## College Physics (4th Edition)

(a) The satellite is a height of $10,400~km$ above the Earth's surface. (b) The satellite's acceleration is $1.42~m/s^2$
(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 $1.42~m/s^2$