Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can use the gravitational force to find the mass $m_p$ of the planet; $F_g = \frac{G~m_p~m_m}{r^2}$ $m_p = \frac{F_g~r^2}{G~m_m}$ $m_p = \frac{(1.1\times 10^{19}~N)(1.5\times 10^8~m)^2}{(6.67\times 10^{-11}~m^3/kg~s^2)(9.4\times 10^{21}~kg)}$ $m_p = 3.95\times 10^{23}~kg$ We can use the equation for the orbital period to find the period in seconds; $T^2 = \frac{4\pi^2~r^3}{G~m_p}$ $T = \sqrt{\frac{4\pi^2~r^3}{G~m_p}}$ $T = \sqrt{\frac{(4)(\pi^2)(1.5\times 10^8~m)^3}{(6.67\times 10^{-11}~m^3/kg~s^2)(3.95\times 10^{23}~kg)}}$ $T = 2.25\times 10^6~s$ We can convert the period to units of days: $T = (2.25\times 10^6~s)(\frac{1~day}{(24~hr)(3600~s/hr)})$ $T = 26.0~days$ The moon's orbital period is 26.0 days.