Answer
The final answer is $7.9*10^3 m/s$.
Work Step by Step
We find the answer using the work-energy theorem :
The increase in the letter's kinetic energy would be equal to the work done by Earth's gravity, thus we get :
$\frac{1}{2}mv^2 = \int_R^0 F dr$
where R is Earth's radius, moreover, v the final speed of the mail :
$\int_R^0Fdr = \int_R^0 (-Kr)dr = \frac{1}{2} KR^2$
Thus : $\frac{1}{2} mv^2 = \frac{1}{2} KR^2$
Which leads us to the following expression of v :
$v = R\sqrt{\frac{K}{m}}$
The only thing left for us to do is to find an expression of K, which we will find using the acceleration of gravity (which is g onn the surface of Earth) given by :
$a_g = \frac{GM}{R^2} = \frac{G \frac{4\pi }{3} R^3 \rho }{R^2}$, where $\rho$ is Earth's average density.
This permits us to write the following :
$\frac{K}{m} = \frac{4 \pi G \rho}{3} = \frac{g}{R}$
Consequently : $v = \sqrt{gR} = \sqrt{9.8 \times 6.37 \times 10^6} = 7.9 \times 10^3 m/s$
