Answer
$T_0 = 5.54\times 10^3~s$
Work Step by Step
We can find the period of the orbits:
$T_0^2 = \frac{4\pi^2~r^3}{GM}$
$T_0 = \sqrt{\frac{4\pi^2~r^3}{GM}}$
$T_0 = \sqrt{\frac{(4\pi^2)(6.37\times 10^6~m+4.0\times 10^5~m)^3}{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}}$
$T_0 = 5.54\times 10^3~s$
