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?
Best Answer
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.