the mass of the sun is 2 x 10^30 kg, and the ...

# Explanation: ## Step1: Convert distance to SI units 1 AU = 1.496×10¹¹ m, so $r = 30\times1.496\times10^{11}\text{ m}=4.488\times 10^{12}\text{ m}$ ## Step2: Recall the formula for orbital - period The formula for the orbital period $T$ of a planet around the Sun is $T = 2\pi\sqrt{\frac{r^{3}}{GM}}$, where $G = 6.67\times10^{- 11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$ is the gravitational constant, $M = 2\times10^{30}\text{ kg}$ is the mass of the Sun, and $r$ is the distance between the planet and the Sun. Substitute the values: \[ \begin{align*} T&=2\pi\sqrt{\frac{(4.488\times 10^{12})^{3}}{6.67\times10^{-11}\times2\times10^{30}}}\\ &=2\pi\sqrt{\frac{4.488^{3}\times10^{36}}{13.34\times10^{-11 + 30}}}\\ &=2\pi\sqrt{\frac{90.7\times10^{36}}{13.34\times10^{19}}}\\ &=2\pi\sqrt{6.8\times10^{17}}\\ &=2\pi\times8.25\times10^{8}\text{ s} \end{align*} \] ## Step3: Convert seconds to years 1 year = 365 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds. So 1 year $=365\times24\times60\times60\text{ s}=3.154\times10^{7}\text{ s}$ $T=\frac{2\pi\times8.25\times10^{8}}{3.154\times10^{7}}\text{ years}\approx164\text{ years}$ # Answer: 164 Earth years