Question

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period in years and average orbital distance in AU 233 years 1.46 AU 5.45 years 3.09 years

Answer 1

Answer:

The value is  $x = 45.99 \ Au$

Explanation:

From the question we are told that

   The period of the  asteroid is   $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

            $R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value  

        $M = 1.99*10^{30} \ kg$

         G  is the gravitational constant with value  $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

           $R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=>       $R = 6.88 *10^{12} \ m$

Generally

         $1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

So

          $R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=>      $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=>       $x = 45.99 \ Au$