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

The value is
`x = 45.99 \ Au`

Step-by-step 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]{ (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]{ (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 = (6.88 *10^(12))/(1.496 *10^(11))`

=>
`x = 45.99 \ Au`