What is the period of the asteroid's orbit? Answer in units of year.The period of the earth around the sun is 1 year and its distance id 150...

We assume that the period of earth and its distance from the sun is T1 and R1.

We assume that the perod of the arteroid is T2 with a distance R2 from the sun.By Newton Kepler's Law,

(T2/T1)^2 = ( R2/R1)^3.

Therefore,

T2 = T1* sqrt[(R2/R1)^3].................(1)

R1 = 150 million km = 150*10^6*10^3= 1.5*10^11 meter. T1 = 1 year. R2 =239million km = 239*10^6*10^3 m = 2.39*10^11 meter. Substituting the values in (1) we get:

T = 1 * sqrt[(239/150)^3]

=2.011221777 year of earth is the asteroid,s period.