Answer
$K = 4.6\times 10^5~J$
Work Step by Step
We can find the speed:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
$v^2 = \frac{GM}{r}$
$v = \sqrt{\frac{GM}{r}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}{(6.37\times 10^6~m)+(5.0\times 10^5~m)}}$
$v = 7620~m/s$
Note that the satellite and pellet both move with this speed but in opposite directions. Thus the speed of the pellet relative to the satellite is $~~15,240~m/s$
We can find the kinetic energy of the pellet in the reference frame of the satellite:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(0.0040~kg)(15,240~m/s)^2$
$K = 4.6\times 10^5~J$
