the mass of the sun is 2 × 10³⁰ kg, and the distance between neptune and the sun is 30 au. what is the…

Answer:

B. 164 Earth years

Explanation:

Step1: Convert distance to meters

1 AU = 1.496×10¹¹ m, so 30 AU = 30×1.496×10¹¹ m = 4.488×10¹² m.

Step2: Use Kepler's third - law

Kepler's third - law is $T^{2}=\frac{4\pi^{2}r^{3}}{GM}$, where $G = 6.67×10^{- 11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$, $M = 2×10^{30}\text{ kg}$, and $r = 4.488×10^{12}\text{ m}$. First, calculate $r^{3}=(4.488×10^{12})^{3}=8.99×10^{37}\text{ m}^{3}$. Then, $GM=6.67×10^{-11}\times2×10^{30}=1.334×10^{20}\text{ N}\cdot\text{m}^{2}/\text{kg}$. $\frac{4\pi^{2}r^{3}}{GM}=\frac{4\times\pi^{2}\times8.99×10^{37}}{1.334×10^{20}}\approx8.37×10^{18}\text{ s}^{2}$. $T=\sqrt{8.37×10^{18}}\text{ s}\approx2.89×10^{9}\text{ s}$. Since 1 year = 365×24×3600 s≈3.154×10^{7} s. $T=\frac{2.89×10^{9}}{3.154×10^{7}}\text{ years}\approx91.6$ years (this is a wrong - unit start approach).

Using the simplified form of Kepler's third - law for the solar system in astronomical units and Earth - years: $T^{2}=a^{3}$, where $a$ is the semi - major axis in AU. Here $a = 30$ AU. $T=\sqrt{30^{3}}=\sqrt{27000}\approx164$ Earth years.