#### Answer

The orbital period is 4.2 hours.

#### Work Step by Step

Let $M$ be the mass of the earth. We can use the satellite's orbital speed to find the radius of the satellite's orbit.
$v = \sqrt{\frac{G~M}{R}}$
$R = \frac{G~M}{v^2}$
$R = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{(5500~m/s)^2}$
$R = 1.32\times 10^7~m$
We then find the time $T$ of one orbit.
$T = \frac{distance}{speed}$
$T = \frac{2\pi~R}{v}$
$T = \frac{(2\pi)(1.32\times 10^7~m)}{5500~m/s}$
$T = 15,080~s = 4.2~hours$
The orbital period is 4.2 hours.