Answer
$v = 5.97\times 10^3~m/s$
Work Step by Step
Let's use 6380 km as the radius of the Earth. Then the satellite's orbital radius $r$ is 6380 km + 4800 km which is 11,180 km.
The force of gravity provides the centripetal force to keep the satellite moving in a circle:
$\frac{m_s~v^2}{r} = \frac{G~m_E~m_s}{r^2}$
$v^2 = \frac{G~m_E}{r}$
$v= \sqrt{\frac{G~m_E}{r}} = \sqrt{\frac{(6.67\times 10^{-11}~N\cdot m^2/kg^2)(5.98\times 10^{24}~kg)}{(1.118\times 10^7~m)}}$
$v = 5.97\times 10^3~m/s$
