What is the orbital period (in years) of an asteroid whose average distance from the sun is 28 A. U.?

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The orbital period is the time that it takes the asteroid to complete an orbit around the sun. To calculate the orbital period we use the third Kepler law. According to this law; for any body that orbits around the Sun, the square of its orbital period is proportional to...

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The orbital period is the time that it takes the asteroid to complete an orbit around the sun. To calculate the orbital period we use the third Kepler law. According to this law; for any body that orbits around the Sun, the square of its orbital period is proportional to the cube of the length of the semimajor axis of its elliptical orbit. This law can be written mathematically as follows:

(T^2)/(R^3) = constant

T, is the orbital period and R is the average distance to the Sun.

So, for any planet in the solar system (the earth for example), and the asteroid we can write the following equality:

(Ta^2)/(Ra^3) = (Te^2)/(Re^3)

Solving for the period of the asteroid, we have:

Ta^2 = (Te^2)(Ra^3)/(Re^3)

Given the values Te=1 year, Re=1 A.U. and Ra=28 A.U., we have:

Ta^2 = (1^2)(28^3)/(1^3)

Ta = sqrt (28^3) 

Ta = 148.16 years

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