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 million km from the sun. An asteriod in a circular orbit around the sun is at a distance 239 million km from the sun.
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It is 2.011221777 year of Earth is the asteroid's period.
I am right.
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.
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.
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