Question

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer 1

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  $T_R = 11.8 \ days$  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  $a_R = 45.0\ AU$

   The semi - major axis of  Planet D is  $a_D = 60 \ AU$

    The orbital  period of planet D is  $T_D = 18.164 \ days$

Generally from Kepler third law

          $T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the  orbital period  while a is the semi major axis

So  

        $\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=>     $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$  

=>     $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=>      $T_R = 11.8 \ days$