Escape velocity at an angle 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?
 A: Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take.
Vague intuition: the vertical speed you estimate is about $60 \frac{m}{s} $ which would be nullified by the gravitational acceleration in $\sim 6 \,s$. But in the meantime the ship would move horizontally by almost $65 \, km$ which is far over the horizon. There the acceleration would be more collinear with the velocity of the ship, so it will have much easier time to escape. 
In search of inspiration, I found this instructive online simulation

Regarding your second question about the orbit. Actually, no the escape velocity will result in escape. To get to the lowest possible orbit (at a distance $\sim 6400 \, km$ from the center, you need only $7.9 \, \frac{km}{s}$ and increasing speed will result in a higher elliptical orbit.
A: In newtonian mechanics the angle does not matter, but in relativity it does.
For example:
An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically.
When you launch it at a distance just above 2GM/c² (the so called Schwarzschild radius) from the center of mass it will escape if launched radially, but fall in if launched horizontally.
For more details on the math see https://physics.stackexchange.com/q/262104
