#### Answer

The speed of the satellite is 1.44 km/s and the altitude above the surface of Mars is $1.71\times 10^7~m$

#### Work Step by Step

We can convert the orbital period $T$ to units of seconds.
$T = (24.8~hours)(3600~s/hr)$
$T = 89,280~s$
We can use the orbital period and the mass of Mars $M_m$ to find the orbital radius $R$.
$T^2 = \frac{4\pi^2~R^3}{G~M_m}$
$R^3 = \frac{G~M_m~T^2}{4\pi^2}$
$R = (\frac{G~M_m~T^2}{4\pi^2})^{1/3}$
$R = (\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(6.42\times 10^{23}~kg)(89,280~s)^2}{4\pi^2})^{1/3}$
$R = 2.05\times 10^7~m$
We can use the radius and the orbital period to find the speed of the satellite.
$v = \frac{distance}{time}$
$v = \frac{2\pi~R}{T}$
$v = \frac{(2\pi)(2.05\times 10^7~m)}{89,280~s}$
$v = 1443~m/s \approx 1.44~km/s$
The radius of the orbit is $2.05\times 10^5~m$. We can use the radius of Mars $R_m$ to find the satellite’s altitude $h$ above the surface of Mars.
$R_m+h = R$
$h = R - R_m$
$h = (2.05\times 10^7~m) - (3.39\times 10^6~m)$
$h = 1.71\times 10^7~m$
The speed of the satellite is 1.44 km/s and the altitude above the surface of Mars is $1.71\times 10^7~m$