homework and exercises - 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.

Answer

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.