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?