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 done

Explanation:

Step1: Use Kepler's third law

Kepler's third law is \(T^{2}=a^{3}\), 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: Substitute the value of \(a\) into the formula

Substitute \(a=5.20\) into \(T^{2}=a^{3}\). So \(T^{2}=(5.20)^{3}\). Calculate \((5.20)^{3}=5.20\times5.20\times5.20 = 140.608\).

Step3: Solve for \(T\)

Take the square root of both sides: \(T=\sqrt{140.608}\). Using a calculator, \(T\approx11.9\).

Answer:

\(11.9\)