Answer
(a) The period of one revolution is 2.93 hours.
(b) The radial acceleration of the satellite is $3.70~m/s^2$.
Work Step by Step
(a) Let $M$ be the mass of the earth. We can use the speed to find the radius of the satellite's orbit.
$v = \sqrt{\frac{G~M}{R}}$
$R = \frac{G~M}{v^2}$
$R = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{(6200~m/s)^2}$
$R = 1.04\times 10^7~m$
We can find the time of one revolution.
$t = \frac{distance}{speed}$
$t = \frac{2\pi~R}{v}$
$t = \frac{(2\pi)(1.04\times 10^7~m)}{6200~m/s}$
$t = 10,540~s = 2.93~hours$
The period of one revolution is 2.93 hours.
(b) We can find the radial acceleration.
$a_c = \frac{v^2}{R}$
$a_c = \frac{(6200~m/s)^2}{1.04\times 10^7~m}$
$a_c = 3.70~m/s^2$
The radial acceleration of the satellite is $3.70~m/s^2$.
