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.

The orbital period of Jupiter around the sun is 11.9 years.To solve this question, we need to understand the concept of Kelper's Third Law and how to use it.What is Kelper's Third Law?Kelper's Third Law posits that the ratio between the square of the orbital period (T) as well as the average distance cube of a planet away from the Sun is constant for all planets circling the Sun.[tex]\mathbf{\dfrac{T^2}{r^3} = constant}[/tex]Suppose, we make a reference to Earth, we have:[tex]\mathbf{\dfrac{T^2_e}{r^3_e} = \dfrac{T^2_j}{r^3_j}}[/tex]Making the orbital period (T_j) of Jupiter as the subject, we have:[tex]\mathbf{{T_j}=\sqrt{ \dfrac{T^2_e \times {r^3_j}}{r^3_e}}}[/tex]where;r_e = avg. distance from earth to sun = 1 au[tex]\mathbf{{T_j}=\sqrt{ \dfrac{(1 \ y) \times {(5.20 \ au) ^3}}{(1 \ au) ^3}}}[/tex][tex]\mathbf{{T_j}=11.9 \ years}[/tex]Learn more about Kelper's Law here: