Answer
(a) The orbital speed is 7410 m/s.
(b) The period of the orbit is 1.71 hours.
Work Step by Step
(a) If the satellite is 890 km above the surface of the earth, the radius $R$ of the orbit is 890 km + 6380 km, which is 7270 km. We can find the speed of the satellite.
$v = \sqrt{\frac{G~M}{R}}$
$v = \sqrt{\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{7.27\times 10^6~m}}$
$v = 7410~m/s$
The orbital speed is 7410 m/s.
(b) We can find the time of one revolution.
$t = \frac{distance}{speed}$
$t = \frac{2\pi~R}{v}$
$t = \frac{(2\pi)(7.27\times 10^6~m)}{7410~m/s}$
$t = 6164~s = 1.71~hours$
The period of the orbit is 1.71 hours.
