Comparing the work required for a mass to escape the Earth's gravity to the necessary initial kinetic energy gives us the escape velocity from the surface of the Earth of around $11\;\mathrm{km\cdot s^{-1}}$.
But assuming no air friction and hence no loss in energy on the axis perpendicular to gravity it shouldn't matter what angle the projectile is projected at.
Is this true?
If there were no air and we fired a projectile at $1^\circ$ to the horizontal at a velocity of $11\;\mathrm{km\cdot s^{-1}}$ should it still escape the Earth's gravity despite such a low ( $11000\sin(01^\circ)\;\mathrm{m\cdot s^{-1}}$) vertical velocity?
Could it be that firing a projectile horizontally at the escape velocity in a frictionless medium would result in a perfect orbit?
Also: Taking escape velocity of $11200\;\mathrm{m\cdot s^{-1}}$ and radius af Earth of $6371000\;\mathrm{m}$ and $g=9.80665\;\mathrm{m\cdot s^{-2}}$. On a flat Earth if you fired an object horizontally at $112000\;\mathrm{m\cdot s^{-1}}$ it would fall $0.039\;\mathrm{m}$ after $1000\;\mathrm{m}$. But using trigonometry the Earth would have curved away by $0.078\;\mathrm{m}$. Twice the value necessary for an orbit. Can someone check this?