in metres. V is therefore in metres per second.

Anything launched from the surface at a greater velocity than this value will not fall back to the ground - it'll escape the planet's gravitational field. It doesn't matter if the object is powered or unpowered - this is a property of the planet itself, not of the object.

See http://www.physlink.com/Education/AskExperts/ae158.cfm for more info.

"Escape velocity" is really a slightly confusing way of summarizing the energy required to lift a body from its present radial distance to an infinite radial distance - if all that energy were imparted in one go, the resulting velocity would be the escape velocity calculated from Dr G's fomula. If you're already some distance from the planet, escape velocity is lower - you just plug in your distance from the centre of the planet as the R term in the formula.
So what you're doing is sneaking up on escape velocity incrementally - your spacecraft is capable of accelerating at more than 1g (otherwise it wouldn't be able to rise from the Earth's surface). As it crawls into the sky it gradually migrates towards regions of lower gravity. Even if you rein back your drive to maintain a steady 20km/hr, eventually you'll reach a distance from the planet at which escape velocity is 20km/hr - only at that point can you say you have escaped from the planet (because this is the first point at which, if you turn off your drive, you won't fall back to the planet), and you've done it by virtue of the large amount of energy you've already expended getting that far from the planet.
I think the idea of escape velocity is one of the more confusing ideas in celestial mechanics, and much confusion would have been avoided if only we spoke about "escape energy".

Grant

mrzee wrote:I'm being told that the escape velocity is a the same for powered craft and non powered craft.

forget it. escape velocity is of course a starting velocity. you need to throw a stone this fast.

Thank's for the reply guy's.
The formula and Grant's excellent description has cleared things up. I must agree, it would make more sense to refer to the escape energy rather than velocity simply from a clarity point of view.
Given the above formula it's interesting to note that an object could be "doomed" to fall back at some pretty far distances.
I guess thats how objects are caught into orbits. They don't have enough energy/velocity to break free.

Looking at the formula and the supplied link (which states it also) this does not include frictional losses due to atmosphere. This implies that the "real" escape velcity would change with shapes,densities, current weather conditions (high pressure/low pressure) etc.

Food for thought.
Thanks,

mrzee wrote:
Looking at the formula and the supplied link (which states it also) this does not include frictional losses due to atmosphere. This implies that the "real" escape velcity would change with shapes,densities, current weather conditions (high pressure/low pressure) etc.

Food for thought.
Thanks,

Not so much the real escape velocity, but the energy needed to achieve it.