## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Let $M$ be the mass of the earth. We can use the satellite's orbital speed to find the radius of the satellite's orbit. $v = \sqrt{\frac{G~M}{R}}$ $R = \frac{G~M}{v^2}$ $R = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{(5500~m/s)^2}$ $R = 1.32\times 10^7~m$ We then find the time $T$ of one orbit. $T = \frac{distance}{speed}$ $T = \frac{2\pi~R}{v}$ $T = \frac{(2\pi)(1.32\times 10^7~m)}{5500~m/s}$ $T = 15,080~s = 4.2~hours$ The orbital period is 4.2 hours.