Answer
The period is $~~93~$ minutes.
Work Step by Step
In part (c), we found that the orbital radius is $~~6.79\times 10^6~m$
We can find the speed:
$K = \frac{1}{2}mv^2 = \frac{GMm}{2r}$
$v^2 = \frac{GM}{r}$
$v = \sqrt{\frac{GM}{r}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}{6.79\times 10^6~m}}$
$v = 7660~m/s$
We can find the period:
$T = \frac{2\pi~r}{v}$
$T = \frac{(2\pi)~(6.79\times 10^6~m)}{7660~m/s}$
$T = 5569.6~s$
$T = 93~min$
The period is $~~93~$ minutes.
