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We know the escape velocity from the Earth is 11.2km/s. Isn't it the velocity required to escape from earth if we go normal to the surface of earth? i.e while we derive the formula for the escape velocity from earth we never consider the slanted motion of the object. So when we launch a rocket we need to use very less value of velocity compare to escape velocity to escape from earth because rocket follows a slanted path so curvature of earth has an effect?

If what I said above is right, then how we can say escape velocity is 11.2 km/s?

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up vote 4 down vote accepted

It is not correct. The meaning of escape velocity is defined the initial kinetic energy in which a particle can go to infinite without going back. That is the kinetic energy have to have the same magnitude as the gravitational potential on Earth given by $mv^2/2=GMm/R$. Since the energy is conserved, it does not matter which direction you are pointing to.

For the rocket, it has no initial KE and it gains KE and PE by consuming its fuel. The reason that a rocket move straight up is to reduce air friction at the beginning. Then it follows a slanted path later is to increase flight time so a only a lower efficiency engine is required.

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The 11.2 km/s is about a generic body that leaves the surface of the earth with that initial velocity, that initial velocity causes it to overcome earth's gravitational pull and escape. If there is a propulsion system (rockets,various other engine types...) then the initial velocity does not need be 11.2 km/s, since earth's gravitational potential is overcome gradually as the rocket does work after launch. But from the law of conservation of energy both launch types will do the same amount of work i.e. equivalent to the gravitational potential of the earth. Just my two pence anyway.

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Of course not. What escape velocity really means that you've to spend sufficient amount of energy (thereby do some work) to escape a massive object which exerts the gravitational force on you. Whatever you do or wherever you go, when you try to escape Earth, you are doing work. Hereby, energy (potential & kinetic) should be conserved. That's all.

If you launch a rocket, the rate of its work done maybe low. But, its doing it continuously. So in all the case, the total energy would still remain constant. Practically, the maximum velocity achieved by rockets are barely some 50,000 km/hr.

If you build a long ladder to somewhere (consider an object in GEO) and you start climbing resting sometime along the path, you are indeed spending energy along your way to the GEO object.

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I believe I've heard Neil deGrasse Tyson say its roughly 17 miles per second to escape Earth's gravitational force in a spacecraft like the Apollo rockets that were used.

So I'm assuming we're basically talking about how fast something has to be accelerating to escape Earth's mass/gravity.

I believe that is correct - 17 miles per second.

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The question asks if going on an angle changes the escape velocity, it is not asking for the numeric value (which it gives in the question itself). – Kyle Kanos Mar 14 '15 at 1:40

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