Projectiles and escape velocity 
Q: The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body is projected at an angle of $45^\circ$ with vertical, the escape velocity will be?

My Approach:
The new vertical velocity will be $u * \sin(45^\circ)$ = $u /\sqrt2$
Calculating expression for escape velocity:
$$\frac{1}{2} mv^2 = \frac{GMm}{r}$$
$$\frac{1}{4} mu^2 = \frac{GMm}{r}$$
$$u = 2\sqrt\frac{GMm}{r}$$
Hence the new escape velocity = $\sqrt2 * 11$km/s.
However, the Correct Answer is 11km/s.
 A: Well, this certainly is an evil trick to play on first year students!
Escape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape "velocity" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from kinetic energy. No direction is presumed, and so it applies any direction.
Hence, v = 11km/s, projected vertically or otherwise.
A: Escape velocity is not dependent on direction. It really should be called "escape speed"
It's the result of the calculation "kinetic energy of object at launch"="potential energy change on leaving Earth's influence"
Kinetic energy isn't a vector. It is $\frac12 m\vec v \cdot \vec v=\frac12 m|\vec v|^2$; a scalar quantity. Note that it is independsnt of the direction of the velocity, only the vector.
Remember, if you throw a ball at 45 degrees, the Earth's gravity will attract it, but it will also speed it up. These two effects "cancel out"
Fun fact: If there was a tunnel through the Earth, and you threw a ball at escape velocity down the tunnel, it would still escape Earth's influence after passing through the tunnel.
A: The (relative) escape velocity of a mass $M$ with respect to a mass $m$ is:
$$ V_{rel} = \sqrt{\frac{2 G(M\!+\!m)}{r} } $$where $r$ is the relative distance of the two bodies.
The escape velocity of a mass $m$ with respect to the inertial center of mass of the two bodies is:
$$ V_m = \sqrt{ \frac{2 G M M}{ r\, (M\!+\!m)} } $$
and the escape velocity of the mass $M$ with respect to the inertial center of mass is:
$$ V_M = \sqrt{ \frac{2 G m m}{ r\, (M\!+\!m)} } $$
The relative velocity is the sum of the two inertial velocities: $ V_{rel} = V_M + V_m $. 
