Answer
The kinetic energy will be $~~1.3\times 10^8~J$
Work Step by Step
We can use conservation of energy to find the kinetic energy at a distance of $4.0\times 10^6~m$:
$U_2+K_2 = U_1+K_1$
$U_2+K_2 = E_1$
$K_2 = E_1-U_2$
$K_2 = (5.0\times 10^7~J)+\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.0\times 10^{23}~kg)(10~kg)}{4.0\times 10^6~m}$
$K_2 = (5.0\times 10^7~J)+(8.34\times 10^7~J)$
$K_2 = 1.3\times 10^8~J$
The kinetic energy will be $~~1.3\times 10^8~J$
