Question

you used a telescope and other mathematics to discover that jupiter is 5.20 au from the sun. use the equation to find its orbital period. round to the nearest tenth of a year. jupiters orbit: years

Explanation:

Step1: Apply Kepler's third - law

$T^{2}=a^{3}$, where $T$ is the orbital period in years and $a$ is the semi - major axis in astronomical units (au). Given $a = 5.20$ au.

Step2: Calculate $T$

$T=\sqrt{a^{3}}=\sqrt{(5.20)^{3}}=\sqrt{140.608}\approx11.9$ years

Answer:

11.9