Question

Using the formula T2=a3, What is the period of Neptune if the semi-major axis is 30.11?

Answer 1

The period of Neptune is 165 years if the semi-major axis is 30.11 using the given formula.

Answer: Option D

Explanation:

The period of any planet revolutions around the sun can be determined using Kepler third law. As that law states that square to the period of one complete revolution of every planet is equal to cube of semi major axis of that planet. So the formula can be derived as  

                             $T^{2}=a^{3}$

Where T is the time taken for completing one revolution and a is the semi-major axis of the planet.

As here Neptune's semi major axis is given as 30.11. So the period of Neptune can be determined as follows:

                      $T^{2}=(30.11)^{3}$

                      $T^{2}=27,298$

Then take square root of this value:

                               $T=\sqrt{27,298}$

                                      T = 165 years

Thus, the Neptune takes 165 years for one complete revolution of the sun.