you observe a small moon orbit a large planet. the moon has a period of 3.9x10^5 s and has a circular orbit…

Explanation:

Step1: Use the formula for the period of a circular orbit

The formula for the period (T) of a circular orbit is (T = 2\pi\sqrt{\frac{r^{3}}{GM}}), where (r) is the radius of the orbit, (G = 6.67\times 10^{- 11}\space N\cdot m^{2}/kg^{2}) is the gravitational constant, (M) is the mass of the planet. We can re - arrange this formula to solve for (M): [M=\frac{4\pi^{2}r^{3}}{GT^{2}}]

Step2: Substitute the given values

Given (r = 8\times 10^{8}\space m), (T = 3.9\times 10^{5}\space s), and (G = 6.67\times 10^{-11}\space N\cdot m^{2}/kg^{2}) [ \begin{align*} M&=\frac{4\pi^{2}(8\times 10^{8})^{3}}{6.67\times 10^{-11}(3.9\times 10^{5})^{2}}\ &=\frac{4\times(3.14)^{2}\times512\times10^{24}}{6.67\times 10^{-11}\times15.21\times10^{10}}\ &=\frac{4\times9.86\times512\times10^{24}}{6.67\times15.21\times10^{-1}}\ &=\frac{20119.68\times10^{24}}{101.4507\times10^{-1}}\ &=\frac{2.011968\times10^{28}}{1.014507\times10^{1}}\ &\approx2.0\times 10^{27}\space kg \end{align*} ]

Answer:

A. (2.0\times 10^{27}\space kg)