Answer
(a) The satellite is a height of $10,400~km$ above the Earth's surface.
(b) The satellite's acceleration is $1.42~m/s^2$
Work Step by Step
(a) We can find use the expression for a satellite's orbital period $P$ to find the orbital radius $R$:
$P = 2\pi~\sqrt{\frac{R^3}{G~M_p}}$
$R = (\frac{P^2~G~M_p}{4\pi^2})^{1/3}$
$R = [~\frac{(21,600~s)^2~(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}{4\pi^2}~~]^{1/3}$
$R = 16,757,760~m$
$R = 16,757.760~km$
We can find the height $h$ above the Earth's surface:
$h = R - 6380~km$
$h = 16,757.760~km - 6380~km$
$h = 10,400~km$
The satellite is a height of $10,400~km$ above the Earth's surface.
(b) We can find the angular speed:
$\omega = \frac{2\pi~rad}{(6.00)(3600~s)}$
$\omega = 0.00029~rad/s$
We can find the acceleration:
$a = \omega^2~R$
$a = (0.000291~rad/s)^2~(16,757,760~m)$
$a = 1.42~m/s^2$
The satellite's acceleration is $1.42~m/s^2$
