# Escape velocity from Earth

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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I +1ed just to get rid of the downvote. It seems like a perfectly fine misconception to me. –  Dimensio1n0 Jul 6 '13 at 15:53

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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